Regarding SQL Expert (1ZO - 047) valid 12 c certification

Similar Questions

• PL/SQL Developer OCA - order of examinations (1ZO-047, 1Z0-144)

  Hello
  "I would like to know if I can get the ' 1ZO-144-Oracle Database 11 g: program PL/SQL ' review * before * the Expert"1Z0-047-Oracle SQL database.
  
  In the 'Page of Certification Oracle' :* http://bit.ly/1oyl6E * I read the diagram and I just looked in this forum, but still have this doubt.
  You would be so kind to share this information, or better yet you can testify your experience doing the * "1Z0-047' in the beginning * of the path?
  
  Thank you in advance,
  concerning
  
  Giuseppe
  
  Published by: zep111 on Sep 3, 2010 10:32

  zep111 wrote:
  Hello
  "I would like to know if I can get the ' 1ZO-144-Oracle Database 11 g: program PL/SQL ' review * before * the Expert"1Z0-047-Oracle SQL database.

  In the 'Page of Certification Oracle' :*http://bit.ly/1oyl6E * I read the diagram and I just looked in this forum, but still have this doubt.
  You would be so kind to share this information, or better yet you can testify your experience doing the * "1Z0-047' in the beginning * of the path?

  Thank you in advance,
  concerning

  Giuseppe

  Published by: zep111 on Sep 3, 2010 10:32

  It is possible and allowed to sit these exams is the order.

  1Z0 - 047 seem relatively difficult to most people and expensive to regain power. So especially if you don't have this form of review before the cheaper and easier (for most people) 1z0-051 is probably a better option.

  But in the end you pays yer money and takes your choice, what works for one individual may not be best for another.

  Rgds - bigdelboy.

• 1Z0-047 Oracle database SQL expert

  Hi all!
  
  I want to prepare for review reference in question.
  How can I make it right?
  
  Is these books sth that I would like to search for? :
  
  Oracle Database 10 g, SQL (Osborne ORACLE Press Series)
  http://www.Amazon.com/GP/product/0072229810/ref=s9subs_c1_14_at1-rfc_g1_si3?pf_rd_m=ATVPDKIKX0DER & pf_rd_s = center-1 & pf_rd_r = 0NY1DZFY480AYW383FNE & pf_rd_t = 101 & pf_rd_p = 463383351 & pf_rd_i = 507846
  
  OCA Oracle database SQL Expert corrected (exam 1Z0-047)
  http://www.Amazon.com/Oracle-database-expert-guide-1Z0-047/DP/0071614214/ref=sr_1_14?ie=UTF8 & s = Books & qid = 1229588476 & SR = 8-14
  
  Are there other options, for example PDF files.
  
  Best regards
  user10679687

  These two references should be useful. You can also refer to:

  Oracle® database SQL Reference 10 g Release 2 (10.2)
  http://download.Oracle.com/docs/CD/B19306_01/server.102/b14200/TOC.htm

  1Z0-047 Oracle database SQL expert
  http://education.Oracle.com/pls/web_prod-PLQ-dad/db_pages.GetPage?page_id=41&p_org_id=1001&lang=us&p_exam_id=1Z0_047

  Sample Questions - 1Z0-047 Oracle database SQL Expert
  http://www.Oracle.com/global/us/education/certification/sample_questions/exam_1z0-047.html

• Oracle SQL Expert review 2015

  Hello world

  Is there someone who take the "Oracle SQL Expert" review recently in 2015? I want to ask your views on review, you can share your opinion about the exams. In fact, I want to know how many Oracle 12 c features included review?

  Can you please share your experience?

  Thanks in advance

  You cannot use other experiences to predict what you'll see, because not everyone gets the same set of questions.  I don't have enough inside information to see how the questions are determined, but based on my experiences, I believe that tests each have a pool question about twice as many questions that the review is designed to mark and the specific questions that each person receives are selected at random.  If this is the case, the number of questions of 12 c receives a given person can vary considerably.

  That said, you can have a topic idea how many questions by themes is possible by comparing the number of subjects to the count of exam question.  There are 79 different topics on 1Z0 - 047 and review will have 75 questions. While this comes out less than one question per subject, 047 review is a bit unusual in that one question may require more knowledge of a subject.  For example a question involving a query that uses a subquery EXISTS and would cover both 'use the operators EXISTS and NOT EXISTS' and 'Write a row and several lines of subqueries' subjects.

  However, it is unlikely to see more than two questions on a given topic, and most will have only one question. With this, you can watch all the listed topics that concern you and come with a relatively precise range of issues you might see on them.

• Certified Oracle SQL Expert Exam - 1Z0-047

  I plan on taking the exam 1Z0-047 and recently (in the last 3 weeks) bought the study guide / review of Kaplan for Oracle 11 g r2 for this test practice (or at least I thought it was up to and including Oracle 11 g r2).

  If I understand correctly what I read, it seems that questions for version 12 of the database are now included in this review.  There is apparently no separate examination that does not include the Oracle 12.  Is this correct, or can I still request a review for 11g r2?

  This is also true for PL/SQL and other exams (must study for any question 12 c)?

  Given that most of the developers I know is not yet up on 12 c, it seems that there should be at least an option of which version of the database you want to testify against.

  Everything / all thoughts are appreciated.

  Barry D.

  If I understand correctly what I read, it seems that the issues for version 12 of the database are now included in this review.  There is apparently no separate examination that does not include the Oracle 12.  Is this correct, or can I still request a review for 11g r2?

