quotes from string trim

Similar Questions

• How to remove tabs, several spaces, quotes from data in the column.

  Hello

  How can we remove legs, several spaces, quotes from data in the column.

  Entry:

  H ope you 'are' Ingo w ell

  Output:

  H ope do you everything - there is a tab in hope and double-spaced in the well. This should be replaced by space. (Tab and multi spaces)

  I found myself with below:

  REGEXP_REPLACE (REGEXP_REPLACE (Replace (col, Chr (9),' '), ' () {2,} ', ' '),'[' ""]', ")

  Can we do more efficiently, your contributions are appreciated.

  I tried

  REGEXP_REPLACE (col,'[^ [a - z, A - Z, 0-9, Chr (0) - Chr (127) [: Space:]]] *', ") but the tabs still exist and that my data have German characters I can't go along with this.

  Thank you

  GVK.

  Hello

  Karthik417 wrote:

  Hello

  Thanks for your suggestion. In the above expression, we still missed to replace the double quotes.

  So thought we must use another regular expression instead of REPLACE.

  Select REGEXP_REPLACE (REGEXP_REPLACE (' H ope you "are" d "Ingo ell w ',' [' |])) CHR (9) | + ',' '),'["" "]') twice;

  I suppose we must use at least 2 REGEXP to achieve and cannot get it with a single expression.

  Thank you

  Sorry, I removed single quotes, not double - quote.  (This shows one of the reasons why having a few lines of sample data is very useful).

  I think you don't need 2 separate functions.  I do not think that the two must be REGEXP_REPLACE; You can use TRANSLATE to remove both types of quotation marks in a single function call:

  TRANSLATE (REGEXP_REPLACE (col

  , '[ ' || CHR (9) | ']+'

  , ' '

  )

  , '?'''"',

  , '?'

  )

  Of course, you can use use REGEXP_REPLACE to remove the quotes, if you really want to.

  Powerful they are, regular expressions are basically only 1 thing, things not 2 or more different.  Sometimes, he can watch as they do several things, because they can operate on character classes, like the series containing the and the and a variable number of characters.  For example, the REGEXP_REPLACE function, I used above is simply changing expression expression x y, where x is 1 consecutive or more characters of a given set.  As we both demonstrated, we can have a separate function that changes from p to q, but I know not all built-in function which, in general, can change x to y at the time that it changes from p to q.  TRANSLATE comes close, but TRANSLATE only works if x, y, p and q are single (or NULL), characters not expressions.

• I'm getting quotes from French style

  Using InDesign CC I get quotes from "French Style" as the attached picture.

  I use American English on my iMac. Works fine in InDesign CS6. Downloaded both programs Adobe Creative Cloud.

  Thanks for any help

  

  Is the file that you have made or did you get it from someone else? It happens to me if the language at the bottom of the character Panel is set to French.

• Generation of single quotes in strings

  Version: 11.2.0.3

  We have a custom stored procedure called GENERATE_STATS and I want to run this procedure for each schema in the comic book as

  exec ('SCOTT', 'BASIC') generate_stats;

  generate_stats exec ('HR', 'BASIC');

  .

  .

  .

  .

  exec ('XXXXXX', 'BASIC') generate_stats;

  -All except XXXXXX is fixed.

  -XXXX is derived rom DBA_USERS

  I want to generate these commands dynamically as

  SQL > SELECT ' exec generate_stats ("' | username |) (', "'BASIC"); from dba_users;

  ERROR:

  ORA-01756: city not properly finished chain

  -I noticed that, to generate a single quotation mark, you need 4 quotes

  SQL > select "' double.

  '

  -

  '

  SQL > select "',"' double.

  '''

  ---

  ','

  -I managed to create so far without much trouble.

  SQL > SELECT ' exec generate_stats ("' | username |) "',"' from dba_users where rownum < 5;

  ' EXECGENERATE_STATS('''||) USER NAME | " ','' '

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

  generate_stats exec ("SYS"

  exec generate_stats ("SYSTEM",")

  exec generate_stats ('OUTLN', ')

  generate_stats exec ('DIP', ')

  It took a lot of trial and error to add the remaining portiton BASIC'); to added. The following finally worked.

