outer join does not give the expected results (missing documents)

Similar Questions

• % ROWCOUNT SQL does not return the expected result

  I have the following function within a package:

  --Update APPERY_JTI_deleted_USERS table:
  FUNCTION fn_updt_app_jti_dlt_usr(
      p_update_co      IN VARCHAR2,
      p_appery_user_id IN APPERY_JTI_deleted_USERS.appery_user_id%TYPE,
      p_outlet_code    IN APPERY_JTI_deleted_USERS.outlet_code%TYPE)
    RETURN NUMBER
  AS
  lv_sql       VARCHAR2(4000);
  lv_rowcount  NUMBER := 0;
  BEGIN
  
  lv_sql := 'UPDATE APPERY_JTI_deleted_USERS SET '||p_update_co||' = 1 WHERE '||p_update_co||' = 0 AND OUTLET_CODE = '''||p_outlet_code||''' AND APPERY_USER_ID = '''||p_appery_user_id||'''';
  
  --EXECUTE IMMEDIATE lv_sql;
  EXECUTE IMMEDIATE 'BEGIN ' || lv_sql || '; :z := sql%rowcount; END; ' USING OUT lv_rowcount ;
  
  RETURN lv_rowcount;
    
  EXCEPTION
  WHEN OTHERS THEN
    RETURN -1;  
  END fn_updt_app_jti_dlt_usr;
  
  

  The function called several times as part of a job. Basically, the main function (Say M), call a few functions (for example A1... A9). Each of these functions is to do something and updated the application using the above function. Therefore, each of the nine functions will call the above function with different parameters.

  The problem that I am facing is:

  First run, invoke only first of all for the above function will return the positive result ($sql rowcount > 0). When I run M the second time, only first and second calls for the above function will be to return positive results and so on.

  How odd, it's that if I change the values again and run a function M, it will start from the beginning. And what strange more, the dynamic update statement is executed correctly and data are updated successfully. It's just sql rowcount % do not return is not the expected results.

  I tried to run execute immediately without (begin, end). I also tried to run it without HELP ON. It's always the same. I'm really confused, what Miss me here.

  I don't know if this is relevant. But for what it's worth, invoking the function above updates the SAME lines in the table appery_jti_deleted_users that the columns are different. So for all the work, the same lines need to be updated. Each function (A1... A9) will call the above function to update a different column of these lines.

  If you REALLY want to help you having US SHOW, not tell us:

  1. WHAT you do

  2. HOW to

  3. WHAT results you get

  4. WHAT results you expect to get

  First run, invoke only first of all for the above function will return the positive result ($sql rowcount > 0). When I run M the second time, only first and second calls for the above function will be to return positive results and so on.

  How odd, it's that if I change the values again and run a function M, it will start from the beginning. And what strange more, the dynamic update statement is executed correctly and data are updated successfully. It's just sql rowcount % do not return is not the expected results.

  Perhaps that the foregoing is true and maybe he's not. We have NO WAY of knowing because you don't SHOW US anything.

  I tried to run execute immediately without (begin, end). I also tried to run it without HELP ON. It's always the same.

  Yet once you showed us NOTHING >

  You do NOT FOLLOW best practices when you use dynamic sql statements: the sql statement real so that you can print:

  1 SEE what's running - make sure there are no syntax errors, and it seems to be what you wanted

  2 EXECUTE/TEST manually - to see if he really runs without error and see what results it really give.

  3. FIX any syntax or other problems and retest it

  Instrument properly your code and you won't have the problems you are having.

  You should know by now what are the parameters are transmitted when you run the function. So, there are at least three sets of these parameters.

  So you should be able to:

  1 run the instrumented manually function three times

  2. see that all three SQL statements are

  3 manually run these instructions

  4. see that the number of lines is after each execution

  We have no idea of what the output of the function is every time you call it. Apparently, you have, if you save the results of the function somewhere, but you didn't post any of this info.

  We need real details to help you - just listen to your story is not enough.

• Why is my zero-order hold to not give the expected result?

  I'm trying to reproduce the functionality of the function of the command discrete so that I can use it without having to have a ' loop control and Simulation ".» The attachment desiredresult.vi shows an example of the function to hold discreet order inside the ' loop control and Simulation ".» The other attachment (zeroorderholdquestion.vi) shows my attempt to achieve the same result. I multiplied all index of the loop of the sampling period for the index table for the output y resampled. I have this based on my understanding of the scheme of discrete command waiting given under http://zone.ni.com/reference/en-XX/help/371894G-01/lvsim/sim_dzoh/#details.

