Group column 2 cfoutput split helps

Similar Questions

• [BEEP 10G] View the report in grouping columns

  I need your help to build a model to display the grouping column for the same data source id:

  For example: a model of (target) associated with this data source:

  id  Value                             

  a                                 X1

  a                                 X2

  a                                 X3

  b                                 Y

  Output:

  ID	Value
  one	X 1 X 2 X 3
  b	THERE

  Best regards

  Just

  Your Assistant model builder (RTF) table has the ability to "Left group" data that will give you the expected results by merging identical values.

• different values on the same grouping column

  Hi all

  I have a requirement to find different recordings of the same grouping columns have value different values in the columns, for example:

  WITH

  Data (G1, COL)

  AS

  (

  SELECT 'A', '456' FROM dual UNION ALL

  SELECT 'A', '345' FROM dual UNION ALL

  SELECT 'A', ' ' FROM dual UNION ALL

  SELECT ' 'B, 123' FROM dual UNION ALL.

  SELECT ' 'B, 123' FROM dual UNION ALL.

  SELECT ' 'B, 123' FROM dual UNION ALL.

  SELECT 'C', "789' FROM dual UNION ALL

  SELECT 'C', "789' FROM dual UNION ALL

  SELECT 'C', "" the DOUBLE

  )

  Select g1, pass data

  Ideally for G1 = A, the NECK should have that a single value across records ' 456 'or ' 345'. So I need to check the values in NOT NULL of the COL where, for any value of G1, we have different values of the colonel G1 = 'B' is perfectly fine, so I don't need. G1 = C is good too and I'm not worried about it. I only need these entries when the neck was non-empty value / NULL not different for the same value of G1.

  So expected out put is

  

  WITH

  Data (G1, COL)

  AS

  (

  SELECT 'A', '456' FROM dual UNION ALL

  SELECT 'A', '345' FROM dual UNION ALL

  SELECT 'A', "double UNION ALL

  SELECT ' 'B, 123' FROM dual UNION ALL.

  SELECT ' 'B, 123' FROM dual UNION ALL.

  SELECT ' 'B, 123' FROM dual UNION ALL.

  SELECT 'C', "789' FROM dual UNION ALL

  SELECT 'C', "789' FROM dual UNION ALL

  SELECT 'C', "the DOUBLE

  )

  Select g1,

  Col

  de)

  Select g1,

  Col,

  Count (distinct trim (col)) at the NTC (g1 partition)

  from the data

  )

  where NTC! = 1

  and trim (col) is not null

  /

  G COL. NO.
  - ---
  A 345
  A 456

  SQL >

  SY.

• Concatenation of the columns with no grouped columns

  Hi all

  Oracle Database 11 g Enterprise Edition Release 11.2.0.3.0 - 64 bit Production

  PL/SQL Release 11.2.0.3.0 - Production

  CORE Production 11.2.0.3.0

  AMT for Linux: Version 11.2.0.3.0 - Production

  NLSRTL Version 11.2.0.3.0 - Production

  I have a normal table,

  Here is a simulated with data table

  {code}

  SELECT "P1" as c1, c2 1 OF DOUBLE UNION ALL

  SELECT "P2", 1 FROM dual UNION ALL

  SELECT "P3", 2 FROM dual UNION ALL

  SELECT "P4", 2 DOUBLE

  {code}

  I need output like this:

  P1 + P3

  P1 + P4

  P2 + P3

  P2 + P4

  LISTAGG tried but it didn't work.

  {code}

  SELECT a GROUP (COMMANDER IN c1) OVER (PARTITION BY c2) FROM LISTAGG(t.c1, '+')

  (SELECT "P1" as c1, c2 1 OF DOUBLE UNION ALL

  SELECT "P2", 1 FROM dual UNION ALL

  SELECT "P3", 2 FROM dual UNION ALL

  SELECT "P4", 2 DOUBLE) t;

  {code}

  How to concatenate based on non grouped column here?

  ThX

  Shan.

  Hello

  Oh!  You want to generate all possible combinations of the different values of c2.

  Forget LISTAGG: it generates a list of a group or partition of lines that all have something in common, for example, they all have the same value of c2.

  You want just the opposite.  You want a list where the items have nothing in common, that is, they all have different values of c2.  CONNECTION BY can do that.

