Query to find the group name for the Business of the employee

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  What is the query to find the name of all applications for all EBS R12.1.3 modules?

  With regard to:

  Mr. Shahzad Saleem

  Try:

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• Query to find the Pages and the list of permissions associated with a particular role in PS

  I want a query to find the Pages and the list of permissions associated with specific roles in PS.

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• SQL query to find the total number of source based nonsource passangersbetween source and destination station and passenger station on the same chekindate

  Hello

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• query to find the dependent task, attached to the task on a response in OIM 11 g

  can someone help me to do a sql query to find the dependent task, attached to the task on a response in OIM 11 g
  
  Published by: user13331347 on Sep 3, 2012 14:09

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• Query to find the coordinates of employee salary

  Hello
  
  Could someone help write the query to find the salary of the employee details.
  
  Thanks in advance.

  This should help you get started:

  SELECT papf.full_name
  papf.email_address
  ppp.proposed_salary_n salary
  OF per_pay_proposals ppp
  per_all_assignments_f ADP
  per_all_people_f women's wear
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  AND paaf.person_id = papf.person_id
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  AND trunc (sysdate) BETWEEN
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• How to find the employee is rehired

  Hello
  
  How to find the employee is rehired?
  
  through Front End and back End
  
  Thank you.

  You're looking to identify if an employee was re-employed, or looking to find all employees who were rehired?

  To determine if an employee was re-employed:
  (1) place you * HRMS Manager > people > enter & maintain
  (2) look at the employee's most recent Start Date field
  (3) compare this to their first Date field engaged in the Services tab

  If the two are different, it means that the employee was re-employed.

  Technically, each period of employment is linked to their period of Service. They will be on the per_periods_of_service table. So, there are 3 rows in per_periods_of_service if an employee was re-employed twice (that is to say, initially engaged and then reembauche twice).

  I hope this helps.

• Query to find the name of the partition

  Hi all

  I'm a little surprised to see that it is not a column I can use to query the name of the partition for a table, in the case of partition of Internval per day...

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  Then, you create a copy of dba_tab_partitions, since you can only convert the LONG data during an insertion type or create the table.

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• VPN access query remote ASA - several group policies for the unique connection profile

  Hi all

  Two quick questions here that I need to help.

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  2. in an ASA-5525 for Clientless Remote access VPN, can pass us the page to connect to an external server? For example, if I have a connection with a URL profile setup: "'https://wyz.vpn.com/ ';" for the LDAP/Radius Authentication, but for https://wyz.vpn.com/data and https://wyz.vpn.com/test I want to HTTP based authentication form and this page needs to be sent to an external server that is to say ASA step will manage this page, but rather the first page for this is served by the external server.

  Scenario: One of our clients is running F5 Firepass to their VPN solution. On the F5 they have pages of configuration such as the https://wyz.vpn.com/ that the F5 shows to the user when they connect via VPN without client; However if the user types https://wyz.vpn.com/data in the browser, the traffic comes to the F5, but F5 redirects this traffic to an external server (with an external url as well). Then it's this external server that transfers the first page of the user requesting authentication for HTTP form based authentication information.

  Thanks in advance to all!

  Hello

  You can have fallback to LOCAL only primary method.

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  HTH

  Averroès.

• Query to find the sum of the different groups in the same column

  Hello
  
  I have a table as follows:
  
  
  
  visitors
  ------------------------------------------------------------------------------------------
  months of amt credit CustID
  --------------------------------------------------------------------------------------------
  001 C 2000-Jan-2012
  
  001D 5000 Feb 2012
  
  001C 3000 Mar-2012
  
  001C 3000-Apr-2012
  
  001D 7000-may-2012
  
  
  
  I have to write a single query to calculate the sum of the credit and the sum of the flow values separately.
  
  
  Thank you and best regards,
  SB2011

  Hello

  Select decode (credit,' it, sum (amt), sum (amt)) in the Group of customers to credit;

  Published by: 937440 on 24 February 2013 07:06

• Query to find the number of attached document

  We've been checking out some of the attachments in the database and identified a number that are are very large, so we want to check what they are.   How can I find the number of invoice/order/request for registration of the ebs if all I have is the info from fnd_documents & fnd_lobs.