  I guess in what above, where you have "now", you want to say 'no '.

  No there is only one exam 1Z0-047.  The number of topics under review that are specific to Oracle 12 c is minimal.  I've updated my page on the contest of Expert SQL include articles on these topics, so you can prepare for them, even if the Kaplan materials do not yet cover them.

  Oracle Certification Preparation: Details of the examination and preparation for 1Z0 - 047 resources

  This is also true for PL/SQL and other exams (must study for any question 12 c)?

  The current reviews of PL/SQL (1Z0-144 and 1Z0-146) have not been updated to 12 c.  In my view, that the two have been validated against ut 12 c. which simply means that nothing they contain is wrong for 12 c - No 12 new c-specific features have been added.

• 11g SQL Fundamentals SQL expert?

  Hi gurus,
  
  I'm getting confused in selecting the proper test. Oracle says that if the exam is difficult then required pass mark will be low compared to other tests.
  
  I heard that the expert SQL (1Z0-047) is a little difficult to compare to 11g SQL Fundamentals(1Z0-051). But for the Expert SQL pass mark is high and also the price of the examination.
  
  Any idea why is there a difference in pricing and passing also. And also suggest me what test I have to attend.
  
  
  Below, I have copied the link.
  
  [http://education.oracle.com/pls/web_prod-plq-dad/db_pages.getpage?page_id=138#2_6]
  

  For example, to maintain a consistent standard for success, a new version of an exam, which contains more dificult questions than the prior version will be assigned a somewhat lower required passing score. For this reason, passing scores may vary between different exams in your certification track, and between later and earlier versions of the same exam.

  Thank you
  Suri

  Oracle says that if the exam is difficult then required pass mark will be low compared to other tests.

  This isn't what the text says, what he says is: "...". a new * version * review, which contains more difficult questions than the * previous version * will be affected a little lower required passing... »

  OE is not 1Z0-051 vs 1Z0-047 because they are not of the different versions of the same test. The statement says that, if they pass a test, do a version update, but find that the new version has more difficult questions, they could slide the pass mark so that it is not more difficult to pass than what it replaces.

  Any idea why is there a difference in pricing and passing also.

  1Z0-051 is available online (i.e. not managed). It's cheaper because this method costs Oracle less to offer. My opinion is that it reduces considerably the value of spend because it is too easy to cheat, but is a completely different topic.

  The only relationship in the two tests is that they both cover SQL. They do so at very different levels of difficulty. Scores between these two tests have no relationship between them. OE sets the pass mark for individual tests at a level designed to allow a certain percentage of candidates to the exam. I don't know what level or how they determine. My conviction is that this score is determined initially as they treat the results of beta testing. If I'm not mistaken, then they take a note of passage based on beta responses which would allow their percentage target of candidates. Maybe it's that they adjust scores of passage early in the life of a review of production if they find that the beta "good percentage" was atypical, but that's just speculation.

  And also suggest me what test I have to attend.

  I wrote an article discussing the two tests and why you might throw on each other a few months ago:

  http://www.gocertify.com/articles/Oracle-SQL-certification-1Z0-051-or-1Z0-047.html

• Help with SQL query for SQL experts!

  Hi all

  For the very expert on SQL.

  ------------------

  create table t1 (STEP number);

  Insert into t1 select rownum from dba_tables where rownum < 11;

  commit;

  create table t2 (the STEP number, name varchar2 (4), varchar2 (1)) of the State;

  insert into t2 values (1, 'TOTO', ' A');

  insert into t2 values (3, 'TOTO', 'C');

  insert into t2 values (4, 'TOTO', ');

  insert into t2 values (1, 'FIFI', ' A');

  insert into t2 values (6, 'FIFI', 'B');

  commit;

  ----------------

  My goal is to list the STATE in the NAME of all the STEPS, even if measures do not exist for a NAME.

  In other words, my query should return a result like this: (query on t1 and t2, of course)

  STEP   1      2             3            4               5            6

  TOTO A C D NO NO NO

  FIFI HAS NO NO NO NO D

  Thank you in advance.

  with the data as)
  Select
  T1. Step step
  name
  State
  Of
  T1, t2
  where
  T1. Step = T2. Step (+)
  )

  Select
  *
  from the data
  Pivot (max (state))
  for step (1,2,3,4,5,6,7,8,9,10)
  )
  where the name is not null

  NAME	1	2	3	4	5	6	7	8	9	10
  TOTO	A	-	C	D	-	-	-	-	-	-
  FIFI	A	-	-	-	-	B	-	-	-	-

  If the column name must be the stage instead of the name

  with the data as)
  Select
  Step 1 T1. Step
  name as step
  State
  Of
  T1, t2
  where
  T1. Step = T2. Step (+)
  )

  Select
  *
  from the data
  Pivot (max (state))
  for step 1 in (1,2,3,4,5,6,7,8,9,10)
  )
  where the step is not null

• JDeveloper does not support the SQL expert, cannot add query clause to view link

  For developers who can know:
  
  I use JDeveloper 10.1.3.4 and I have two tables, we're STUDENTS, the other is the PRICE, with a one-to-many relationship (a student may have several awards), and ID is the foreign key:
  

  STUDENTS
  ------------------------
  ID    NAME    DOB
  ------------------------
  9999  JOHN    01/02/1990
     |  
     |  
     |  
     |           AWARDS
     |           ------------------------
     |           ID      DESC      AMOUNT
     |           ------------------------
     ----------- 9999    PELL      500
                 9999    PERKINS   800
                 9999    LOAN      900

  In the application, when a student (9999, JOHN) connects, the information above is displayed. I need to add up all the awards for the student and to display not only the elements of price, but also the total amount of all awards. To do this, I created another object to view named TotalAwards in the object entity the price just to get the total amount. The SQL code of the VO's
  

  SELECT SUM(Awards.AMOUNT) AS TOTAL
  FROM AWARDS Awards

  Then, a link is created between the students and the TotalAwards, and use of the new link to the view is added to the application module. TotalAwards is now available in the data control palette to be drag-and - drop to pages.
  