  If c2 is always consecutive integers (no gap) from 1, you can do this:

  SELECT LTRIM (SYS_CONNECT_BY_PATH (c1, '+'))
  , '+'
  ) AS combination
  T
  WHERE CONNECT_BY_ISLEAF = 1
  START WITH c2 = 1
  CONNECT BY PRIOR c2 + c2 = 1
  ;

  If c2 does not meet these requirements, you can generate c2_num this fact:

  WITH got_c2_num AS
  (
  SELECT c1, c2
  DENSE_RANK () OVER (ORDER BY c2) AS c2_num,
  T
  )
  SELECT LTRIM (SYS_CONNECT_BY_PATH (c1, '+'))
  , '+'
  ) AS combination
  OF got_c2_num
  WHERE CONNECT_BY_ISLEAF = 1
  START WITH c2_num = 1
  CONNECT BY c2_num = 1 + PRIOR c2_num
  ;

  Whatever it is, you need not to know how many different values of c2 will take place.

• Is it possible to group several columns individually without the help of the union?

  Hi, if I have a table with a column numebr I want group, for example, sex, nationality, age and I want to present a summary of this kind:
  
  County group
  10 male
  23 women
  British 19
  14 Irish
  18 10
  
  ...
  etc.
  
  Or even better:
  County group
  10 men
  Sex-wife 23
  British Nationaility 19
  14 Irish
  10 to 18 years
  
  Is it possible without using loads of trade unions? I just want to do the best way and most effective the table involved has > 100 million lines.
  
  Thank you very much
  
  Jen :)

  Hey, Jen,

  Here is another way, which could be faster because it does not unpivot data:

  SELECT    CASE  0
             WHEN  GROUPING (gender)       THEN  'Gender'
             WHEN  GROUPING (nationality)  THEN 'Nationality'
             WHEN  GROUPING (age)         THEN 'Age'
         END                    AS label
  ,       COALESCE ( gender
                 , nationality
               , TO_CHAR (age)
               )           AS val
  ,       COUNT (*)          AS cnt
  FROM       t
  GROUP BY  GROUPING SETS ( (gender)
                            , (nationality)
                 , (age)
                 )
  ORDER BY  label
  ,            val
  ;
  

  Output, using the data of knapen:

  LABEL           VAL                    CNT
  --------------- --------------- ----------
  Age             25                       2
  Age             30                       2
  Age             40                       2
  Age             45                       2
  Age             50                       2
  Gender          F                        5
  Gender          M                        5
  Nationality     BRITISH                  4
  Nationality     INDIAN                   3
  Nationality     IRISH                    3
  

• help with query - where the multiples in the Group column

  4 columns in the table named RRID1
  
  RRI_ID
  USER_ID
  PROCESS_TYPE
  FUNCTION
  
  don't know how to make this request.
  
  Need to shoot where all RRI_ID PROCESSTYPE = 'INFO' and PROCESSTYPE = "RESOLUTION".
  for this RRI_ID.
  This same user, same RRI_ID and 2 processes the records, 1 INFO and RESOLUTION 1
  so unless there is double info, only if there is info and resolution.
  
  That's what some of the data looks like to...
  
  PROCESSTYPE | USER_ID. FUNCTION | RRI_ID
  
  3668 30287 817 INFO
  3748 30287 817 INFO
  4714 30287 817 INFO
  4875 30287 817 INFO
  4882 30287 817 INFO
  4885 30287 817 INFO
  3748 30288 818 INFO
  4714 30288 818 INFO
  4716 30288 818 INFO
  4882 30288 818 INFO
  4885 30288 818 INFO
  819 30289 3668 INFO
  819 30289 4716 INFO
  819 30289 4875 INFO
  819 30289 4882 INFO
  819 30289 4885 INFO
  3668 30290 820 INFO
  4716 30290 820 INFO
  4875 30290 820 INFO
  4882 30290 820 INFO
  4885 30290 820 INFO
  3668 30291 821 INFO
  4714 30291 821 INFO
  4716 30291 821 INFO
  4875 30291 821 INFO
  4882 30291 821 INFO

  If all you need returned is rri_idand rri_id is unique in rrid1, then:

  SELECT r1.rri_id
    FROM rrid1 r1
    JOIN rrid1 r2 ON r1.rri_id = r2.rri_id AND r2.processtype = 'RESOLUTION'
   WHERE r1.processtype = 'INFO'
  

  If rri_id is not unique to rrid1, then the following:

  SELECT UNIQUE r1.rri_id
    FROM rrid1 r1
    JOIN rrid1 r2 ON r1.rri_id = r2.rri_id AND r2.processtype = 'RESOLUTION'
   WHERE r1.processtype = 'INFO'
  

  Published by: Starlight Rider October 1, 2010 11:23
  Corrected my stupid mistake!

