One record per person, either the min year or year max

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  I really need help here.  I do a fusion of data and that you have selected the options to make more than one record per page, but instead, it is to its own page, merge.  Can someone tell me what I'm missing?

  CS6 seems to have some difficulty to know when there is enough space. The simple solution is to the margins on the right and down the 0 value in the layout options.  And several record preview is known to be buggy. It is best to take a leap of faith and the merger without preview or cancel after you have previewed and before the actual merger.

• InDesign, fusion and mailing: change of the function "record per page.

  I am a new user of InDesign and have a problem that I spent several days trying to correct.  In short, we are 'attempt' create our product catalog by importing our QuickBooks data (such as a csv or txt file).

  I think I've isolated the problem to the installation of "records per page.

  For some reason any, when I open the .csv or .txt file (window > utility > data merge).

  Since the Panel to merge the data, I select the source file, and the box 'Create a merged Document' begins.

  However, the ID is defaulting to one record per page.  When I open the presentation of Multiple recording tab I get... "optiion disabled because only the record is selected... »

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  Thanks in advance for your time and attention.

  Hi Jeffrey,.

  Thank you for taking the time to put video capture whole.  Unfortunately, he is suspended until about a third of the way through...

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  So, I wonder if I have a problem with my installation ID or some corrupted how a configuration or default file... Hmmmm

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• Data merge, its is possible to display the 5 records per page?

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  Basically, set you up a single instance of the placeholders field on your page, then select several records per page in the merge options. You cannot specify a number directly, but ID will be reprduce as many records on the page all items using the spacing settings you set up, then you want to add an empty block size required to get your five instances with zero spacing and group them with field labels.

• Script to place the image folder to 100%, one image per page?

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  Kind regards

  Cognet

• Definition of ethnicity to group a rec per person based on several conditions

  Hello
  
  The final results will be used in a report and the table. A new category for ethnicity has been added to the database.
  For this reason, the field that we use to report is no longer available. I've implemented a query
  identify a record per person for ethnicity codes. The rules are if the person has a record ID as
  Hispanic who use little matter that they have more than one record. If the person is Hispanic,
  Then identify which category they are. If they have only a record (or multiple records in the same category),
  Code for the main category. If they have more than one record in all categories (except Hispanic), then count
  them as multi-ethnic.
  
  Because the table is large and is joined with other tables for the various reports, I would like to know
  If there is a way more efficient and parallel to write the sql statement. In addition, for the future, they
  intend to request a report that will decompose the multiethnic in different categories to display a chart.
  I thought that I would need pivot 'horizontally', or could happen "vertically," the way that the
  underlying table is.
  
  In the sample data:
  1st person has two records Hispanic and European; count under "Hispanic."
  2nd person has two white records and Western Europe; count as 'White '.
  3rd person has two records Indian and African; count under "multiethnic".
  4th person has a record of the Sioux; count under "American Indian or Alaska Native".
  
  
  Here is a table of sample data:

  CREATE TABLE ETHTEK
  (
    ID             VARCHAR2(5 CHAR),
    ETHCODE        VARCHAR2(1 CHAR),
    ETHGROUP_CODE  VARCHAR2(3 CHAR),
    ETHGROUP       VARCHAR2(30 CHAR)
  );
  
