(For each) components in the OSB

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• Type a table defined clusters to hold configuration data - definition of default values for each element of the array

  Hello

  I was wondering if I could get some information and opinions on the use of an array of type defined clusters to store configuration data.  I am creating a program to test several EHR and wanted to have a control of type defined for each HAD with the information needed to create the DAQmx tasks for all signals for it must HAVE.  I am eager to do so that the data are encoded in hard and not in a file that the user might spoil.

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  JDeveloper 12.1.3.0.0

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  Hi all

  Help me on the writing of SQL query to retrieve a single record for each employee of the following table? preferably a standard SQL.

  CREATE TABLE xxc_contact)

  empnum NUMBER,

  alternatecontact VARCHAR2 (100),

  relationship VARCHAR2 (10),

  phtype VARCHAR2 (10),

  Phone NUMBER

  );

  insert into xxc_contact values (123456, 'Rick Grimes', 'SP', 'Cell', 9999999999)

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  insert into xxc_contact values (123456, 'Maggie Greene', 'CH', 'Home', 9999999999)

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  123456 rick Grimes SP cell 9999999999

  Daryl Dixon EN work 7777777777 123457

  Home 123458 Maggie Greene CH 6666666666

  Thanks in advance.

  994122 wrote:

  Thank you all, that I got a result

  http://www.orafaq.com/Forum/m/620305/#msg_620305

  By Lalit Kumar B...

  Specifically, the two simple solutions provided were:

  1 using the row_number, entitled Oracle ranking based on descending order of the inside telephone each empnum group. And finally selects the lines which has least rank (of least since that order is descending for phone).

  SQL > column alternatecontact format A20;

  SQL >

  SQL > SELECT empnum, alternatecontact, relationship, phtype, phone

  2 from (SELECT a.*, row_number() over r (PARTITION BY empnum ORDER BY phone / / DESC))

  3 FROM xxc_contact one)

  4. WHEN r = 1

  /

  EMPNUM ALTERNATECONTACT RELATIONSHIP PHTYPE PHONE

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

  123456 rick Grimes SP cell 9999999999

  Daryl Dixon EN work 7777777777 123457

  Home 123458 Maggie Greene CH 6666666666

  2. with the help of MAX, Oracle automatically assigns the maximum phone for all the rows in each group of empnum. And finally selects the rows with the maximum phone. Order by clause is omitted here intentionally. You can find out why.

  SQL > SELECT empnum, alternatecontact, relationship, phtype, phone

  2 (SELECT a.*, MAX (phone) over (PARTITION BY empnum) rn FROM xxc_contact one)

  3 WHERE phone = rn

  4.

  EMPNUM ALTERNATECONTACT RELATIONSHIP PHTYPE PHONE

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

  123456 rick Grimes SP cell 9999999999

  Daryl Dixon EN work 7777777777 123457

  Home 123458 Maggie Greene CH 6666666666

  Kind regards

  Lalit

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• Many storage spaces and data for each quarter of the year files

  Hello

  I am junior DBA and I have a question about tablespace partitioning or data file sharing.

  I have a database of 10g in many storage spaces, for example:

  KER_DATA_2009_Q1

  KER_DATA_2009_Q2

  KER_DATA_2009_Q3

  KER_DATA_2009_Q4

  KER_INDX_2009_Q1

  KER_INDX_2009_Q2

  KER_INDX_2009_Q3

  KER_INDX_2009_Q4

  There is also a data file for each tablespace that is mentioned.

  But for this year 2013, only two big storage space I can get one and data KER_DATA, KER_INDX files.

  All tables and indexes in the whole database are KER_DATA, KER_INDX, tablespaces.

  I don't know any cut in two storage spaces and means of data files. Is this something like tablespace partitioning? This can be done automatically or someone needed to do it manually on the end of each quarter?

  This should be done to improve performance, several small files data instead a large?

  Can some admin explain this to me?

  The reason for this separation of database/tablespace was partitioning. There are a lot of tables partitioned by date column, and there are partitions for each quarter of the year. There are tablespace with a data file for each quarter (partition).

• Ungroup a group and keep the scriptlabel for each item in the Group

  Hello

  does anyone know how to ungroup a group of rectangles with a certain tag (example groupA) give one every rectangle a clean label (the same text groupA)

  I can find the Group and ungroup it but I can't label the rectangles...

  var oPageItems = app.activeDocument.allPageItems;

  for (var j = oPageItems.length - 1;)  j > = 0; d-) {if (oPageItems [j] .label == ("groupA")) {oPageItems [j] .ungroup ()))}}

  ...???

  Help, please

  Hello

  front

  oPageItems [j] .ungroup ();

  go through the loop and set a label for each item within the Group:

  for (var k; k)< opageitems[j].length;="">

  oPageItems [j] [k] .label = "groupA";

  assuming that your oPageItems [j] is a group indeed.

