Format of column with the percent sign (%)

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• help me create a column with the time format

  Hi all
  
  Please help me find the query to create a table with 3 columns and insert the following data
  
  ENTITYID | CHGDATE | CHGTIME
  
  ======= ==================== ========
  
  4214984 | 2 OCTOBER 06 12.00.00.000000 AM | 07:23:57
  
  4214985 | 12.00.00.000000 11-SEP-08 AM | 10:52:19
  
  and update of the first line of the above table as follows
  ---------------------
  4214984 | 2 OCTOBER 06 07.23.57.000000 AM | 07:23:57
  ------------------------
  you will need to update the CHGDATE column with the time of the CHGTIME column.

  Sorry, we need a new format around the to_date

  select to_date(to_char(chgdate, 'mm/dd/yyyy') || ' ' || to_char(chgtime, 'hh24:mi:ss'), 'mm/dd/yyyy hh24:mi:ss') from etime3;
  

  This should telll Oracle to the format of the generated string. If this isn't the case, run select with the to_char two as separate columns to see that get us the data that we expect.

• [8i & 10g XE] How to compare a column with the text of a column with numbers

  This is probably a simple question, but I can't seem to find the answer. I tried variations on use to_number to_char and interpreters, all nothing will do.
  
  The real problem is that I have a table with a column of numbers in a database 8i (datatype = NUMBER (3)) and a column with the text (datatype = CHAR (3)), and I need to compare two values.
  
  This table has thousands of lines, but here's an example of the data in these two columns:

  TXT     CHR
  -----------
  001     1
  001     2
  002     2
  XXX     1
  003     3

  What I want to do is select the lines where the TXT and CHR VALUES do not match, then it would be (given my example):

  TXT     CHR
  -----------
  001     2
  XXX     1

  But, I'm having difficulties in comparing two columns, because they are not the same type of data.
  
  I tried to work on a simplified version of the problem in the 8i database both my 10g XE database:

  SELECT     *
  FROM     (
       SELECT     1.000               AS nbr
       ,     TO_CHAR(1.000,'000')     AS txt
       ,     CAST('001' AS CHAR(3))     AS chr
       FROM     dual
       )
  WHERE     txt = chr
  ;

  But this also returns any line, and it isn't even the "XXX" in the text column to treat in this example.
  
  Can someone help me understand what it is that I'm missing?
  
  Thank you!

  Hello

  user11033437 wrote:
  This is probably a simple question...

  Simple if you know the secret; maddening if you do not have.

  ... I tried to work on a simplified version of the problem in the 8i database both my 10g XE database:

  SELECT     *
  FROM     (
       SELECT     1.000               AS nbr
       ,     TO_CHAR(1.000,'000')     AS txt
       ,     CAST('001' AS CHAR(3))     AS chr
       FROM     dual
       )
  WHERE     txt = chr
  ;
  

  Try:

  TO_CHAR (1.000, 'FM000')
  

  By default, TO_CHAR leaves place the beginning of the string to a sign less, in which case he should ever one, TO_CHAR (1,000, '000') returns makes 4 characters, not 3. "FM" in the format said TO_CHAR do not add a space.

  If it does not, after the release of DUMP (txt) for the few lines that you can't match, so that we can see exactly what is in them.

• I have a column with two values, separated by a space, in each line. How to create 2 new columns with the first value in a column, and the second value in another column?

  I have a column with two values, separated by a space, in each line. How do I create 2 new columns with the first value in one column and the second value in another column?

  Add two new columns after than the original with space separated values column.

  

  Select cell B1 and type (or copy and paste it here) the formula:

  = IF (Len (a1) > 0, LEFT (A1, FIND ("", A1) −1), ' ')

  shortcut for this is:

  B1 = if (Len (a1) > 0, LEFT (A1, FIND ("", A1) −1), ' ')

  C1 = if (Len (a1) > 0, Member SUBSTITUTE (A1, B1 & "", ""), "")

  or

  the formula of the C1 could also be:

  = IF (Len (a1) > 0, RIGHT (A1, LEN (A1) −FIND ("", A1)), "")

  Select cells B1 and C1, copy

  Select cells B1 at the end of the C column, paste

• How to take a column of duplicate names and fill a different column with the same names, excluding duplicates?