  INSERT INTO ETHTEK
  (ID, ETHCODE, ETHGROUP_CODE, ETHGROUP)
  Values
  ('11111', '5' ,'SEU', 'Southern European');
  INSERT INTO ETHTEK
  (ID, ETHCODE, ETHGROUP_CODE, ETHGROUP)
  Values
  ('11111', 'H', 'SAM', 'South American');
  INSERT INTO ETHTEK
  (ID, ETHCODE, ETHGROUP_CODE, ETHGROUP)
  Values
  ('11121', '5', 'WEU', 'Western European');
  INSERT INTO ETHTEK
  (ID, ETHCODE, ETHGROUP_CODE, ETHGROUP)
  Values
  ('11121', '5', '5', 'White');
  INSERT INTO ETHTEK
  (ID, ETHCODE, ETHGROUP_CODE, ETHGROUP)
  Values
  ('11122', '3', 'AF', 'African');
  INSERT INTO ETHTEK
  (ID, ETHCODE, ETHGROUP_CODE, ETHGROUP)
  Values
  ('11122', '2', 'IND', 'Indian');
  INSERT INTO ETHTEK
  (ID, ETHCODE, ETHGROUP_CODE, ETHGROUP)
  Values
  ('11159', '1', 'SIO', 'SIOUX');

  Note: for do not need to create another table, I've hardcoded the translation of the description in the box instructions
  in the following query. The actual query uses a lookup table.
  
  Here's the sql code: (for some reason, in my preview, the upper/lower that the symbols are not, so put a comment on the lines that should be not equal in case they don't show up when posting)

  WITH e AS                                                       --Ethnicity
          (SELECT DISTINCT id,
                           ethcode, 
                           ethgroup_code
                           ethgroup
                      FROM ethtek
          ),
          h AS                                                        --Hispanic
          (SELECT id, ethcode
             FROM e
            WHERE ethcode = 'H'),
          nhm AS                                  --Non Hispanic or Multi Ethnic
          (SELECT DISTINCT id, ethcode
                      FROM e
                     WHERE ethcode <> 'H'),             -- Not Equal 'H'
          nh AS                                                   --Non Hispanic
          (SELECT id,
                  CASE
                     WHEN COUNT (*) OVER (PARTITION BY id) >
                                                               1
                        THEN 'M'
                     ELSE ethcode
                  END AS ethcode
             FROM nhm
            WHERE ethcode <> 'H')                     -- Not Equal 'H'
     SELECT id, ethcode,
            CASE
               WHEN ethcode = 'H'
                  THEN 'Hispanic or Latino'
               WHEN ethcode = '5'
                  THEN 'White'
               WHEN ethcode = 'M'
                  THEN 'Multi-Ethnic'
               WHEN ethcode = '1'
                  THEN 'American Indian or Alaskan Native'
               ELSE ethcode
            END AS ethnic_desc,
            CASE
               WHEN ethcode = 'H'
                  THEN '4'
               WHEN ethcode = '4'
                  THEN '5'
               WHEN ethcode = '5'
                  THEN '6'
               ELSE ethcode
            END AS old_ethnicity_code
       FROM (SELECT DISTINCT h.*
                        FROM h
             UNION
             SELECT DISTINCT nh.*
                        FROM nh LEFT OUTER JOIN h ON nh.id =
                                                                    h.id
                       WHERE h.id IS NULL) eth;

  What is the result of the query:

  Row#     ID     ETHCODE     ETHNIC_DESC     OLD_ETHNICITY_CODE
  1     11111     H     Hispanic or Latino               4
  2     11121     5     White                         6
  3     11122     M     Multi-Ethnic                    M
  4     11159     1     American Indian or Alaskan Native     1

  Thus, on 7 disks, we only have 4 people once.
  
  This is the result of summary query (not included):

       Group               Total
  American Indian/Alaskan Native     1
  Hispanic               1
  White                    1
  Multi-Ethnic               1
  Total                    4

  Hello

  Thanks for posting the CREATE TABLE and INSERT.
  The main reason for this is to allow people who want to help you recreate the problem and test their ideas. If you post statements that don't work, it defeats the purpose.
  Please check all the code before you post. Each of the INSERT statements has a lack of parentheses or quotation (usually two).