  You can collect the size of each by running this script:

  Select fad.entity_name, sum (DBMS_LOB. GETLENGTH (FL.file_data)) SIZE_BYTES;

  Count (*)

  of fnd_documents fnd_lobs fl, fnd_documents_tl fdtl, fd, fnd_attached_documents ADF

  where fd.media_id = fl.file_id

  and fd.document_id = fdtl.document_id

  and fd.document_id = fad.document_id

  and fl.program_name is null

  and fl.expiration_date is null

  and fd.datatype_id = 6

  and fdtl.language = 'en '.

  and (fad.entity_name like '% po2'

  OR fad.entity_name like '% REQ')

  Fad.entity_name group;

  You can find requisitions affected by the following:

  SELECT THE PH. SEGMENT1, PH. PO_REQUISITION_LINES_ALL PL, PH PO_REQUISITION_HEADERS_ALL REQUISITION_HEADER_ID

  WHERE PL. REQUISITION_HEADER_ID = PH. REQUISITION_HEADER_ID

  AND PH. REQUISITION_HEADER_ID in)

  SELECT PK1_VALUE FROM FND_ATTACHED_DOCUMENTS

  WHERE ENTITY_NAME = 'REQ_HEADERS');

  SELECT THE PH. SEGMENT1, PL. REQUISITION_LINE_ID
  OF PO_REQUISITION_LINES_ALL PL,
  PO_REQUISITION_HEADERS_ALL PH
  WHERE PL. REQUISITION_HEADER_ID = PH. REQUISITION_HEADER_ID
  AND PL. REQUISITION_LINE_ID in)

  SELECT PK1_VALUE FROM FND_ATTACHED_DOCUMENTS

  WHERE ENTITY_NAME = 'REQ_LINES');

  Please mark answers 'Proper' or 'Useful', as appropriate, to facilitate solutions to the other participants of the RTO.

  If you do not see these buttons, open the chat in a new window/tab by right clicking on the object and choose "open in a new tab.

  You should see Correct and helpful icons below each post.

• Query to find the previous activity in STM

  I have a data base to reflect a State of Transition (STM) Machine. It contains 3 tables: activity, node, and Transition.
  
  Tables of knots and Transition are the configuration of the STM. They define the different possible States and the possible transitions between States.
  
  The way that I take through the STM is recorded in the activity Table. A record is created for each visited node.
  
  I need a query that uses the id of my current activity to find my previous activity, i.e. the activity that precedes me directly. The following rules apply:
  -My Id is meaningless
  -It should work for different configurations of the transition and the node tables, but:
  -The starting node is always the node "A".
  -There is no recursive transition, for example ('B', 'B');
  -There is no circular transitions, for example ('B', 'C') and ('C', 'B')
  -Transition and node tables will never change between the creation of the first activity and the execution of the query.
  -The path reflected in the activity table is always a valid path.
  -My current activity is always the last activity.
  
  For all data below, find the previous activity of activity with id = 4. The correct answer is "C".
  

  DROP TABLE Transition;
  DROP TABLE Activity;
  DROP TABLE Node;
  
  CREATE TABLE Node
  (
       Id VARCHAR2(1) PRIMARY KEY
  );
  
  CREATE TABLE Activity
  (
       Id Number(8,0) PRIMARY KEY,
       Node_Id VARCHAR2(1)
  );
  
  CREATE TABLE Transition
  (
       FromNode_Id VARCHAR2(1),
       ToNode_Id VARCHAR2(1)
  );
  
  ALTER TABLE Activity
  ADD FOREIGN KEY
  (Node_Id) REFERENCES Node(Id);
  
  ALTER TABLE Transition
  ADD FOREIGN KEY
  (FromNode_Id) REFERENCES Node(Id);
  
  ALTER TABLE Transition
  ADD FOREIGN KEY
  (ToNode_Id) REFERENCES Node(Id);
  
  INSERT INTO Node VALUES ('A');
  INSERT INTO Node VALUES ('B');
  INSERT INTO Node VALUES ('C');
  INSERT INTO Node VALUES ('D');
  