  How to take a column of duplicate names and fill a different column with the same names, excluding duplicates?

  I find easier to use this copy separate Automator Service (download Dropbox).

  To install in your numbers > Services, double-click menu just the package downloaded .workflow and if necessary give permissions in system preferences > security & privacy.

  To use, just:

  1. Select the cells in the column with duplicate names.
  2. Choose separate copy in numbers > Services menu.
  3. Click once in the upper cell where you want the deduplicated values appear.
  4. Command-v to paste.

  SG

• Querying the oracle table that has a column with the name of "FILE".

  Hi all

  I need to have an oracle table that has the column with the name "FILE".

  I can query all columns with the query "select * from table".

  But I'm not able to use the query "select the table file.

  This table is converted from ms access to oracle and I need to have this column with the name "FILE".

  Any suggestions?

  Thank you

  Abdellaoui

  Hello

  FILE is a keyword from Oracle, so it's not a good name,

  Use FILEDATE, or DATE_FILED, or something else that is not in V$ RESERVED_WORDS. KEYWORD as the name column.

  If you need to use the FILE, then you must use quotation marks.

• An update on an index column with the same value generates an index to the top

  An update on an index column with the same value generates an update of the index?
  
  
  Thank you

  In addition to my previous answer, see also

  http://orainternals.WordPress.com/2010/11/04/does-an-update-statement-modify-the-row-if-the-update-modifies-the-column-to-same-value/

  Riyaj Shamsudeen has this to say:
  "+ We have an index on this column v1 and we update this column indexed too." Oracle was updating the indexed column? N ° if the values match the level of the indexed column, then the code of RDBMS isn't up-to-date index, a feature for optimization again. Only the row of table is updated, and the index is not updated. + "

  Hemant K Collette

• Columns with the same name but different data or precision length.

  DB version: 10 gr 2
  
  In a diagram, I need to find all of the columns with the same name but different or precision data length (if the column is of type number). I know I have to use the USER_TAB_COLS view for this. But I need to find a logic to compare the names of columns and its lengths.

  something like

  select t1.table_name
       , t1.column_name
       , t1.data_type
       , t1.data_length
       , t2.table_name
       , t2.column_name
       , t2.data_type
       , t2.data_length
    from user_tab_columns t1
       , user_tab_columns t2
   where t1.table_name != t2.table_name
     and t1.column_name = t2.column_name
     and t1.data_type = t2.data_type
     and t1.data_length != t2.data_length
  

  ... could be a starting point.

• Documentation about adding column with the DEFAULT value.

  Hello
  
  http://download.Oracle.com/docs/CD/B19306_01/server.102/b14200/statements_3001.htm#i2198241
  
  < quote >
  If you add a column, then the initial value of each row in the new column is NULL unless you specify the DEFAULT clause. In this case, Oracle database updates each row in the new column with the value specified for the DEFAULT value. This update operation, in turn, triggers AFTER UPDATE triggers defined on the table.
  < quote >
  
  I am not able to understand the emphasis on the part AFTER UPDATE while the column with DEFAULT values addition triggers both BEFORE and AFTER triggers defined on UPDATE.
  
  According to the documents of 11 g
  
  http://download.Oracle.com/docs/CD/B28359_01/server.111/b28286/statements_3001.htm#i2133105
  
  n.m. (u) r only changes with the NON NULL columns, but focus on AFTER UPDATE is still there. No trigger defined on the fire of the update in this case.
  
  This insistence is intended? If not, IMO, it should be changed.
  
  Kind regards

  Hi Sissi. After further discussion, we have added some information about a change in behavior. Here is what says the next version of the doc:

  When you add a column, the initial value of each row in the new column is null.

  * If you specify the DEFAULT clause for a column NOT NULL, then the default value is stored as metadata, but the column itself is not populated with data. However, the following queries that specify the new column are rewritten so that the default value is returned in the result set.

  This optimized behavior differs from earlier versions, when as part of operation ALTER TABLE Oracle database updated every line in the newly created with the default column and then fired defined update triggers on the table. In this release, no trigger is triggered because the default value is stored only in the form of metadata.