  However, I have difficulties to get the frequency of the wave before and after resampling to match to the top. (See attachment for more clarity on what I mean by that). Any suggestions on what I am doing wrong?

  Here, I wrote something similar perhaps to simulate a multi channel, mux, taste and will hold with crosstalk

  http://forums.NI.com/T5/LabVIEW/how-can-I-create-and-sample-and-hold-circuit-in-LabVIEW/m-p/2407256#...

  If time samplerate and hold have a clean report you just need help vi

  Another simple version:

  

• Oracle text index does not return the expected results.

  Hello
  
  My project requires a generic search on 5 different columns and all the VARCHAR2 data type.
  As performance is very slow with AS % search, I try to use for the multi_column_datastore that is recommended for this.
  
  Here's what I did.
  
  1 - Create table.

  CREATE TABLE BILLING_SCHEMA.NAMES_TABLE ( 
      ID                  NUMBER(9,0) NOT NULL,
      ENTITY_NUMBER         VARCHAR2(30) NOT NULL,
      GIVEN_NAMES         VARCHAR2(50) NULL,
      SURNAME             VARCHAR2(30) NULL,
      ORGANISATION_NAME     VARCHAR2(60) NULL,
      DISPLAY_NAME        VARCHAR2(100) NULL 
      )
  /
  
   

  My search must be made on the columns ENTITY_NUMBER, GIVEN_NAMES, surname, ORGANISATION_NAME and DISPLAY_NAME.
  
  2. - insert values into the table

  INSERT INTO NAMES_TABLE VALUES(1, 'GH', NULL, NULL, NULL, 'GROUP OF HOSPITAL')
  /
  INSERT INTO NAMES_TABLE VALUES(2, 'DH', NULL, NULL, NULL, 'DEPARTMENT OF HEALTH')
  /
  INSERT INTO NAMES_TABLE VALUES(3, 'REC', NULL, NULL, NULL, 'DEBTOR FOR RECEIPTS')
  /
  INSERT INTO NAMES_TABLE VALUES(4, 'UN_REC', NULL, NULL, NULL, 'DEBTOR FOR UNMATCHED RECEIPTS')
  /
  INSERT INTO NAMES_TABLE VALUES(5, 'SAM_FUND', NULL, NULL, NULL, 'THE S FUND')
  /
  
   

  3 - create context multi_column_datastore index
  

  BEGIN
  CTX_DDL.CREATE_PREFERENCE('NAMES_TABLE_MULTI', 'MULTI_COLUMN_DATASTORE');
  CTX_DDL.SET_ATTRIBUTE('NAMES_TABLE_MULTI', 'COLUMNS', 'ENTITY_NUMBER, GIVEN_NAMES, SURNAME, ORGANISATION_NAME, DISPLAY_NAME');
  END;
  
  CREATE INDEX NAMES_TABLE_IDX
  ON NAMES_TABLE(ENTITY_NUMBER)
  INDEXTYPE IS   CTXSYS.CONTEXT
  PARAMETERS ('DATASTORE NAMES_TABLE_MULTI')
  
   

  4 - create the trigger that should update the column ENTITY_NUMBER if everything is inserted/updated

  CREATE OR REPLACE TRIGGER NAMES_TABLE_TRIGGER
  BEFORE INSERT OR UPDATE OF GIVEN_NAMES, SURNAME, ORGANISATION_NAME, DISPLAY_NAME  ON NAMES_TABLE
  FOR EACH ROW
  BEGIN
    :NEW.ENTITY_NUMBER := :NEW.ENTITY_NUMBER;
  END
   

  5. the search query
  

  SELECT *
  FROM NAMES_TABLE
  WHERE  CONTAINS(ENTITY_NUMBER,'%TH%') > 0
   

  I am so very confused here because the result returned is only
  

  ID     ENTITY_NUMBER     GIVEN_NAMES     SURNAME     ORGANISATION_NAME     DISPLAY_NAME         
   2      DH                  (null)        (null)         (null)          DEPARTMENT OF HEALTH 
   

  When I wait for him that the 5th inserted record is returned too as it contains "THE S FUND" that begins with "TH".
  
  When I use the search string '% %' and %, no results are returned, although I expect at least two records in each case.
  
  I would appreciate help with this.
  