  INSERT INTO Transition VALUES ('A','B');
  INSERT INTO Transition VALUES ('B','C');
  INSERT INTO Transition VALUES ('B','D');
  INSERT INTO Transition VALUES ('C','D');
  
  INSERT INTO Activity VALUES (1,'A');
  INSERT INTO Activity VALUES (2,'B');
  INSERT INTO Activity VALUES (3,'C');
  INSERT INTO Activity VALUES (4,'D');

  Desired output:

  ID
  -
  C

  Hello

  Assuming that all Activity_id tells us nothing on the way, but all lines of activity together to form a single path, we can get results this way:

  WITH     all_paths    AS
  (
       SELECT     'A' || SYS_CONNECT_BY_PATH (a.Node_id, '/')
                || '/'     AS node_id_path
       FROM     Transition  t
       JOIN     Activity    a  ON  a.Node_Id  = t.ToNode_Id
       WHERE     LEVEL             = (
                              SELECT  COUNT (*)
                           FROM     activity
                           ) - 1
       START WITH  t.FromNode_Id  = 'A'
       CONNECT BY  t.FromNode_Id  = PRIOR t.ToNode_Id
  )
  SELECT       REGEXP_SUBSTR ( node_id_path
                   , '([^/]+)/' || (
                                      SELECT  Node_Id
                               FROM    activity
                               WHERE   Id     = 4  -- target_id
                                  )
                           || '/'
                 , 1
                 , 1
                 , NULL
                 , 1
                 )     AS prev_id
  FROM       all_paths
  ;
  

  This means, there is some character (I used "/" above) that we know never appears in Activity.Node_Id.

  Since there are no loops in Transition, it cannot be more than 1 way that involves all activity lines. If there are N lines in operation, this full path will be one that extends to LEVEL = N - 1. (It is not N, N - 1, because the join between the activity and the Transition will always leave 1 line in activity with Node_Id = 'A'). As soon as we have the full path (which is node_id_path in 1 row, produced by all_paths), we just need to guess what was the Node_Id, which corresponds to the id of the target (4 in this example) and find the previous of the delimited node_id_path node_id.

• Query to find the nth highest salary...

  Hi guys,.
  
  I can't understand this query that we got a Web site. It is used to find the nth highest salary of the employee... Can someone explain to me please each and every part of it...
  
  The Charly is:
  
  Select distinct (a.salary)
  EMP a
  where 1 = (select count (distinct (salary))
  b emp
  where a.salary < = b.salary)
  
  The "BOLD" part which is the return value... is my main question about this query...
  
  Help, please

  It is called a correlated sub query. The inner query is executed for each iteration of the outer query. For example, if the table emp has 14 rows and then for each salary 14 the inner query will run.
  The part

  select count(distinct(sal))
  from emp b
  where a.sal<=b.sal
  

  actually counts the number of treatments that are less than or equal to the treatment of the outer query.
  Clearly for the King, who gets the highest salary is greater than or equal to 1 (himself). It returns the highest salary.

  SQL> select *
    2  from emp a
    3  where 1 >= (select count(distinct(sal))
    4  from emp b
    5  where a.sal<=b.sal)
    6  order by sal desc;
  
       EMPNO ENAME                               JOB              MGR HIREDATE         SAL       COMM     DEPTNO
  ---------- ----------------------------------- --------- ---------- --------- ---------- ---------- ----------
        7839 KING                                PRESIDENT            17-NOV-81       5000                    10
  

  Salary of the first two.

  SQL> ed
  Wrote file afiedt.buf
  
    1  select *
    2  from emp a
    3  where 2 >= (select count(distinct(sal))
    4  from emp b
    5  where a.sal<=b.sal)
    6* order by sal desc
  SQL> /
  
       EMPNO ENAME                               JOB              MGR HIREDATE         SAL       COMM     DEPTNO
  ---------- ----------------------------------- --------- ---------- --------- ---------- ---------- ----------
        7839 KING                                PRESIDENT            17-NOV-81       5000                    10
        7902 FORD                                ANALYST         7566 03-DEC-81       3000                    20
        7788 SCOTT                               ANALYST         7566 09-DEC-82       3000                    20
  