  I hope that...

• Select a record for each Member of the Group

  Hello
  will have a table where in I'll take data for so many members of the group. I need to retrieve data for a particular group whose number of rows of data may be more in number (but I don't want only one row of data for each Member of the Group)
  
  Here is the query to retrieve all rows of data
  

  select RI.RESOURCE_NAME,TR.MSISDN,TR.ADDRESS1_GOOGLE, TR.MSG_DATE_INFO, FROM TRACKING_REPORT TR, RESOURCE_INFO RI 
  WHERE TR.MSISDN IN (SELECT MSISDN FROM RESOURCE_INFO WHERE GROUP_ID ='1' AND COM_ID=2 ) AND RI.MSISDN=TR.MSISDN 
   order by MSG_DATE_INFO

  This query result is...
  >
  DDD 12345 13 March 10 19:43:03
  EEE 54321 Tamilnadu, India 13 March 10 19:39:48
  DDD 12345 13 March 10 19:32:58
  EEE 54321 Tamilnadu, India 13 March 10 19:30:07
  DDD 12345 13 March 10 19:23:08
  EEE 54321 Tamilnadu, India 13 March 10 19:20:14
  FFF 98765 March 13 10 19:19:22
  DDD 12345 13 March 10 19:13:01
  EEE 54321 Tamilnadu, India 13 March 10 19:09:50
  DDD 12345 13 March 10 19:02:56
  EEE 54321 tn, ind March 13, 10 18:59:49
  DDD 12345 13 March 10 18:53:08
  EEE 54321 tn, ind March 13, 10 18:49:50
  DDD 12345 13 March 10 18:42:56
  EEE 54321 tn, ind March 13, 10 18:39:50
  DDD 12345 13 March 10 18:33
  EEE 54321 tn, ind March 13, 10 18:29:50
  DDD 12345 13 March 10 18:22:54
  EEE 54321 tn, ind March 13, 10 18:19:50
  DDD 12345 13 March 10 18:12:56
  EEE 54321 tn, ind March 13, 10 18:09:50
  DDD 12345 13 March 10 18:02:54
  EEE 54321 tn, ind March 13, 10 18:00:02
  FFF 98765 Tamilnadu, India March 13, 10 17:59:26
  FFF 98765 Tamilnadu, India March 13, 10 17:54:26
  DDD 12345 13 March 10 17:52:56
  EEE 54321 tn, ind March 13, 10 17:49:50
  FFF 98765 Tamilnadu, India March 13, 10 17:49:25
  FFF 98765 Tamilnadu, India March 13, 10 17:44:26
  DDD 12345 13 March 10 17:42:56
  
  >
  
  This output, I only want a new album for each member(ddd,eee,fff). That is to say
  >
  DDD 12345 13 March 10 19:43:03
  EEE 54321 Tamilnadu, India 13 March 10 19:39:48
  FFF 98765 March 13 10 19:19:22
  >
  
  How to change the query to do this...?

  Ok. I looked more carefully at your sample and it looks like you are looking for:

  SELECT  RI.RESOURCE_NAME,
          TR.MSISDN,
          MAX(TR.ADDRESS1_GOOGLE) KEEP(DENSE_RANK LAST ORDER BY TR.MSG_DATE_INFO),
          MAX(TR.MSG_DATE_INFO)
    FROM  TRACKING_REPORT TR,
          RESOURCE_INFO RI
    WHERE TR.MSISDN IN (
                        SELECT  MSISDN
                          FROM  RESOURCE_INFO
                          WHERE GROUP_ID ='1'
                            AND COM_ID=2
                       )
      AND RI.MSISDN = TR.MSISDN
    GROUP BY  RI.RESOURCE_NAME,
              TR.MSISDN
  / 
  

  SY.

• Creation of 5 rows in table B for each row in the table has

  create table A
  (col1 number,
  col2 varchar2(20)
  )
  /
   
  create table B
  (col1 number,
  col2 varchar2(20)
  )
  /
  
  insert into A values (1,'Jane');
  insert into A values (2,'Kate');
  insert into A values (3,'Stomp');
  
  
  SQL> select * from a;
  
        COL1 COL2
  ---------- -----------------
           1 Jane
           2 Kate
           3 Stomp

  For each row in the table, I want 5 rows to be created in the table b. table B should look like

      COL1   COL2
  ---------- -----------------
           1 Jane
           2 Jane
           3 Jane
           4 Jane
           5 Jane
           6 Kate
           7 Kate
           8 Kate
           9 Kate
           10 Kate
           11 Stomp
           12 Stomp
           13 Stomp
           14 Stomp
           15 Stomp

  How can I do this?