  Problem occurs when the application is invoked. An error is caught before the application can begin. The error message indicates that "the ends or the link view do not specify an attribute.
  
  It seems that the TOTAL 'SUM (Awards.AMOUNT) AS TOTAL' is not recognized as an attribute. I have reviewed the query clause in the display link to see if it is there, such as:
  

           Attribute        Bind Variable
           ------------     --------------
           Students.ID      :Bind_Id
  
  
  WHEHE    :Bind_Id = Awards.ID

  There is no such clause in the query clause page. And I can not add a no more, because the page is grayed out and disabled. I think maybe it's for the same reason: TOTAL is not recognized as an attribute of the price.
  
  How to get around the problem? Or are there other ways to summarize the total amount and view it in the browser?
  
  Thank you very much for help!
  
  
  Newman

  Hello

  Here you have a solution for the TOTAL:
  http://www.freewebalbum.com/blogs/faces/bjanko/blogs.jsp?blog=bjanko20070725180020

  Kind regards

  Branislav

• In Oracle XE limitations regarding SQL

  Hi all,
  
  Is that there is no limitation in queries SQL Oracle XE?
  
  Because I run a SQL query in Oracle Standard Edition which is clothed with results while I'm not able to run the same query against the Oracle XE with the same table structure and data.
  
  In Oracle XE, it runs the error as ERROR at line 31:
  ORA-00904: "EMP". "" EMPLOYEE_CODE ": invalid identifier
  