  Thanks in advance,
  Marilyn
  
  Published by: user5615580 on 08/03/2011 21:35
  
  Published by: user5615580 on 08/04/2011 16:33
  

• Request does not give the same result according to the computer

  Hi all

  My problem is very unclear to me, so I'll try to explain it the best I can...

  So I'm almost finished my project, and I want to try it on another computer.

  In my project, I generate a txt through several file datas, can I use this file to display graphical information on a graph. It works very well.

  So I made a kit of Distribution following this tutorial: http://www.howtocvi.net/2011/12/how-to-make-distribution-kit.html

  I installed the .exe on another computer, and I launched the program.

  My problem is that, for the same data, I used previously, the program is not the same behavior.

  My tracks on the chart do not look like they should be.

  I don't know what are the causes that both programs use the same algorithm...

  If you have an idea, I'll take it!

  Thanks for your answers,

  Kind regards

  Jerome.

  Well, the example was just stressing that debug and release can behave differently, I thought not that it could directly apply to your situation.

  If I were you, I'd start checking the inizialization of the variables, since it is one of the biggest differences between these two scenarios.

  After that, I will add a few DebugPrintf to print relevant during the life of the program variables: DebugPrintf output can be seen in the window Debug Output while developing the program and in utilities like DebugView when running the executable, so you can compare his production in two situations.

  Normally, after some trial and error clarifies the situation and the appropriate corrections can be made to the code.

• Outer join does not

  Hello
  
  Pls help me my request. I tried the following, but it does not give the expected results.
  
  Tab1
  EmplId, RepDt, Code, Hrs
  1/100,1/2009,199,8
  1/100,1/2009,200,4
  1/100,1/2009,255,3
  200,1/1/2009,200,4
  100.5/1/2009,199,8
  
  
  Tab2
  EmplId, RepDt, Code, Hrs
  1/100,1/2009,200,6
  200,1/1/2009,200,3
  
  I need output like this - 4 rows - need of all the rows from TAB1 when EMPLID, REPDT match TAB2, field values: need at a time when Code is also, when emlid, correspondence of the date, the Code missing so need A hrs, display ZERO as Tab2 hours.
  
  1/100,1/2009,199,8,0 - did not exist is not in tab2, so hrs Tab2 is ZERO
  1/100,1/2009,200,4,6 - Emplid, RepDt, Code match existence in Tab2, Tab1 Hrs 8, 6 Hrs Tab2
  1/100,1/2009,255,3,0 - did not exist is not in Tab2, so Tab2 hrs is equal to ZERO
  200,1/1/2009,200,4,3 - Emplid, Repdt, Code Match existed in Tab2, 4 Hrs of Tab1, Tab2 3 Hrs
  
  We are in 10g, Oracle.
  
  I tried the following
  T1. EmplId = T2.emplid AND T1.repdt = T2.repdt AND T1.code, T2.code = (+) - returns the unique corresponding lines, 2 rows.
  
  Pls help.
  
  Thanks in advance.
  
  Published by: NL 23 February 2009 09:20

  Hello

  See the Boneist first message in this thread:

  select t1.emplid, t1.repdt, t1.code, t1.hrs, nvl(t2.hrs, 0)
  from   tab1 t1,
         tab2 t2
  where  t1.emplid = t2.emplid (+)
  and    t1.repdt = t2.repdt (+)
  and    t1.code = t2.code (+)
  and    (t1.emplid, t1.repdt) in (select emplid, repdt
                                   from   tab2)
  order by repdt, emplid, code;
  

  As Boneist said, this fact corresponds to option (b)
  "(b) do just an outer join, with a condition EXISTS (or IN) in the WHERE clause to find corresponding repdts.
  The

  (+)
  

  an outer join are signs.

• Why "bgp bestpath missing-as-worst med" command does not produce the desired result?

  Can dear all, someone tell me why "bgp bestpath missing-as-worst med" command does not produce the desired result?
  And I use GNS3 to this practice, the IOS Version: 12.4 (3)
  TKS...