• grouping column in the same table based on the State of the other column

  Hello
  
  MyTable (eid, ename, status, done);
  MyTable2 (did, dname);
  
  Status column have given in 1/2.
  is the key to Mytable2 forieng.
  
  I want o/p as:
  
  DNAME, County (eid with status = 1), count (eid with status = 2).
  
  Please help me to write the query.
  
  Kind regards
  Krishna

  A possibility (untested):

  SELECT      DNAME
  ,     COUNT((CASE WHEN STATUS=1 THEN 1 ELSE NULL END)) CNT_STATUS_1
  ,     COUNT((CASE WHEN STATUS=2 THEN 1 ELSE NULL END)) CNT_STATUS_2
  FROM     MyTABLE
  JOIN     MyTABLE2 ON MyTABLE.DID = MyTABLE2.DID
  GROUP BY DNAME
  

• [OBIEE 11 g] Create customized to group columns in a table view column header

  Hello world

  I got a requirement to group certain columns in the table view with a custom column header name (for example to create a two-level header).

  The layout I have now:

  [Jan]     [February]     [Mar]     [Apr]     [May]     [June]     [Jul]     [August]     [Total sales]

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

  1          2          3            4          5           6           7          8           36

  The layout I want to achieve:

  [                  Q1                    ] [                   Q2                   ]    

  [Jan]     [February]     [Mar]     [Apr]     [May]     [June]     [Jul]     [August]     [Total sales]

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

  1          2          3            4          5           6           7          8           36

  I use Table View, is it possible to get the above provision? Any comments appreciated. Thank you.

  See you soon,.

  Kayip

  Hello

  It is a simple parameter, it is unpacking already...

  

  The property "of the topic folder" of your columns allows you to enter the required text.

  PS: Do you really have columns in your field with the name of the month? Is it not something better managed with a hierarchy of real time and using a pivot to display the month columns? In this case the grouping of quarters is already there...

• Tab crossed with the total group columns is not printing zero if no data exists

  Hi all
  
  Please help me question.
  
  I have obligation to prepare a report in xml editor
  -----------------------------------
  Operator:
  BRAATEN, DON
  NAME all Standard distributions
  Before 7 0 0 0
  07:00 08:00 0 0 0
  08:00 09:00 0 0 0
  09:00 10:00 1 0 1
  10:00 11:00 0 0 0
  11:00 12:00 0 0 0
  12:00 13:00 0 0 0
  13:00 14:00 0 0 0
  14:00 15:00 0 0 0
  After 15:00 0 0 0
  1 0 1 totals
  
  ---------------------------------------------
  If no data exists for the operator to a particular name, we print zeros.
  now the problem is with the total of the columns. If no data exists that is simply not printing total column as zero.
  
  our production is illustrated below...
  
  -----------------------------------------------------------
  Operator:
  BRAATEN, DON
  NAME all Standard distributions
  Before 7 0 0 0
  07:00 08:00 0 0 0
  08:00 09:00 0 0 0
  09:00 10:00 1 0 1
  10:00 11:00 0 0 0
  11:00 12:00 0 0 0
  12:00 13:00 0 0 0
  13:00 14:00 0 0 0
  14:00 15:00 0 0 0
  After 15:00 0 0 0
  1 1 totals
  ---------------------------------------------------
  
  Please, any body help me on this issue...
  