  
  My query is
  
  SELECT "EMP". "" EMPLOYEE_CODE, "
  "EMP". "" NAME, "
  "EMP". "' BIRTH_DATE ',.
  "EMP". "" FH_NAME, "
  "EMP". "" JOINING_DATE, "
  "EMP". "" DESIGNATION. "
  "EMP". "" PAN_NO, "
  "DEPT". "" DESCRIPTION. "
  "EMP". "" DIVISION_CODE, "
  "DIV". "" DESCRIPTION. "
  "LOC". "" DESCRIPTION. "
  "EMP". "" PF_NO, "
  "EMP". "" ESI_NO, "
  "EMP". "" CURRENT_STATUS, "
  "L"." CL_BALANCE,"
  "L"." EL_BALANCE,"
  "L"." ML_BALANCE,"
  "A"." ATTENDED_DAYS,"
  "A"." PUBLIC_HOLIDAYS,"
  "A"." WEEKLY_HOLIDAYS,"
  (SELECT SUM (CASE WHEN "LEAVE". "" LEAVE_CODE "="CL"AND"EXIT".". CASUAL_LEAVE END OF LEAVE_DAYS") LEAVE WHERE PERIOD = 88 AND EMPLOYEE_CODE = EMP. EMPLOYEE_CODE) AS CASUAL_LEAVE,
  (SELECT SUM (CASE WHEN "LEAVE". "" LEAVE_CODE "= 'EL' AND 'EXIT'." EARNED_LEAVE END OF LEAVE_DAYS") LEAVE WHERE PERIOD = 88 AND EMPLOYEE_CODE = EMP. EMPLOYEE_CODE) AS EARNED_LEAVE,
  "SICK_LEAVE,"
  "ON_DUTY,"
  "ESI_LEAVE,"
  (SELECT SUM (CASE WHEN "LEAVE". "" LEAVE_CODE "="LP"AND THEN"EXIT".". LOSS_OF_PAY_DAYS END OF LEAVE_DAYS") LEAVE WHERE PERIOD = 88 AND EMPLOYEE_CODE = EMP. EMPLOYEE_CODE) AS LOSS_OF_PAY_DAYS,
  'PAID_DAYS', 'DIV '. "" ADDRESS2 "| " '||" """DIV"."" ADDRESS3"AS THE ADDRESS,
  (SELECT PLATE OF)
  SELECT A.EMPLOYEE_CODE, SUM (A.AMOUNT) AS AMOUNT OF PAYROLL A, B PAY_ELEMENT
  WHERE A.PAY_ELEMENT_CODE = B.PAY_ELEMENT_CODE AND
  A.PERIOD = 88 AND A.EMPLOYEE_CODE = EMP. EMPLOYEE_CODE AND B.PAY_ELEMENT_TYPE = 'E'
  AND B.PAY_ELEMENT_CODE NOT IN ('FC')
  A.EMPLOYEE_CODE GROUP
  UNION
  SELECT A.EMPLOYEE_CODE, AMOUNT OF FOOD_COUPON A, PAY_ELEMENT B
  WHERE EMPLOYEE_CODE = EMP. EMPLOYEE_CODE
  AND B.PAY_ELEMENT_CODE = A.PAY_ELEMENT_CODE
  AND B.PAY_ELEMENT_TYPE = 'E'
  AND PERIOD = 88) D
  WHERE EMPLOYEE_CODE = EMP. EMPLOYEE_CODE) AS REMUNERATION.
  (SELECT PLATE OF)
  SELECT A.EMPLOYEE_CODE, SUM (A.AMOUNT) AS AMOUNT OF PAYROLL A, B PAY_ELEMENT
  WHERE A.PAY_ELEMENT_CODE = B.PAY_ELEMENT_CODE AND
  A.PERIOD = 88 AND A.EMPLOYEE_CODE = EMP. EMPLOYEE_CODE AND B.PAY_ELEMENT_TYPE = A '
  AND B.PAY_ELEMENT_CODE NOT IN ('CDF')
  A.EMPLOYEE_CODE GROUP
  UNION
  SELECT A.EMPLOYEE_CODE, AMOUNT OF FOOD_COUPON A, PAY_ELEMENT B
  WHERE EMPLOYEE_CODE = EMP. EMPLOYEE_CODE
  AND B.PAY_ELEMENT_CODE = A.PAY_ELEMENT_CODE
  AND B.PAY_ELEMENT_TYPE = A '
  AND PERIOD = 88) D
  WHERE EMPLOYEE_CODE = EMP. EMPLOYEE_CODE) AS DEDUCTIONS.
  NVL ((SELECT AMOUNT OF COMPENSATION WHERE PAY_ELEMENT_CODE = 'LICENSE' AND EMPLOYEE_CODE = EMP. (EMPLOYEE_CODE), 0) AS BASIC,.
  NVL ((SÉLECTIONNEZ MONTANT PAYE WHERE PAY_ELEMENT_CODE = 'HRA' AND EMPLOYEE_CODE = EMP.)) (EMPLOYEE_CODE), 0) AS HRA,.
  NVL ((SELECT AMOUNT OF COMPENSATION WHERE PAY_ELEMENT_CODE = 'SP2' AND EMPLOYEE_CODE = EMP. (EMPLOYEE_CODE), 0) AS SPECIALPAY2,
  NVL ((SELECT NVL(AMOUNT,0) WHERE PAY_ELEMENT_CODE = 'OA' AND EMPLOYEE_CODE = EMP COMPENSATION. (EMPLOYEE_CODE), 0) AS OTHERALLOW,
  NVL ((SELECT AMOUNT OF COMPENSATION WHERE PAY_ELEMENT_CODE = 'TA' AND EMPLOYEE_CODE = EMP. (EMPLOYEE_CODE), 0) AS TRANSALLOW,
  NVL ((SELECT AMOUNT OF COMPENSATION WHERE PAY_ELEMENT_CODE = 'FC' AND EMPLOYEE_CODE = EMP. (EMPLOYEE_CODE), 0) AS FOODCOUPONS
  "EMPLOYEE" "EMP",.
  "DEPARTMENT" "DEPT"
  'DIVISION', 'DIV ',.
  'LOCATION' 'LOC ',.
  "LEAVE_BALANCE" 'L ',.
  'PAY. '
  "PRESENCES", "A".
  WHERE ("EMP". "" DEPARTMENT_CODE "="DEPT ". (' "DEPARTMENT_CODE") and
  ("EMP". "" DIVISION_CODE "="DEPT ". (' "DIVISION_CODE") and
  ("EMP". "" DIVISION_CODE "="DIV ". (' "DIVISION_CODE") and
  ("EMP". "" LOCATION_CODE "="LOC ". (' "LOCATION_CODE") and
  ("EMP". "' EMPLOYEE_CODE ' = 'L '. (' "EMPLOYEE_CODE") and
  ("DIV". "' DIVISION_CODE ' = 'L '. (' "DIVISION_CODE") and
  ( "L"." EMPLOYEE_CODE"="PAYROLL. " (' "EMPLOYEE_CODE") and
  ("PAYROLL". "' DIVISION_CODE ' = 'L '. (' "DIVISION_CODE") and
  (("EMP". ("' CURRENT_STATUS ' = 'C') AND
  ("PAYROLL". ("" PAY_ELEMENT_CODE "="BSC") AND
  ("PAYROLL". ("' PERIOD ' = 88) AND
  ("EMP". ' EMPLOYEE_CODE '= 'A'. (' "EMPLOYEE_CODE") AND
  ("A"." PERIOD "(=88))"
  /
  
  
  
  The above query returs ranks in Standard edition but get the error in XE with the same table structure and data.
  
  Kindly help me in this regard
  
  Delphine K

  I'm not sure if it works in normal 10.2 but as I see you have leave the comparison "employee_code = emp.employee_code" inside the view on the "D" line for where condition outside the view inline (lines 49 and 69 in the reformatted below statement):