  If you have a parent-child relationship within the gorrace; some lines are parents of other lines, and you want the children to inherit traits from their parents.
  What you posted is the usual way to the relational database for handling that: a self-join. I don't see any obvious ways to make the self-join more quickly.
  You can avoid the self-join in your query if you are willing (and able) to denormalize the table and copy the traits inherited from their parents to their children. This looks like the kind of table that won't change very often, in order to maintain the standard table shouldn't be a big project. However, if the table has only 56 lines, so whatever you do with it, maybe not a great impact on the overall performance of the query.

  If you want to improve performance, a more scientifically to go about this is to know what that bottlenecks in the existing query and remedy.
  See these discussions on how to report a tuning issue:
  When your query takes too long...
  HOW to: Validate a query of SQL statement tuning - model showing
  Then start a new thread, setting just about the request.

• Number of records per column in each table

  Hi all
  
  I need to get the number of records by column for each table in a schema.
  
  Is there a way I get the list of columns in each table in a schema and get the number of records per column where the count > 0?
  
  We have oracle 10g
  
  Thank you

  Smidreb wrote:
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  the script below gives me a correct result.

  If it works well for you, then it works OK for me.

  Realize that Oracle is that I/O block-level.
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• FETCH ONE RECORD IN THE SECOND TABLE OF CORRELATED SUB QUERY

  Hi all

  I have provided the script below, I want to single fecth record in the second table in the join query,

  based on the example below, I want to go get one record of the table emp2 what matches with the emp_id of table emp1, please note emp2 may contain more record for the emp_id emp1 which respects

  all records can be selected in the table emp2.

  DROP TABLE emp1.

  CREATE TABLE emp1 (emp_id NUMBER);

  INSERT INTO emp1 VALUES (1);

  INSERT INTO emp1 VALUES (2);

  COMMIT;

  DROP TABLE emp2.

  CREATE TABLE emp2 (emp_id NUMBER, emp_name VARCHAR2 (100));

  INSERT INTO emp2 VALUES (1, 'Name1');

  INSERT INTO emp2 VALUES (2, 'Name2');

  INSERT INTO emp2 VALUES (1, 'Name3');

  INSERT INTO emp2 VALUES (2, 'Conjoint4');

  COMMIT;

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  SELECT * from emp2.

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  FROM EMP1, EMP2 T2 T1

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  GROUP T1. EMP_ID;

  My output should be the same as the result set of query above, but I don't want this logic, please provide the solution by using a different logic, thanks in advance.

  2811876 wrote:

  Thanks for your comments :-)

  My business logic will change to 'Fetch N second timeline table', that's the reason why I asked for a different approach, if I use max, min to achieve this does not allow me to evolve dynamically.

  Although logic has not been expressed at all in your original question, so good job I asked.

  You could do something like:

  SQL > ed
  A written file afiedt.buf

  1 with emp1 (select 1 as the emp_id of union double all the)
  2. Select 2 double
  3               )
  4, emp2 (select 1 as emp_id, 'name 1' as emp_name double union all
  5. Select "name 2' Union double every 2
  6 select 1, 'name' 3' from dual union all
  7. Select 2, 'name 4' double union all.
  8 select 1, 'name 5' from dual union all '.
  9 select 2, 'name 6' from dual '.
  10               )
  11-
  12. end of test data
  13-
  14 select emp_id, emp_name
  15 of)
  16 select t1.emp_id, t2.emp_name
  17, row_number() over (partition by order of t2.emp_name t1.emp_id) rn
  emp1 t1 18
  19 join t2 emp2 (t1.emp_id = t2.emp_id)
  20       )
  21 * where rn<=>
  SQL > /.

  Enter the value for rows_required: 1
  21 Alumni: where rn<=>
  21 news: where rn<=>

  EMP_ID EMP_NA
  ---------- ------
  1 name 1
  2 name 2

  SQL > /.
  Enter the value for rows_required: 2
  21 Alumni: where rn<=>
  21 news: where rn<=>

  EMP_ID EMP_NA
  ---------- ------
  1 name 1
  1 name 3
  2 name 2
  2 name 4

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  Hello
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