  Hello

  A way

  insert into b
  select rownum*5  , col2 from a
  union all
  select rownum*5+1, col2 from a
  union all
  select rownum*5+2, col2 from a
  union all
  select rownum*5+3, col2 from a
  union all
  select rownum*5+4, col2 from a
  

  Concerning
  Anurag

• Select the last value for each day of the table

  Hello!
  
  I have a table that contains several measures for each day. I need two queries on this table, and I'm not sure how to write them.
  
  The table stores the rows (sample data)
  

  *DateCol1                 Value       Database*
  27.09.2009 12:00:00       100           DB1
  27.09.2009 20:00:00       150           DB1
  27.09.2009 12:00:00       1000          DB2
  27.09.2009 20:00:00       1100          DB2
  28.09.2009 12:00:00       200           DB1
  28.09.2009 20:00:00       220           DB1
  28.09.2009 12:00:00       1500          DB2
  28.09.2009 20:00:00       2000          DB2

  Explanation of the data in the sample table:
  We measure the size of the data files belonging to each database to one or more times a day. The value column indicates the size of the files of database for each database at some point (date in DateCol1 European model).
  
  
  What I need:
  Query 1:
  The query must return to the last action for each day and the database. Like this:

  *DateCol1       Value      Database*
  27.09.2009        150          DB1
  27.09.2009       1100          DB2
  28.09.2009        220          DB1
  28.09.2009       2000          DB2

  Query 2:
  The query should return the average measurement for each day and the database. Like this:

  *DateCol1       Value      Database*
  27.09.2009       125          DB1
  27.09.2009      1050          DB2
  28.09.2009       210          DB1
  28.09.2009      1750          DB2

  Could someone please help me to write these two queries?
  
  Please let me know if you need further information.
  
  Published by: user7066552 on September 29, 2009 10:17
  
  Published by: user7066552 on September 29, 2009 10:17

  Why two queries when it suffice ;)

  SQL> select dt
    2       , db
    3       , val
    4       , avg_val
    5    from (
    6  select dt
    7       , val
    8       , db
    9       , row_number () over (partition by db, trunc (dt)
   10                                 order by dt desc
   11                            ) rn
   12       , avg (val) over (partition by db, trunc (dt)) avg_val
   13    from test)
   14   where rn = 1
   15  order by dt
   16  /
  
  DT        DB           VAL    AVG_VAL
  --------- ----- ---------- ----------
  27-SEP-09 DB2         1100       1050
  27-SEP-09 DB1          150        125
  28-SEP-09 DB2         2000       1750
  28-SEP-09 DB1          220        210
  

• iTunes separating artist for each song by the same artist

  This is something that concerns me for a long time. After you add a few Radiohead songs to the collection, it wouldn't put them in the same section of the artist. In this case a very large number. So I thought that any other time in this case, I have just remove the artist artist name and album for all of the music of this artist and works in general.

  Not this time. Instead, what it's done is divide the artist for each song. Strangely, if I switch to the display of the album, each song is where it should be. I tried re-entry sorting and grouping information, but nothing seems to work.

  Hello winningbagoafx,

  I understand that iTunes is unable to well to bundle a number of songs together under the same artist. The article linked below, although written for a slightly different situation, details information that can also be applied to a similar situation involving the artist rather than the album.

  Why the songs with the same album art are not grouped together?
  https://support.Apple.com/en-us/HT204538

  All my best.

• Are there separate libraries for each Member in the family plan?

  We want to start the Apple family of music Plan in our House, but everyone in my family wants to add their own music and to keep it separate from the rest of the members of the plan.

  I have 2 questions about it:

  1. every member gets their own library of music that they can sort and customize without interfering with the music of other members?

  2. each member individually has the limit of 100,000 of the song, or is it 100 000 songs for the whole family?

  We will get the plan until these issues are addressed, if anything like that would be useful!

  Hi Josie1374,

  I understand that you have a few questions about the sharing of family in Apple music. I have some information for you which should be able to help answer these questions. First of all, you should know that each family member in the family of the Apple music-sharing plan will be themselves, individual account, Apple's music:

  A subscription to Apple's music for
  each Member of your family.

  With a family membership of Apple's music, up to 6 people in your family can enjoy unlimited access to music from Apple on their devices. Everyone gets full access and individual music library of Apple, recommendations of experts, radio world-wide 24/7 with 1 beats and radio on demand. This is the world of music for the creations of any

  iCloud - family sharing - Apple

  https://www.Apple.com/icloud/family-sharing/

  As is the case, this information applies to each individual account, which should solve your second question:

  How music Apple manages your downloaded songs

  • You can have up to 100 000 songs in your music library. Songs you purchased from the iTunes Store do not count against this limit.

  Add music to your collection to your music library Apple - Apple Support

  https://support.Apple.com/en-us/HT204925

  Be vigilant and thank you for visiting the communities of Apple Support.