  My xml file
  ----------------
  < CCSDRPRCR >
  < LIST_G_TIME >
  < G_TIME >
  < RESOURCE_NAME > BRAATEN, DON < / RESOURCE_NAME >
  < NAME > do all repair < / NAME >
  < TIME_SEQ > 00001 < / TIME_SEQ >
  < TIME > 7 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > BRAATEN, DON < / RESOURCE_NAME >
  < NAME > do all repair < / NAME >
  < TIME_SEQ > 00002 < / TIME_SEQ >
  < TIME > 07:00 08:00 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > BRAATEN, DON < / RESOURCE_NAME >
  < NAME > do all repair < / NAME >
  < TIME_SEQ > 00003 < / TIME_SEQ >
  < TIME > 08:00 09:00 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > BRAATEN, DON < / RESOURCE_NAME >
  < NAME > do all repair < / NAME >
  < TIME_SEQ > 00004 < / TIME_SEQ >
  < TIME > 09:00 10:00 < / TIME >
  < TIMECOUNT > 1 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > BRAATEN, DON < / RESOURCE_NAME >
  < NAME > do all repair < / NAME >
  < TIME_SEQ > 00005 < / TIME_SEQ >
  < TIME > 10:00 11:00 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > BRAATEN, DON < / RESOURCE_NAME >
  < NAME > do all repair < / NAME >
  < TIME_SEQ > 00006 < / TIME_SEQ >
  < TIME > 11:00 12:00 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > BRAATEN, DON < / RESOURCE_NAME >
  < NAME > do all repair < / NAME >
  < TIME_SEQ > 00007 < / TIME_SEQ >
  < TIME > 12:00 13:00 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > BRAATEN, DON < / RESOURCE_NAME >
  < NAME > do all repair < / NAME >
  < TIME_SEQ > 00008 < / TIME_SEQ >
  < TIME > 13:00 14:00 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > BRAATEN, DON < / RESOURCE_NAME >
  < NAME > do all repair < / NAME >
  < TIME_SEQ > 00009 < / TIME_SEQ >
  < TIME > 14:00 15:00 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > BRAATEN, DON < / RESOURCE_NAME >
  < NAME > do all repair < / NAME >
  < TIME_SEQ > 00010 < / TIME_SEQ >
  TIME < after 15:00 > < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > BRAATEN, DON < / RESOURCE_NAME >
  Repair Standard < NAME > < / NAME >
  < TIME_SEQ > 00001 < / TIME_SEQ >
  < TIME > 7 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > BRAATEN, DON < / RESOURCE_NAME >
  Repair Standard < NAME > < / NAME >
  < TIME_SEQ > 00002 < / TIME_SEQ >
  < TIME > 07:00 08:00 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > BRAATEN, DON < / RESOURCE_NAME >
  Repair Standard < NAME > < / NAME >
  < TIME_SEQ > 00003 < / TIME_SEQ >
  < TIME > 08:00 09:00 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > BRAATEN, DON < / RESOURCE_NAME >
  Repair Standard < NAME > < / NAME >
  < TIME_SEQ > 00004 < / TIME_SEQ >
  < TIME > 09:00 10:00 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > BRAATEN, DON < / RESOURCE_NAME >
  Repair Standard < NAME > < / NAME >
  < TIME_SEQ > 00005 < / TIME_SEQ >
  < TIME > 10:00 11:00 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > BRAATEN, DON < / RESOURCE_NAME >
  Repair Standard < NAME > < / NAME >
  < TIME_SEQ > 00006 < / TIME_SEQ >
  < TIME > 11:00 12:00 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > BRAATEN, DON < / RESOURCE_NAME >
  Repair Standard < NAME > < / NAME >
  < TIME_SEQ > 00007 < / TIME_SEQ >
  < TIME > 12:00 13:00 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > BRAATEN, DON < / RESOURCE_NAME >
  Repair Standard < NAME > < / NAME >
  < TIME_SEQ > 00008 < / TIME_SEQ >
  < TIME > 13:00 14:00 < / TIME >
  < TIMECOUNT > 1 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > BRAATEN, DON < / RESOURCE_NAME >
  Repair Standard < NAME > < / NAME >
  < TIME_SEQ > 00009 < / TIME_SEQ >
  < TIME > 14:00 15:00 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > BRAATEN, DON < / RESOURCE_NAME >
  Repair Standard < NAME > < / NAME >
  < TIME_SEQ > 00010 < / TIME_SEQ >
  TIME < after 15:00 > < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > Baran, Rick < / RESOURCE_NAME >
  < NAME > do all repair < / NAME >
  < TIME_SEQ > 00001 < / TIME_SEQ >
  < TIME > 7 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > Baran, Rick < / RESOURCE_NAME >
  < NAME > do all repair < / NAME >
  < TIME_SEQ > 00002 < / TIME_SEQ >
  < TIME > 07:00 08:00 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > Baran, Rick < / RESOURCE_NAME >
  < NAME > do all repair < / NAME >
  < TIME_SEQ > 00003 < / TIME_SEQ >
  < TIME > 08:00 09:00 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > Baran, Rick < / RESOURCE_NAME >
  < NAME > do all repair < / NAME >
  < TIME_SEQ > 00004 < / TIME_SEQ >
  < TIME > 09:00 10:00 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > Baran, Rick < / RESOURCE_NAME >
  < NAME > do all repair < / NAME >
  < TIME_SEQ > 00005 < / TIME_SEQ >
  < TIME > 10:00 11:00 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > Baran, Rick < / RESOURCE_NAME >
  < NAME > do all repair < / NAME >
  < TIME_SEQ > 00006 < / TIME_SEQ >
  < TIME > 11:00 12:00 < / TIME >
  < TIMECOUNT > 2 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > Baran, Rick < / RESOURCE_NAME >
  < NAME > do all repair < / NAME >
  < TIME_SEQ > 00007 < / TIME_SEQ >
  < TIME > 12:00 13:00 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > Baran, Rick < / RESOURCE_NAME >
  < NAME > do all repair < / NAME >
  < TIME_SEQ > 00008 < / TIME_SEQ >
  < TIME > 13:00 14:00 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > Baran, Rick < / RESOURCE_NAME >
  < NAME > do all repair < / NAME >
  < TIME_SEQ > 00009 < / TIME_SEQ >
  < TIME > 14:00 15:00 < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < G_TIME >
  < RESOURCE_NAME > Baran, Rick < / RESOURCE_NAME >
  < NAME > do all repair < / NAME >
  < TIME_SEQ > 00010 < / TIME_SEQ >
  TIME < after 15:00 > < / TIME >
  < TIMECOUNT > 0 < / TIMECOUNT >
  < / G_TIME >
  < / LIST_G_TIME >
  < / CCSDRPRCR >
  