• Query to find the setting and add it to the minimum value

  Hi, I need a query to run under operation
  I have a table as shown below
  
  chg tax code Bill rec_no r_cost d_cost
  12345 1 114,35 100.30 1.00 1.00 234
  2 12345 114,35 2.00 1.00 1.00 201
  3 12345 114,35 14.00 1.00 1.00 600
  4 23412 120.90 110,40 1.00 1.05 700
  5 23412 120.90 12,50 1.00 1.05 900
  
  
  In this table, I need to find the setting for each rec_no as
  (r_cost + chg + taxes)-sum (d_cost) for rec_no = 12345
  (114.35 + 1.00 1.00)-(100.30+2.00+14.00) plus 0.05
  This 0.05 must be added to the minimum (code) for the rec_no 12345.
  If the d_cost of the code, with rec_no = 12345 201 should be (2.00 + 0.05 = 2.05).
  
  and,
  (120.90 + 1.00 1.05)-(110.40+12.50) plus 0.05
  This 0.05 must be added to the minimum (code) for the rec_no 23412.
  If the d_cost of the code, with rec_no = 23412 700 should be (110,40 + 0.05 = 110.45).
  
  Kindly help me in writing of an oracle application to perform the above operation.

   with tab as (
      select 1 invoice,12345 rec_no,114.35 r_cost,100.30 d_cost,1.00 chg,1.00 tax,234 code from dual union all
      select 2 invoice,12345 rec_no,114.35 r_cost,2.00 d_cost,1.00 chg,1.00 tax,201 code from dual union all
      select 3 invoice,12345 rec_no,114.35 r_cost,14.00 d_cost,1.00 chg,1.00 tax,600 code from dual union all
      select 4 invoice,23412 rec_no,120.90 r_cost,110.40 d_cost,1.00 chg,1.05 tax,700 code from dual union all
      select 5 invoice,23412 rec_no,120.90 r_cost,12.50 d_cost,1.00 chg,1.05 tax,900 code from dual
  )
   SELECT tab.*,
    CASE
      WHEN DENSE_RANK() OVER (partition BY rec_no ORDER BY code)=1
      THEN r_cost + chg +tax-SUM(d_cost) over (partition BY rec_no)+MIN(d_cost) keep (DENSE_RANK FIRST
     ORDER BY code) over (partition BY rec_no)
      ELSE d_cost
    END adj
     FROM tab
  ORDER BY invoice
  

  Find out what...

  Ravi Kumar

• query to find the number of dates

  Hi all

  I need assistance in creating a query. Ask the details I have given below.

  Table:

  Create the table emp_record (emp_id number, emp_join_date date);

  insert into emp_record values (1, 1 June 14 ');

  insert into emp_record values (2, 26 May 14 ');

  insert into emp_record values (3, 16 May 14 ');

  insert into emp_record values (4, 14 May 14 ');

  Suppose I ran the query with sysdate, we need to subtract the date of sysdate, then a report should be prepared in function number how many have subtracted values sysdate below 0 to 3 days, 3-6 date.

  Output should be like

  0 - 3 3-6 6-9 days more than 9 days

  1              0             1                    2

  Please help to write this query. Thank you

  Hello

  Do not essentially repeat the same CASE expression with the same types of calculations for each column.  You can do the job with a CASE expression, like this:

  WITH got_group_id AS

  (

  SELECT THE CHECK BOX

  WHEN emp_join_date< trunc="" (sysdate)="" -="" 8 ="" then ="">

  WHEN emp_join_date< trunc="" (sysdate)="" -="" 5 ="" then ="">

  WHEN emp_join_date< trunc="" (sysdate)="" -="" 2 ="" then ="">

  0 OTHERWISE

  Group_id END AS

  Of emp_record

  WHERE emp_join_date<= sysdate ="" --="" if="">

  )

  SELECT *.

  OF got_group_id

  PIVOT (COUNT (*)

  FOR group_id IN (0 '0-2 days'

  , 3 AS a "3-5 days.

  6 as a "6-8 days"

  , 9 as "9 or more days.

  )

  )

  ;

  It will be much easier to maintain when needs change.

  ED mentioned, using a string, 1 June 14 ' in a place where we expect a date is just a matter of trouble.  2-digit years are too.