  SQL> get t1
    1  SELECT "EMP"."EMPLOYEE_CODE",
    2         "EMP"."NAME",
    3         "EMP"."BIRTH_DATE",
    4         "EMP"."FH_NAME",
    5         "EMP"."JOINING_DATE",
    6         "EMP"."DESIGNATION",
    7         "EMP"."PAN_NO",
    8         "DEPT"."DESCRIPTION",
    9         "EMP"."DIVISION_CODE",
   10         "DIV"."DESCRIPTION",
   11         "LOC"."DESCRIPTION",
   12         "EMP"."PF_NO",
   13         "EMP"."ESI_NO",
   14         "EMP"."CURRENT_STATUS",
   15         "L"."CL_BALANCE",
   16         "L"."EL_BALANCE",
   17         "L"."ML_BALANCE" ,
   18         "A"."ATTENDED_DAYS",
   19         "A"."PUBLIC_HOLIDAYS",
   20         "A"."WEEKLY_HOLIDAYS",
   21         (SELECT SUM(CASE WHEN "LEAVE"."LEAVE_CODE"='CL' THEN "LEAVE"."LEAVE_DAYS" END ) CASUAL_LEAVE
   22          FROM LEAVE WHERE PERIOD=88 AND EMPLOYEE_CODE=EMP.EMPLOYEE_CODE) AS CASUAL_LEAVE,
   23         (SELECT SUM(CASE WHEN "LEAVE"."LEAVE_CODE"='EL' THEN "LEAVE"."LEAVE_DAYS" END) EARNED_LEAVE
   24          FROM LEAVE WHERE PERIOD=88 AND EMPLOYEE_CODE=EMP.EMPLOYEE_CODE) AS EARNED_LEAVE,
   25         "SICK_LEAVE",
   26         "ON_DUTY",
   27         "ESI_LEAVE",
   28         (SELECT SUM(CASE WHEN "LEAVE"."LEAVE_CODE"='LP' THEN "LEAVE"."LEAVE_DAYS" END) LOSS_OF_PAY_DAYS
   29          FROM LEAVE WHERE PERIOD=88 AND EMPLOYEE_CODE=EMP.EMPLOYEE_CODE) AS LOSS_OF_PAY_DAYS,
   30         "PAID_DAYS",
   31         "DIV"."ADDRESS2"||' '||"DIV"."ADDRESS3" AS ADDRESS,
   32         (SELECT SUM(AMOUNT)
   33          FROM (SELECT A.EMPLOYEE_CODE, SUM(A.AMOUNT) AS AMOUNT
   34             FROM   PAYROLL A,
   35                 PAY_ELEMENT B
   36             WHERE  A.PAY_ELEMENT_CODE=B.PAY_ELEMENT_CODE
   37             AND       A.PERIOD= 88 AND B.PAY_ELEMENT_TYPE='E'
   38             AND       B.PAY_ELEMENT_CODE NOT IN('FC')
   39             GROUP BY A.EMPLOYEE_CODE
   40             UNION
   41             SELECT A.EMPLOYEE_CODE, AMOUNT
   42             FROM   FOOD_COUPON A,
   43                 PAY_ELEMENT B
   44             WHERE  B.PAY_ELEMENT_CODE=A.PAY_ELEMENT_CODE
   45             AND       B.PAY_ELEMENT_TYPE='E'
   46             AND       PERIOD=88
   47            )D
   48           WHERE EMPLOYEE_CODE=EMP.EMPLOYEE_CODE
   49           and   employee_code = emp.employee_code
   50          ) AS EARNINGS,
   51         (SELECT SUM(AMOUNT)
   52          FROM (SELECT A.EMPLOYEE_CODE, SUM(A.AMOUNT) AS AMOUNT
   53             FROM   PAYROLL A,
   54                 PAY_ELEMENT B
   55             WHERE  A.PAY_ELEMENT_CODE=B.PAY_ELEMENT_CODE
   56                AND    A.PERIOD= 88
   57             AND       B.PAY_ELEMENT_TYPE='D'
   58             AND       B.PAY_ELEMENT_CODE NOT IN('FCD')
   59             GROUP BY A.EMPLOYEE_CODE
   60             UNION
   61             SELECT A.EMPLOYEE_CODE, AMOUNT
   62             FROM   FOOD_COUPON A,
   63                 PAY_ELEMENT B
   64             WHERE  B.PAY_ELEMENT_CODE=A.PAY_ELEMENT_CODE
   65             AND       B.PAY_ELEMENT_TYPE='D'
   66             AND PERIOD=88
   67            )D
   68          WHERE EMPLOYEE_CODE=EMP.EMPLOYEE_CODE
   69          and   employee_code = emp.employee_code) AS DEDUCTIONS,
   70         NVL((SELECT AMOUNT
   71           FROM     PAY
   72           WHERE     PAY_ELEMENT_CODE='BSC'
   73           AND     EMPLOYEE_CODE=EMP.EMPLOYEE_CODE),0)AS BASIC,
   74         NVL((SELECT AMOUNT
   75           FROM     PAY
   76           WHERE     PAY_ELEMENT_CODE='HRA'
   77           AND     EMPLOYEE_CODE=EMP.EMPLOYEE_CODE),0)AS HRA,
   78         NVL((SELECT AMOUNT
   79           FROM     PAY
   80           WHERE     PAY_ELEMENT_CODE='SP2'
   81           AND     EMPLOYEE_CODE=EMP.EMPLOYEE_CODE),0)AS SPECIALPAY2,
   82         NVL((SELECT NVL(AMOUNT,0)
   83           FROM     PAY
   84           WHERE     PAY_ELEMENT_CODE='OA'
   85           AND     EMPLOYEE_CODE=EMP.EMPLOYEE_CODE),0)AS OTHERALLOW,
   86         NVL((SELECT AMOUNT
   87           FROM     PAY
   88           WHERE     PAY_ELEMENT_CODE='TA'
   89           AND     EMPLOYEE_CODE=EMP.EMPLOYEE_CODE),0)AS TRANSALLOW,
   90         NVL((SELECT AMOUNT
   91           FROM     PAY
   92           WHERE     PAY_ELEMENT_CODE='FC'
   93           AND EMPLOYEE_CODE=EMP.EMPLOYEE_CODE),0)AS FOODCOUPONS
   94  FROM  "EMPLOYEE" "EMP",
   95        "DEPARTMENT" "DEPT",
   96        "DIVISION" "DIV",
   97        "LOCATION" "LOC",
   98        "LEAVE_BALANCE" "L",
   99        "PAYROLL" ,
  100        "ATTENDANCE" "A"
  101  WHERE ( "EMP"."DEPARTMENT_CODE" = "DEPT"."DEPARTMENT_CODE" ) and
  102        ( "EMP"."DIVISION_CODE" = "DEPT"."DIVISION_CODE" ) and
  103        ( "EMP"."DIVISION_CODE" = "DIV"."DIVISION_CODE" ) and
  104        ( "EMP"."LOCATION_CODE" = "LOC"."LOCATION_CODE" ) and
  105        ( "EMP"."EMPLOYEE_CODE" = "L"."EMPLOYEE_CODE" ) and
  106        ( "DIV"."DIVISION_CODE" = "L"."DIVISION_CODE" ) and
  107        ( "L"."EMPLOYEE_CODE" = "PAYROLL"."EMPLOYEE_CODE" ) and
  108        ( "PAYROLL"."DIVISION_CODE" = "L"."DIVISION_CODE" ) and
  109        ( ( "EMP"."CURRENT_STATUS" = 'C' ) AND
  110          ( "PAYROLL"."PAY_ELEMENT_CODE" = 'BSC' ) AND
  111          ( "PAYROLL"."PERIOD" = 88) AND
  112          ("EMP"."EMPLOYEE_CODE"="A"."EMPLOYEE_CODE") AND
  113          ("A"."PERIOD"=88)
  114*        )
  115  /
  