  Back to the beginning:
  192.168.23.0/24 192.168.12.0/24
  RA(F0/0)-(f0/0) RB (s2/0) - RC (s2/0)
  [AS1]                    [AS2]                    [AS3]

  Configuration:
  * RA *:
  A #do sh run (config - router). b r b
  router bgp 1
  no synchronization
  The log-neighbor BGP-changes
  1.1.1.0 netmask 255.255.255.0
  192.168.12.2 neighbor remote-2
  setmed map of nearby route 192.168.12.2 out
  No Auto-resume

  IP route 1.1.1.0 255.255.255.0 Null0

  setmed allowed 10 route map
  the metric value 20

  * RB *:
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  router bgp 2
  no synchronization
  BGP always-compare-med
  The log-neighbor BGP-changes
  BGP bestpath missing-as-worst med
  neighbor 192.168.12.1 distance-1
  neighbour 192.168.23.3 distance-3
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  * RC *:
  C (config - router) #do sh run | b r b
  router bgp 3
  no synchronization
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  neighbour 192.168.23.2 distance-2
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  IP route 1.1.1.0 255.255.255.0 Null0

  But when I checked out table RB bgp, as below:

  B (config - router) #do sh ip bgp
  BGP table version is 2, local router ID is 192.168.23.2
  Status codes: deleted, cushioning d s, history of h, * valid, > best, i - internal.
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  Network Next Hop path metrics LocPrf weight
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  *                          192.168.12.1            20                         0  1 i

  The prefix is assigned a MED of 0, but also the best path value when I use "bgp bestpath missing-as-worst med" in how to configure bgp... :-(

  Please take a look at the discussion on the configuration of drugs to an eBGP neighbor:

  https://supportforums.Cisco.com/thread/343397?TSTART=0

  Note that BGP MED is set to 0 even if the injected BGP route from a connected route.

  The first time I came across this problem sending MED was there in a picture of 5-6 years service provider. Routes were injected into BGP with the command 'network', has been validated by OSPF routes or static to null0. Involuntary consignment of drugs was mess with load balancing between 2 links of our inbound traffic from the same upstream service provider.

  Also note that if a route is learned via iBGP, border router removes MED before the road to advertising for an eBGP peers (i.e. the above comments apply a when a route is injected locally into the border and then router BGP sent to an eBGP peer).

  I don't know what the cause of the problem that you said in your last post. It delivers last for a long time or that it does not change after a while?

  p.s. Sorry, I edited post because I said "iBGP" instead of "eBGP" at some point.

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• Outer join does not not as expected left

  Hi all

  I have it here are three tables with values. Mentioned the under outer join query does not and behave like the inner query.

  CREATE TABLE RET_FUND_FEE

  (

  NPTF VARCHAR2 (8 CHAR),

  TPART VARCHAR2 (4 CHAR)

  );

  CREATE TABLE PART_PTF

  (

  Mf_Id VARCHAR2 (6 CHAR) NOT NULL,

  TPARTS VARCHAR2 (4 CHAR) NOT NULL

  );

  CREATE TABLE TFC_FUNDS

  (

  NPTF VARCHAR2 (8 CHAR) NOT NULL,

  MULTIFONDS_ID VARCHAR2 (6 CHAR)

  );

  INSERT INTO RET_FUND_FEE VALUES('111','A');

  INSERT INTO RET_FUND_FEE VALUES('111','D');

  INSERT INTO RET_FUND_FEE VALUES('111','E');

  INSERT INTO PART_PTF VALUES ('MF1', 'A');

  INSERT INTO PART_PTF VALUES ('MF1', 'B');

  INSERT INTO PART_PTF VALUES('MF1','C');

  INSERT INTO TFC_FUNDS VALUES('111','MF1');

  INSERT INTO TFC_FUNDS VALUES('111','MF1');

  INSERT INTO TFC_FUNDS VALUES('111','MF1');

  SELECT A.TPART, B.TPARTS, A.NPTF, B.Mf_Id, C.MULTIFONDS_ID, C.NPTF

  OF RET_FUND_FEE A, PART_PTF B, TFC_FUNDS C

  WHERE A.NPTF = C.NPTF

  AND C.MULTIFONDS_ID = B.Mf_Id

  AND A.TPART = B.TPARTS (+)

  AND C.MULTIFONDS_ID = 'MF1 '.

  AND C.NPTF = '111'

  Here, I expect all records in the RET_FUND_FEE table that I am using outer join.

  But I'm only corresponding chronogram RET_FUND_FEE, PART_PTF as an inner join. Can you get it someone please let me know what lack us.

  Is my version of oracle 11g

  SELECT

  A.TPART, B.TPARTS, A.NPTF, B.Mf_Id, C.MULTIFONDS_ID, C.NPTF

  Of

  PART_PTF B

  Join

  C TFC_FUNDS

  on (C.MULTIFONDS_ID = B.Mf_Id

  AND C.MULTIFONDS_ID = 'MF1 '.