  
  Thanks in advance.
  NGO

  http://winrichman.blogspot.com/2009/11/bi-Publisher-crosstabtotal.html

  http://winrichman.blogspot.com/search/label/cross%20tab

• SDO_RELATE slow when you use the Group of... Help, please!

  Hi all
  
  I'm pretty new on the space side of Oracle please go easy on me.
  
  I have 2 tables (a and b), each containing line geometries. There is a one (a) to many (b) relationship between the two.
  
  I want to write a query that selects the information of table b where geometries in table b are fully covered by the geometries in the table has, but the return info is grouped by rows in table a.
  
  Here is my code:
  
  Select
  a.Route_ID,
  a.Step,
  Max (b.gritter_status),
  Sum (b.salt_usage)
  Of
  tbl_routedirs a, tbl_routelinks b
  where
  a.Route_ID = b.route_id and
  sdo_covers (a.geoloc, b.geoloc) = 'TRUE '.
  Group
  a.Route_ID, a.step
  
  The application is extremely slow, taking up to 15 minutes to run... The thing that confuses me is the group by clause of the statement runs in less than half a second.
  
  I checked all the indexes, etc - are all valid. I tried to re - order tables in the clause that ive read it can be a factor and certainly I have little knowledge about what to Exchange where but it seemed to have no impact on speed.
  
  Any suggestions would be warmly welcome... Thank you.
  
  Published by: user8760008 on August 27, 2009 07:08 - I copied the script and do not add functions of aggregation on the selected columns...

  If this is the case, please try:

  Select / * + leading (a) use_nl (a and b) index (your_index_on_b.geoloc b) * /.
  a.Route_ID,
  a.Step,
  Max (b.gritter_status),
  Sum (b.salt_usage)
  Of
  tbl_routedirs a, tbl_routelinks b
  where
  a.Route_ID = b.route_id and
  sdo_coveredby (b.geoloc, a.geoloc) = 'TRUE '.
  Group
  a.Route_ID, a.step

  that is to have nested loops join, then take a small table as external.
  Note that you should not all three tips to get the nested loops join plan.