  no rows selected
  
  SQL> 
  

• Query regarding SQL logs

  I'm currently working on database forensics, if I do some manual changes in the database through some sql queries there no entry in the logs to prove...

  Can anyone suggest me some tools that can give the desired result. (i.e. pulled the log of all manual queries)

  PS I am using SQL Server 2005

  Any help would be appreciated

  Thank you.

  Hello

  Your question is beyond the scope of this community.

  Please repost your question in the SQL Server TechNet Forums.

  https://social.technet.Microsoft.com/forums/SQLServer/en-us/home?category=SQLServer

  See you soon.

• PL/SQL, help in the validation of the data exists in a table.

  Greetings,

  I'm still new to PL/SQL and try to create a stored procedure that would allow me to check if a user exists in the connection table.  The final objective is to have a connection of the user to an ASP.net web application and have a stored procedure validate the user exists in the table user.

  To start, I just want to test the SP only from the DB.  So, I'm wondering how I can configure the parameter with a value to imitate an application by passing a value to this parameter.

  Here is my code for the SP that was compiled without error.

  CREATE OR REPLACE PROCEDURE SP_LOGIN_CHK

  (

  P_USRNAME IN VARCHAR2

  ) IS

  v_login_id VARCHAR2 (20);

  BEGIN

  Select login_id

  in v_login_id

  of login_user where v_login_id = p_usrname;

  END SP_LOGIN_CHK;

  Here is what I use to run the stored procedure...

  exec sp_login_chk('Chris');

  I get an error "no data found".  However, when I run this query on the database, I get 1 line returned.

  Select login_id login_user where login_id = "Chris";

  Hello

  cjpicc11 wrote:

  Greetings,

  I'm still new to PL/SQL and try to create a stored procedure that would allow me to check if a user exists in the connection table.  The final objective is to have a connection of the user to an ASP.net web application and have a stored procedure validate the user exists in the table user.

  To start, I just want to test the SP only from the DB.  So, I'm wondering how I can configure the parameter with a value to imitate an application by passing a value to this parameter.

  Here is my code for the SP that was compiled without error.

  CREATE OR REPLACE PROCEDURE SP_LOGIN_CHK

  (

  P_USRNAME IN VARCHAR2

  ) IS

  v_login_id VARCHAR2 (20);

  BEGIN

  Select login_id

  in v_login_id

  of login_user where v_login_id = p_usrname;

  END SP_LOGIN_CHK;

  Here is what I use to run the stored procedure...

  exec sp_login_chk('Chris');

  I get an error "no data found".  However, when I run this query on the database, I get 1 line returned.

  Select login_id login_user where login_id = "Chris";

  At the same time that you run the query, the local variable v_login_id has the value NULL, then the condition:

  where v_login_id = p_usrname;

  won't be true.  I bet you want to use the column in the table in this condition, not the local variable, like this:

  where login_id = p_usrname;

• You place your order by comparing - SQL Experts pls help

  Hi all
  
  I want to measure the closest points according to the percentage between these points and order.
  in other words, I want to show the relationship between the points according to the percentage value.
  

   
  drop table temp_value; 
  create table temp_value(id number(10),location varchar2(20), percentage number(9)); 
  
  
  insert into temp_value values  (1,'LOC 1,2',30); 
  insert into  temp_value values (2,'LOC 1,3',0); 
  insert into  temp_value values (3,'LOC 1,4',0); 
  insert into  temp_value values (4,'LOC 1,5',0); 
  insert into  temp_value values (5,'LOC 1,6',50); 
  insert into  temp_value values (6,'LOC 2,3',0); 
  insert into  temp_value  values(7,'LOC 2,4',0); 
  insert into  temp_value values (9,'LOC 2,5',0); 
  insert into  temp_value values (10,'LOC 2,6',10); 
  insert into  temp_value values (11,'LOC 3,4',90); 
  insert into  temp_value values (12,'LOC 3,5',80); 
  insert into  temp_value values (13,'LOC 4,5',0); 
  insert into  temp_value values (14,'LOC 4,6',0); 
  insert into  temp_value values (15,'LOC 5,6',0); 