  AND C.NPTF = '111'

  )

  right outer join

  RET_FUND_FEE HAS

  on (A.TPART = B.TPARTS

  and A.NPTF = C.NPTF) - added as correction

  TPART	TPARTS	NPTF	MF_ID	MULTIFONDS_ID	NPTF
  A	A	111	MF1	MF1	111
  A	A	111	MF1	MF1	111
  A	A	111	MF1	MF1	111
  E	-	111	-	-	-
  D	-	111	-	-	-

  or

  SELECT A.TPART, d.TPARTS, A.NPTF, d.Mf_Id, d.MULTIFONDS_ID, d.NPTF

  OF RET_FUND_FEE HAS

  , (

  Select

  *

  PART_PTF b, TFC_FUNDS C

  where B.Mf_Id = C.MULTIFONDS_ID

  AND C.MULTIFONDS_ID = 'MF1 '.

  AND C.NPTF = '111'

  ) d

  WHERE A.TPART = D.TPARTS (+)

  and A.NPTF = D.NPTF (+) - added as a correction

  Sorry had to correct the syntax oracle solution.

  The first one was bad because it would return also B lines that have no match in C.

  Sorry a correction more on these two approaches, missed the second predicate.

• Left Outer Join does not work why?

  Hello
  
  I have a question I want to return all the id group grp_id table and count and give me a count of all "active" members in the Member table. There are a few members who are not active in the table of members who belong to groups in the grp_id table. I still want a record returned for these grpid, but since there is no active members for these groups, I would like a zero as the mbr_count. An active mbr is the one whose eff_dt is less than the date of the day and whose exp_dt is greater than today's date.
  
  I have listed the query below, which returns active records mbr, but unfortunately does not return a grpid (as in my example of data) is not an active member (grpid: has ""). I have also provided the DDL for both tables as well as some examples of data that gives me the (Incorrect) results, that I posted. I have also posted what I need (the 'Correct' results).
  
  Thanks for any help... (I use PL/SQL)
  
  THE QUERY I USE:
  =============
  SELECT
  G00.grpid,
  Count (M00.grpid) mbr_count
  Of
  g00 grp_id
  LEFT OUTER JOIN
  m00 MBR on
  G00.grpid = m00.grpid
  WHERE
  M00.eff_dt < sysdate
  AND
  M00.exp_dt > sysdate
  GROUP BY
  G00.grpid
  
  It gives me results like:
  
  INCORRECT RESULTS:
  ===============
  GRPID MBR_COUNT
  A 2
  B 2
  C 1
  
  I want to see is:
  
  CORRECT RESULTS
  ===============
  GRPID MBR_COUNT
  A 2
  B 2
  C 1
  D 0
  
  
  Here's the DOF and the sample data:
  
  create table grp_id (grpid varchar (1))
  
  insert into grp_id values ('A')
  insert into grp_id values ('B')
  insert into grp_id values ('C')
  insert into grp_id values (')
  
  Commit
  
  create table mbr (mbr_name varchar (10), varchar (1) grpid, eff_dt date, exp_dt date)
  
  insert into values of mbr ('BRAND', 'A', to_date (January 1, 2011 ',' DD-MM-YYYY '), to_date (July 1, 2011 ',' DD-MM-YYYY '))
  insert into values of mbr ('BRAND', 'A', to_date (July 1, 2011 ',' DD-MM-YYYY '), to_date (1 January 2012 ',' DD-MM-YYYY '))
  
  insert into values of mbr ('MARTY', 'A', to_date (January 1, 2011 ',' DD-MM-YYYY '), to_date (July 1, 2011 ',' DD-MM-YYYY '))
  insert into values of mbr ('MARTY', 'A', to_date (July 1, 2011 ',' DD-MM-YYYY '), to_date (1 January 2012 ',' DD-MM-YYYY '))
  
  insert into values of mbr ('FRANK', 'B', to_date (January 1, 2011 ',' DD-MM-YYYY '), to_date (July 1, 2011 ',' DD-MM-YYYY '))
  insert into values of mbr ('FRANK', 'B', to_date (July 1, 2011 ',' DD-MM-YYYY '), to_date (1 January 2012 ',' DD-MM-YYYY '))
  
  insert into values of mbr ('MARY', 'B', to_date (January 1, 2011 ',' DD-MM-YYYY '), to_date (July 1, 2011 ',' DD-MM-YYYY '))
  insert into values of mbr ('MARY', 'B', to_date (July 1, 2011 ',' DD-MM-YYYY '), to_date (1 January 2012 ',' DD-MM-YYYY '))
  