• Dependency service or group do not reach the HELP of star

  When I put in cmd tfc/scannow it says error help please

  Hello

  1. What is the version of the operating system installed on your computer?

  2. do you have the problem when connecting to Internet/network?

  3. when exactly you receive this error message?

  4. What is the exact full error message?

  5. what exactly you are trying to do?

  You can see the following article on how to scan sfc...

  http://support.Microsoft.com/kb/929833

  Note: Collect and send the log CBS.log files:

  Download the CBS logs located in the "C:\Windows\Logs\CBS\" and copy all of the files 'CBS' then follow the instructions to download.

  Use SkyDrive download collected files and post screen/image capture

  Hope the above information is helpful.

• Group of startup programs in Help and Support

  I have a long list of startup programs that appears in the help and support after the judgment. They are MS and program updates are working, but some of them are weird and some I do not use. It says 12520850 is a Trojan or a virus. How can I get rid of these programs?

  Information in information system, software environment startup programs, startup programs (or msinfo32) can show more things that it should and can be quite intimidating.
  When you see a great list you describe, the process of collecting the information went haywire somewhere along the way, but does not seem to present other problems (next to the particular quest).  Most of the people never look as well as the article, so there are probably a lot of systems that have the problem and he is not even noticed.  It certainly should work fine but.
  You can fix it so it makes more sense and is less scary research with a small VB Script from the site of Ramesh.
  You can record and play back the script, if you care to know what it does and how it works.  You can do everything yourself by hand, but the script is easier.
  The script will make some registry changes, so if you choose to run the script, it won't hurt to back up your registry first.
  This is a popular free registry backup tool:
  http://www.SnapFiles.com/get/ERUNT.html
  Here is a link with more information on scripts and you can also download the Riou script to clean the list of startup programs.
  http://WindowsXP.MVPs.org/msinfofix.htm
  Thank you, Ramesh.
  Please report your results.
  If you still think you may have a malware infection, I could do this:

  Download, install, update and do a full scan with these free malware detection programs:

  Malwarebytes (MMFA): http://malwarebytes.org/

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• Select all lines under a special value of columns... Help.

  Hello
  
  Sorry for posting my query again...
  
  Knani replied, but I have little more to change out put...
  
  
  Table structure:
  

  CREATE TABLE T_20 (CONTROL NUMBER(10) NOT NULL,
                         PO NUMBER(10) NOT NULL,
                         AMENDNO NUMBER(3) NOT NULL,
                         FACTOR VARCHAR2(3) NOT NULL,
                         COMMENT_X VARCHAR2(40),
                         FLAG VARCHAR2(1))
   
   
  SQL> alter table t_20 add constraint t_20_pk primary key (control, po, amendno, factor);

  INSERT orders:
  

  SQL> INSERT INTO T_20 VALUES(101,1000,01,'MSC','NO COMMENT','Y')
    2  / 
   
  1 row created.
   
  SQL> INSERT INTO T_20 VALUES(101,1000,02,'MSC','NO COMMENT','Y')
    2  / 
   
  1 row created.
   
  SQL> INSERT INTO T_20 VALUES(101,1001,00,'NDP','NO COMMENT','Y')
    2  / 
   
  1 row created.
   
  SQL> INSERT INTO T_20 VALUES(101,1001,01,'NDP','NO COMMENT','Y')
    2  / 
   
  1 row created.
   
  SQL> INSERT INTO T_20 VALUES(102,1002,00,'ABC','NO COMMENT','Y')
    2  / 
   
  1 row created.

  out put:
  

  SQL> SELECT * FROM T_20
    2  / 
   
     CONTROL         PO    AMENDNO FAC COMMENT_X                                F 
  ---------- ---------- ---------- --- ---------------------------------------- - 
         101       1000          1 MSC NO COMMENT                               Y 
         101       1000          2 MSC NO COMMENT                               Y 
         101       1001          0 NDP NO COMMENT                               Y 
         101       1001          1 NDP NO COMMENT                               Y 
         102       1002          0 ABC NO COMMENT                               Y

  Now, I want to select only the control number that have several acht.
  
  In above example only control 101 with several PO 2 PO with 1000 and 1001
  So I need to select only this 101 related lines...
  
  for 102 having in. sinple folders then it will have to be removed... because I have millions of records in the primary table...
  