  I want the output like this, the ranking of the point like this

   
  4 
  3 
  5 
  6 
  1 
  2 

  or like this

    6 
  1 
  2 
  4 
  3 
  5 

  If you ask me why I say this ranking only

   
  
  4    [90 percent between point 3,4 that mean can be 3,4 or 4,3] 
  3 
  5    [80 percent between 3,5 that mean can be 5 after 3,4] 
  6    [50 percent between 1,6 that mean can be 1,6 or 6,1 ] 
  1 
  2    [ 30 percent between 1,2 that mean can be 1,2 or 2,1] 

  concerning
  
  Published by: Isabelle October 6, 2012 10:18
  
  Published by: Isabelle October 6, 2012 10:43
  
  Published by: Isabelle October 6, 2012 20:04

  SQL> select loc
    2  from(
    3  select loc,rownum rn
    4  from(
    5  select percentage,location,regexp_substr(location,'\d+',1,rn) loc
    6  from temp_value a,
    7       (select 1 rn from dual union all
    8        select 2 from dual ) b
    9  order by percentage desc,to_number(regexp_substr(location,'\d+',1,rn)) asc
   10  ))
   11  group by loc
   12  order by min(rn);
  
  LOC
  --------------------
  3
  4
  5
  1
  6
  2
  
  6 rows selected.
  

• The question regarding SQL intervals

  I'm writing a piece of SQL for a report that will caclulate the total number of active sessions. The name of table I in draught is called "sessionhistory. This table has a field named "sessioninstantiationtime" for the start of the session and one called "sessiondestructiontime" for when it ends. I want counts the number of active sessions and calculate the bandwidth for each interval. Currently, my SQL works, but it only beginning resembles the session (sessioninstantiationtime). I need to change so that I count the sessions began, but have not yet finished. Here's the monetary SQL, can someone point me in the right direction? Any help is greatly appreciated...
  
  --
  
  SELECT
  To_char (sessioninstantiationtime, 'HH24') | ':' || To_char (Floor (TO_NUMBER (to_char (sessioninstantiationtime, 'MI')) / 15) * 15, "FM00") interval_value,.
  Count (case when to_char(sh.sessioninstantiationtime,'D') = '1', THEN '1' ELSE null END) as Sun_Streams,
  Sum (case when to_char(SH.sessioninstantiationtime,'D') = '1' Then sh.tsdownstreambw else 0 end) / 1000/1000 as Sun_Bandwidth,
  Count (case when to_char(sh.sessioninstantiationtime,'D') = '2', THEN '1' ELSE null END) as Mon_Streams,
  Sum (case when to_char(SH.sessioninstantiationtime,'D') = '2' Then sh.tsdownstreambw else 0 end) / 1000/1000 as Mon_Bandwidth,
  Count (case when to_char(sh.sessioninstantiationtime,'D') = '3', THEN '1' ELSE null END) as Tue_Streams,
  Sum (case when to_char(SH.sessioninstantiationtime,'D') = '3' Then sh.tsdownstreambw else 0 end) / 1000/1000 as Tue_Bandwidth,
  Count (case when to_char(sh.sessioninstantiationtime,'D') = '4', THEN '1' ELSE null END) as Wed_Streams,
  Sum (case when to_char(SH.sessioninstantiationtime,'D') = '4' Then sh.tsdownstreambw else 0 end) / 1000/1000 as Wed_Bandwidth,
  Count (case when to_char(sh.sessioninstantiationtime,'D') = '5', THEN '1' ELSE null END) as Thu_Streams,
  Sum (case when to_char(SH.sessioninstantiationtime,'D') = '5' Then sh.tsdownstreambw else 0 end) / 1000/1000 as Thu_Bandwidth,
  Count (case when to_char(sh.sessioninstantiationtime,'D') = '6', THEN '1' ELSE null END) as Fri_Streams,
  Sum (case when to_char(SH.sessioninstantiationtime,'D') = '6' Then sh.tsdownstreambw else 0 end) / 1000/1000 as Fri_Bandwidth,
  Count (case when to_char(sh.sessioninstantiationtime,'D') = '7', THEN '1' ELSE null END) as Sat_Streams,
  Sum (case when to_char(SH.sessioninstantiationtime,'D') = '7' Then sh.tsdownstreambw else 0 end) / 1000/1000 Sat_Bandwidth
  Of
  sessionHistory sh
  WHERE
  SH.sessioninstantiationtime BETWEEN TO_DATE('01-aug-09', 'dd-mon-yy') - 6 AND TO_DATE (1 August 09 ',' dd-mon-yy') + 1
  GROUP BY
  To_char (sessioninstantiationtime, 'HH24') | ':' || To_char (Floor (TO_NUMBER (to_char (sessioninstantiationtime, 'MI')) / 15) * 15, "FM00")
  ORDER BY
  To_char (sessioninstantiationtime, 'HH24') | ':' || To_char (Floor (TO_NUMBER (to_char (sessioninstantiationtime, 'MI')) / 15) * 15, "FM00");

  Hello

  If you want to specify the 7th day of the period of 7 days instead of 1 day and then subtract 6 in the two places where this setting is used.
  If you want a variable number of periods / hour instead of 4, then replace the three occurrences of the 'magic number' 4 with the number of periods per hour (60 /: period_length, where: length of the period is the number of minutes, for example 5, 15, 30 or 60.)