  insert into values of mbr ('JOHN', 'C', to_date (January 1, 2011 ',' DD-MM-YYYY '), to_date (July 1, 2011 ',' DD-MM-YYYY '))
  insert into values of mbr ('JOHN', 'C', to_date (July 1, 2011 ',' DD-MM-YYYY '), to_date (1 January 2012 ',' DD-MM-YYYY '))
  
  insert into values of mbr ("NOAM" 'd', to_date (January 1, 2011 ',' DD-MM-YYYY '), to_date (July 1, 2011 ',' DD-MM-YYYY ') ")

  Thanks for the script, shame on the missing semicolon :)

  Your date restrictions are implemented as "filter predicates" rather than "join predicates.
  that is the difference between a place ANDS extra o in the JOIN clause.

  SQL> SELECT
    2  g00.grpid,
    3  count(m00.grpid) mbr_count
    4  FROM
    5  grp_id g00
    6  LEFT OUTER JOIN
    7  mbr m00 on
    8  g00.grpid = m00.grpid
    9  WHERE
   10  m00.eff_dt < sysdate
   11  AND
   12  m00.exp_dt > sysdate
   13  GROUP BY
   14  g00.grpid
   15  /
  
  G  MBR_COUNT
  - ----------
  A          4
  B          4
  C          2
  
  SQL> SELECT
    2  g00.grpid,
    3  count(m00.grpid) mbr_count
    4  FROM
    5  grp_id g00
    6  LEFT OUTER JOIN
    7  mbr m00 on
    8  g00.grpid = m00.grpid
    9  AND
   10  m00.eff_dt < sysdate
   11  AND
   12  m00.exp_dt > sysdate
   13  GROUP BY
   14  g00.grpid;
  
  G  MBR_COUNT
  - ----------
  D          0
  A          4
  B          4
  C          2
  
  SQL> 
  

• Want 4504: Hp Envy 4504 does not give the impression of color

  Hello

  My Hp Envy 4504 gives color impressions.

  Its keep giving only black & white or gray scale images.

  Also, I couldn't find the printing color profiles or presets under the printer settings.

  I only 'Black and White' and ' black and white - gray ' presets.

  What should I do to get the color print.

  I have two hp color and black ink cartridges installed.

  Help, please.

  Thank you

  Hi @sivak,

  Welcome to the Forums of HP Support!

  I understand that you are unable to print in color on your HP Envy 4504 printer. I will definitely do my best to help you.

  You are able to print a test highlighted report? The printer prints a blank Page or does not black print or ink color for the HP ENVY 4500 and Deskjet Ink Advantage 3540 e-all-in-One Printer Series.

  The problem only occurs when printing from Mac? It seems that there is a problem with the printer driver. Please see these instructions by my colleague Anti-double-U-di.

  Publish:

  10.10 yosemite, foot, make sure you have the latest updates, including Apple updates as well as updates from HP by Apple Software Update (ASU), currently HP printer software update Ver3.0

  Look for the updates (ASU) Apple
  1. click on the Apple in the left menu bar
  2. click on "about this Mac".
  3. click the button of "software updates... »
  4. make sure that the "Updates" icon is selected, click on 'Store' in the top menu
  5. click on 'reload the Page.
  6. install the HP updates (look for the last 'HP Printer Software Update"by the version number)

  Note: If you can't print or scan if he is an all-in-one HP printer, and then follow the steps below.

  Delete the print queue HP printer
  1. click on System Preferences
  2. click on printers & Scanners
  3. click on your printer HP to highlight, then click on the "-" symbol below to remove it from the queue
  4. click on the button "Delete printer" to confirm
  
  Add the HP printer in the print queue
  1. click on the symbol '+' at the bottom left
  2. If a drop down list with the printers menu appears, just click on "Add printer or Scanner..." »
  3 select your printer HP printer or series (depends USB/Ethernet/WLAN connection type, so choose your HP printer with the correct connection, IE Hello for printers single function and Hello to an MFP multifunction)
  4. after the lower filled, click on the drop down menu to "use:" and select your HP printer series
  5. click on the ' Add ' button.

  Try to print in coiour.

  If the steps above resolved your problem, you don't have to continue troubleshooting.  If you're still having problems, please continue to the next step.

  Before continues, the following steps will remove all printers to printers and Scanners, you may need your software and drivers for the printers to re - install if it is not part of the Apple software updates.