  * If you specify the DEFAULT for a nullable column clause, then the default value is added to existing lines under this ALTER TABLE statement, and any update triggers defined on the table are activated. This behavior also means if you change a NOT NULL column with a default value to be nullable.

  Hope that helps to clarify the matter further.

  Kind regards
  Diana

• column with the only format of loading time

  My DB, 11 2 on OEL 5.5 GR
  Hi all
  I have a problem with the loading of the external table. I have a txt file with a format of columns HH24:MI:SS

  START_TIME
  00:10:20
  01:15:11
  02:11:10

  And I don't know which data type use in external table because in the future, I'll have to do some calculations on this column, so I should load as a DATE, but miss me "mm/dd/yy" to the front. Any suggestion how to do properly?

  You could use a data type CHAR or date for the outer table and do your math with this column correctly.

  The key is that convert a time string to a date data type will add the current date to it. For example:

  SQL> alter session set nls_date_format='dd/mm/yyyy hh24:mi:ss';
  
  Session altered.
  
  SQL> select to_date('00:10:20', 'hh24:mi:ss') from dual;
  
  TO_DATE('00:10:20',
  -------------------
  01/02/2011 00:10:20
  

  Some examples of code for both cases, by using the DATE and CHAR data type. I tested it on 11.2.0.1:

  CREATE OR REPLACE DIRECTORY ext_tab_dir AS '/tmp';
  
  -- /tmp/info.txt
  1,00:10:20
  2,01:15:11
  3,02:11:10
  
  alter session set nls_date_format='dd/mm/yyyy hh24:mi:ss';
  

  (1) using the DATE data type

  drop table tab_time_date;
  
  CREATE TABLE tab_time_date (
    key VARCHAR2(5),
    start_time date)
    ORGANIZATION EXTERNAL (
       TYPE ORACLE_LOADER
       DEFAULT DIRECTORY ext_tab_dir
       ACCESS PARAMETERS
        (FIELDS TERMINATED BY ','
         (key char(5),
          start_time CHAR(8) DATE_FORMAT DATE MASK "HH24:MI:SS"
         )
        )
        LOCATION ('info.txt'));
  
  select * from tab_time_date;
  
  KEY   START_TIME
  ----- -------------------
  1     01/02/2011 00:10:20
  2     01/02/2011 01:15:11
  3     01/02/2011 02:11:10
  

  (2) using the CHAR data type

  CREATE TABLE tab_time_char (
    key VARCHAR2(5),
    start_time char(8))
    ORGANIZATION EXTERNAL (
       TYPE ORACLE_LOADER
       DEFAULT DIRECTORY ext_tab_dir
       ACCESS PARAMETERS
        (FIELDS TERMINATED BY ','
        )
        LOCATION ('info.txt'));
  
  select key, to_date(start_time, 'hh24:mi:ss') from tab_time_char;
  
  KEY   TO_DATE(START_TIME,
  ----- -------------------
  1     01/02/2011 00:10:20
  2     01/02/2011 01:15:11
  3     01/02/2011 02:11:10
  

  Kind regards.

• Average of column with the format of date/time in the response of BI

  Hello world
  
  We working on the response of BI and training to get the average of the hours of work of departments. Work hours column is the Date in the database format, so we tried to converted to CHAR through the Cast function to modify columns in BI answers formula, and this gave us date "2007/09/01' as a result! We have also tried to throw the working hours for:
  
  CAST (CHARACTER of WORK HOURS FORMAT AS 'yyyyMMdd-HHmmss')
  
  NOTE: In working hours, we have null values...
  
  work real hours to appear in the picture of the response is "01/07/2009-06:13.
  
  
  How to get the column of working hours average on response BI?
  
  Appreciate your help
  Concerning
  
  Published by: user817525 on July 21, 2009 23:15

  No, u do not attend the general tab. you need to go to the Data Type and click the button change. you will see a list of logical column. In your case, you will need to find the logical column when u used to store your WorkingHour and click on the button with three love (close your logical column). Now you have expression builder where you put formula: EXTRACTS ("Ain_Access" TIME. "" "" Fact_Access '. "" Hours of work)

  I hope this helps?

  all best
  Phoenix
  Belgrade

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