  VARIABLE  report_end_date     VARCHAR2 (20)
  EXEC      :report_end_date     := '03-Jul-2009';
  
  VARIABLE  period_length          NUMBER
  EXEC     :period_length          := 5;
  
  WITH     all_periods     AS
  (
       SELECT  TO_DATE (:report_end_date, 'DD-Mon-YYYY') + ( (LEVEL - 1)
                                              / (24 * 60 / :period_length)
                                         )
                                       - 6     AS period_start
       ,     TO_DATE (:report_end_date, 'DD-Mon-YYYY') + (  LEVEL
                                              / (24 * 60 / :period_length)
                                         )
                                       - 6     AS next_period_start
       FROM    dual
       CONNECT BY     LEVEL <= 7 * 24 * 60 / :period_length
  )
  SELECT     TO_CHAR (ap.period_start, 'HH24:MI')     AS period_start_time
  ,     COUNT    (CASE WHEN TO_CHAR (ap.period_start, 'Dy') = 'Fri' THEN sh.tsdownstreambw END)     AS fri_streams
  ,     NVL (SUM (CASE WHEN TO_CHAR (ap.period_start, 'Dy') = 'Fri' THEN sh.tsdownstreambw END), 0)
             / (1000 * 1000)                                             AS fri_bandwidth
  ,     COUNT (   CASE WHEN TO_CHAR (ap.period_start, 'Dy') = 'Sat' THEN sh.tsdownstreambw END)     AS sat_streams
  ,     NVL (SUM (CASE WHEN TO_CHAR (ap.period_start, 'Dy') = 'Sat' THEN sh.tsdownstreambw END), 0)
             / (1000 * 1000)                                             AS sat_bandwidth
  FROM           all_periods     ap
  LEFT OUTER JOIN  sessionhistory     sh     ON    sh.sessioninstantiationtime <  ap.next_period_start
                                    AND   sh.sessiondestructiontime   >= ap.period_start
  WHERE       TO_CHAR (ap.period_start, 'HH24:MI')     LIKE '01:__'     --     *****  FOR TESTING ONLY!  *****
  GROUP BY  TO_CHAR (ap.period_start, 'HH24:MI')
  ORDER BY  TO_CHAR (ap.period_start, 'HH24:MI')
  ;
  

  Note that the main request is completely unchanged.

  Out with the above parameters and the set of two rows of sample data:

  .           FRI_      FRI_       SAT_      SAT_
  PERIO    STREAMS BANDWIDTH    STREAMS BANDWIDTH
  ----- ---------- --------- ---------- ---------
  01:00          0      .000          0      .000
  01:05          0      .000          0      .000
  01:10          0      .000          0      .000
  01:15          0      .000          0      .000
  01:20          0      .000          0      .000
  01:25          0      .000          0      .000
  01:30          2     7.500          0      .000
  01:35          2     7.500          0      .000
  01:40          1     3.750          0      .000
  01:45          1     3.750          0      .000
  01:50          1     3.750          0      .000
  01:55          0      .000          0      .000
  

  The query will cover a period of 7 days consecutive ending with (and including): report_end_date.
  'Saturday' in this example means June 27, 6 days before Friday.

• How SQL Workshop of Apex valid DDL?

  I was getting a runtime error to install to an Apex application scripts, and I went down to this piece of code:
  

  CREATE TABLE dummy
     (     "CREATED_BY" VARCHAR2(20) DEFAULT sys_context('USERENV','CLIENT_IDENTIFIER'));

  If I run it via SQL * more (Instant Client 11g or one with XE), it is not report any error.
  If I run it through Home > workshop SQL > SQL commands, it errors with a "ORA-01401: inserted value too large for the column.
  It's (kind of) true, such as a SYS_CONTEXT peut return a longer string of 20 characters. If I do the column 500 characters, or make a substring so no error is reported and the table is created.
  
  I just can't work on how it gets the error in the creation of the table when SQL * more additional operators.
  Back-end database is XE (10.2.0.1)

  Myers,

  If you are running

  select sys_context('USERENV','CLIENT_IDENTIFIER') from dual;
  

  Apex-> SQL workshop, you can see it returns 'USERNAME:SESSION_VALUE' but if you run the same SQL * more ot TOAD, then returns a NULL value.

  It is obvious that when you set the value "default" for any column, oracle will check if the defined length will be sufficient to contain the value 'default '. So, when you run it from SQL * more, it returns only NULL if no problem, but when you run from APEX, it returns user_name:session_value through your defined length!

  See you soon,.
  Hari

• 2 year + old gateway machine is suddenly popping up messages regarding Windows 7 is not valid not - how can I prove it my installation is valid?

  I get a message on a bridge that I bought 2.5 years ago at Frys in Renton, WA - all of a sudden the messages began to appear on Windows 7 is not invalid.  I had the computer here in Italy for 2 years - it has been working well. The sticker on the machine said 'Win 7 Home Prem OA - Acer Group'.  I tried to use the "windows genuine advantage" forum on social.microsoft.com, but the help guys wasn't able to resolve the problem (see http://social.microsoft.com/Forums/en-US/genuinewindows7/thread/7ca496e8-f998-4e1b-bf48-328db7864dc7)

  Any suggestions?

  No problem - except that Support MS once again failed miserably to even understand the problem and even less to repair.

  As I said in the other place, you can call back and see if they can do a good job this time :)