  Reset the printing system:
  1. open System Preferences
  2. open Printers and scanners
  3. on the left side where would be referred to as printers, right click and select "reset printing system...". »
  4. click on the "Reset" button when asked to confirm
  5. Enter your password, and then click OK (if prompted)
  6. Once completed, (notifications cannot pop up) you will simply see the Virgin 'printers & Scanners', just close

  Remove the software from HP/Hewlett-Packard:
  1. go on 'Finder '.
  2. go in "Applications".
  3. Locate and open the folder "Hewlett-Packard".
  4. If you see an "Uninstall HP program" listed here then run the uninstall program, otherwise, go to step 8
  5. click on the continue"" button.
  6. Enter your password, and then click OK (if prompted)
  7. When you are finished, click "exit".
  8 remove the HP folder or Hewlett-Packard (if still listed after uninstall) drag the folder to the trash
  9. right click on the Recycle Bin in the dock, click on "Empty" - and then confirm to empty the trash

  Repair disk permissions
  1. in the Finder, click Applications
  2. open the Utilities folder
  3. double-click on the disk utility
  4. on the left side, click on the hard drive which has Yosemite 10.10 installed to highlight
  5. click on the "repair disk permissions" nearly half lower the window (time may vary to finish)
  6. what repairs completed, a message near the bottom window 'Fix full permissions', now you can close the disk utility
  7. click on the Apple in the upper left corner, click "Restart" to restart your Mac

  Manually the drivers of HP printer install v3.0
  1. this link will take you to the apples for v3.0 HP printer drivers download
  2. click on the button 'Download' (remember where your download is if your downloads are not on the dock)
  3. open the "HPPrinterDrviers3.0.dmg".
  4. double-click on the "HewlettPackardPrinterDrivers.pkg".
  5. click on the continue"" button.
  6. After reading the license agreement, click the continue"" button.
  7. click on the button "I agree" If you agree
  8. click on the drive that Yosemite 10.10 is installed on and click on continue
  9. click on the 'Install' button (this will install the drivers for HP to your Mac)
  10. Enter your password, and then click the button 'Install Software' (if prompted)
  11. click on the "Close" button when the message "installation succeeded."

  Add a printer to print the queue
  1. open System Preferences
  2. open Printers and scanners
  3. click on the symbol ' + '.
  4. click on "Add a printer or a Scanner..." »
  5. click on "HP Envy printer 4504.
  6. click on the drop down menu to "use:" and choose "HP Envy 4504 Series."
  7. click on the ' Add ' button.

  Try to print in color.

  Please reply to this message with the result of your troubleshooting. I can't wait to hear strings attached good luck!

• I have run this query to get the result as below, but, even if my query is running fine, that I do not get the expected result.

  I'm looking for only column compare to my same target table as a table source.

  My query:
  

  
  

  Select case when column_name_s is null and column_name_t is not null

  then "alter table GRADE_CONVERSION drop | column_name_t | ';'

  When column_name_s is not null and column_name_t is null

  then "alter table GRADE_CONVERSION add | column_name_s | ' ' || data_type_s | « ; »

  else 'alter table modify GRADE_CONVERSION | column_name_t | ' ' || data_type_t | « ; »

  alterations of the end

  from (select s.column_name column_name_s, t.column_name column_name_t,

  s.data_type data_type_s, t.data_type data_type_t

  (select column_name, column_id, data_type

  of all_tab_cols@database

  where owner = 'erhan.

  and table_name = "GRADE_CONVERSION."

  + 1

  full outer join

  (select column_name, column_id, data_type

  of all_tab_cols@database

  where owner = 'sarigul.

  and table_name = "GRADE_CONVERSION."

  + 6

  on s.column_name = t.column_name

  )

  
  

  
  

  
  

  Tables:

  
  

  
  

  Target table: table GRADE_CONVERSION in sarigul@database

  
  

  LETTER_GRADE		VARCHAR2 (2)
  GRADE_POINT		NUMBER (3.2)
  MAX_GRADE		NUMBER (3)
  MIN_GRADE		NUMBER (3)

  
  

  
  

  
  

  Table source: Table GRADE_CONVERSION in erhan@database

  LETTER_GRADE		VARCHAR2 (2)
  GRADE_POINT		NUMBER (3.2)
  MAX_GRADE		NUMBER (3)
  MIN_GRADE		NUMBER (3)
  CREATED_BY		VARCHAR2 (30)
  CREATED_DATE		DATE
  MODIFIED_BY		VARCHAR2 (30)
  MODIFIED_DATE		DATE