  If one of the members (knapen in this forum) suggested query below:
  

  select *
    from (
  select t.*, count(distinct po) over(partition by control) cnt
    from t20_ t
        )
   where cnt > 1

  This gives the right output select only controls that have several PO - this satisfied requirment...
  
  Yet again others want to dig the out put with the condition as:
  
  Select only controls with multiple PO AND WHERE EACH '' HAVE SEVERAL AMENDNO
  
  So I modified the query knani as
  

  select *
    from (
  select t.*, count(distinct po,amendno) over(partition by control) cnt
    from t20_ t
        )
   where cnt > 1

  But I get the error message: invalid number of arguments...
  
  any help please...
  
  
  Thanks in advance
  Prasanth

  Like this?

  WITH tbl AS (select *
                 from (
                       select t.*, count(distinct po) over(partition by control) cnt
                       from t_20 t
                      )
                 where cnt > 1
                )
  SELECT * FROM
             (SELECT tbl.*, COUNT(amendno) OVER (PARTITION BY control,po) cnt2
              FROM tbl
             )
  WHERE cnt2>1;
  

• split the data belonging to the same group at two different levels

  Hello
  
  I have data in the same group I want to split into two levels.
  
  For example: If the XML is:
  
  < data >
  Taxable < Type > < / type >
  < value1 > one < / value1 >
  b < value2 > < / value2 >
  < / data >
  < data >
  Taxable < Type > < / type >
  e < value1 > < / 1 >
  f < value2 > < / value2 >
  < / data >
  < data >
  Taxable < Type > < / type >
  g < value1 > < / 1 >
  h < value2 > < / value2 >
  < / data >
  < data >
  Taxable < Type > < / type >
  u < value1 > < / 1 >
  v < value2 > < / value2 >
  < / data >
  < data >
  Taxable < Type > < / type >
  o < value1 > < / 1 >
  < value2 > x < / value2 >
  < / data >
  
  
  The output using the RTF model should be:
  
  Taxable
  a and b
  e f
  g h
  
  Not taxable
  u v
  l x
  
  I can't change the query to add a group.
  
  Kindly, if anyone can help
  
  Published by: user10606061 on 25/06/2012 01:05

  
  

  delete a table with 2 columns and two rows. In the second column of row 1
  
  in column 2
  

  table below
  

  PS: your must be within a root element

• Processing of the steps of 40 + "cases" manage subscriptions to e-mail group, desired look instead of single list column

  I am creating a custom subscription management page and have about 40 groups of emails I would like users to be able to manage their preferences on. I want to format e-mail groups box to check and be in more than one column instead of a long list.

  I am running into trouble. First, I brought in all the e-mail groups as single contact fields check box. This turns the page in a super long list of e-mail - not desirable groups. I can set up the processing steps so that it works to subscribe or unsubscribe to groups of e-mail users, however, I want to look more aesthetic where the e-mail group options would be in the columns.

  

  

  I am wanting a solution for the boxes of the e-mail group to fall into the columns. However, when I use the approach of adding a custom field group > > column three custom > > then download a list of all the e-mail groups, can't do the processing steps allowing subscribes and cancels his subscription. In the steps of treatment, that it looks like the entire entity boxes wants to correspond with the group just email. I need the solution allowing each group of e-mail under the boxes to be able to map to a different electronic group at Eloqua and SFDC.

  

  Can someone help me?

  I want to be able to have my email subscription group text boxes in a list of multi-coloumn aesthetic, instead of one single long list. It is essential that the solution still allows processing steps to unsubscribe or subscribe users to this group of specific coordination with the box e-mail they either checked or disabled.

  Here, the answer is in the code.  What you need to do is to use CSS to create fake columns in your layout.  So what you need is a container wrapped around your input box element (and label).  The necessary code is likely to vary, but the example of bootstrap below is a good reference to highlight the code.  In this case, the container with a 'field-group' class is used to contain the entire box input element.

  
   Email address
  
 

  In this example, the form-group class is what you want to style.  In the CSS, you will need to use the following text:

  . Form-group {width: 33%; float: left ;}}

  This style will cause the width of the form group confined to a fixed width of 33% and the left float attribute will allow the element to be removed from the flow of the document and create a type column in your page look.