  
  

  
  

  want to see output that is similar to this * (please ignore the names of column here it's just a clear example :))

  
  

  ALTER table Target_table change BOOK_ID Varchar2 (4);

  ALTER table Target_table I addSBN_10 Varchar2(13), null;

  ALTER table drop TITLE Target_table;

  Erhan_toronto wrote:

  1.I used src.nullable src_nullable and tgt.nullable tgt_nullable but only show Yes as below: but want to see the result as not null or null

  ALTER table TEST_TARGET change the NUMBER of MAX_GRADE (3, 2) Yes

  Ok. So it's all about the Yes and the no decoding to Default Null or Not Null, isn't it?

  So, to test, change one of the table of sample for NOT NULL columns in the source table, and then run the following query:

  with src as
  (
    select src.table_name src_table_name, src.column_name src_col_name, src.data_type src_data_type, src.data_length src_data_len, src.data_precision src_data_precision, src.data_scale src_data_scale,
           src.nullable src_nullable
      from user_tab_columns src
     where table_name = 'TEST_SOURCE'
  ),
  tgt as
  (
    select tgt.table_name tgt_table_name, tgt.column_name tgt_col_name, tgt.data_type tgt_data_type, tgt.data_length tgt_data_len, tgt.data_precision tgt_data_precision, tgt.data_scale tgt_data_scale,
           tgt.nullable tgt_nullable
      from user_tab_columns tgt
     where table_name = 'TEST_TARGET'
  ),
  col_details as
  (
    select src.src_table_name, nvl(tgt.tgt_table_name, first_value(tgt_table_name) over(order by tgt_table_name nulls last)) tgt_table_name,
           src.src_col_name, src.src_data_type, src.src_data_len, src.src_data_precision, src.src_data_scale, src.src_nullable,
           tgt.tgt_col_name, tgt.tgt_data_type, tgt.tgt_data_len, tgt.tgt_data_precision, tgt.tgt_data_scale, tgt.tgt_nullable
      from src
      left outer join tgt
        on (
            src.src_col_name = tgt.tgt_col_name
           )
  )
  select *
    from (
          select case
                  when tgt_data_type != src_data_type or tgt_data_len != src_data_len or tgt_data_precision != src_data_precision or tgt_data_scale != src_data_scale or src_nullable != tgt_nullable
                    then 'alter table ' || tgt_table_name || ' modify ' || tgt_col_name || ' ' || src_data_type || ' (' ||
                    case when src_data_type in ('DATE') then null
                         else
                              case
                                when src_data_type in ('VARCHAR', 'VARCHAR2')
                                  then nvl(to_char(src_data_len), ' ') || ') '
                                else  decode(nvl(src_data_precision, -1), -1, null, nvl(to_char(src_data_precision), ' ') || ', ' || nvl(to_char(src_data_scale), ' ') || ')')
                              end
                    end
                    || decode(src_nullable, 'NO', ' NOT NULL', ' DEFAULT NULL')
                  when tgt_col_name is null
                    then 'alter table ' || tgt_table_name || ' add ' || src_col_name || ' ' || src_data_type ||
                    case when src_data_type in ('DATE') then null
                         else
                              case
                                when src_data_type in ('VARCHAR', 'VARCHAR2')
                                  then nvl(to_char(src_data_len), ' ') || ') '
                                else  decode(nvl(src_data_precision, -1), -1, null, nvl(to_char(src_data_precision), ' ') || ', ' || nvl(to_char(src_data_scale), ' ') || ')')
                              end
                    end
                    || decode(src_nullable, 'NO', ' NOT NULL', ' DEFAULT NULL')
                 end alter_statement
            from col_details
          )
  where alter_statement is not null;
  

  Erhan_toronto wrote:

  2. when I run below under user sarigul and erhan I get the result as OWNER, TABLE_NAME, COLUMN_NAME DATA_TYPE... I have a link between two users. They have access to two tables.

  • Select * from all_tab_columns

  where owner = 'erhan' and table_name = "TEST_SOURCE."

  • Select * from all_tab_columns

  where owner = 'sarigul' and table_name = "TEST_TARGET."

  Alright. This means that you both users are on the same database. Only change, you will have to do in the above query is so change user_tab_columns to all_tab_columns and add the OWNER predicate respectively with the clause.