  SQL > SELECT ' exec generate_stats ("' | username |) (', "BASIC");' from dba_users where rownum < 5;

  ' EXECGENERATE_STATS('''||) USER NAME | " (', "BASIC"); "

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

  generate_stats exec ('SYS', 'BASIC');

  exec ('SYSTEM', 'BASIC') generate_stats;

  generate_stats exec ('OUTLN', 'BASIC');

  exec ('DIVE', 'BASIC') generate_stats;

  To generate the string 'BASIC' . I didn't have to use single quotes 4 as I once did.

  It generated when I placed between her with just two single quotes like that "BASIC");'

  There are a few basic rules I should keep in mind when you try to generate strings with single quotes?

  Hello

  In a string literal, single quotes normally indicate the beginning or the end of the literal.

  If you want to specify a single quote, is to use 2 consecutive single quotes in the literal.  For example, the name of 9 characters O'Higgins could be written in a literal string like this

  'O' Higgins

  ICT that there are 4 single quotes above: 1 to indicate the beginning of the literal string, 2 to represent a single quotation mark in the literal and 1 at the end of the literal.

  If you want a string of characters-1 that contains only a single quote, and then, as you noticed, it was released as 4 of them in a row

  ''''

  If you want to represent the 12 string

  ","BASIC");"

  in a string literal, then you have to put everything in a pair of simple and douuble quotes all quotes that occur in the sting, like this:

  (', "BASIC"); "

  When you asked if there was a rule of THUMB , you were doing a play on words?

  There is a rule of practice. When you use this type of string literal, single quotes always occur in pairs.  You always need a pair to mark the beginning and the end of the literal.  For each single quote representation, you need a pair.  If the string that represents the 12 above string literal contains 4 pairs of single quotes.  I'll write this liteal once again, with the numbers in pairs:

  (', "BASIC"); "

  122 33 44 1

  1 pair surrounds the literal; pairs of 2, 3 and 4 represent single quotes in the literal.

  If ever you have an odd number of single quotes using this notation, then you know that something is wrong.

  It is the older way of the use of single quotes in the string literal.  From Oracle 10, you have the possibility to use Q-notaation as well:

  Q'<', 'basic');="">'

  When you use Q-rating, you can have literals well trained with an odd number of single quotes.  The literal above happens to have 5, 2 that surround the literal.

  See the manual of the SQL language for more info on Q-rating:

  Literals

• quote from the escape

  I have an update like this statement.
  
  Update table_name
  Set query_column ='select column_names from table1, table2, etc. '
  WHERE condition;
  
  I have to update a column that stores a query inside with a very large query containing so many quotes (') in it.
  
  I use Q '[' but it always gives me error "string literal too long."
  
  Please help me in this.
  
  Thank you
  Vinod

  910575 wrote:
  The query I will carry out is as...

  Update order_report
  Set report_query = Q'[select Order_id, Order_Name... etc Order_mstr, Order_details, etc.] »
  where Report_id = "ORDER_REPORT";

  The data type of the column 'report_query' is CLOB.

  Your UPDATE statement is a SQL and the string used in the SQL can be character 4000 max. That's what I'm trying to tell you. You should try something like this.

  create table t(sql_str clob)
  /
  declare
    type tbl is table of varchar2(4000) index by pls_integer;
  
    lSql tbl;
    lSqllob clob;
  begin
    lSql(1) := q'[select e.* from ]';
    lSql(2) := q'[emp e join dept d ]';
    lSql(3) := q'[on e.deptno = d.deptno ]';
  
    dbms_lob.createtemporary(lSqlLob, true);
  
    for i in 1..lSql.count
    loop
      dbms_lob.writeappend(lSqllob, length(lSql(i)), lSql(i));
    end loop;
  
    insert into t (sql_str) values (lSqllob);
  end ;
  /
  select * from t
  /
  

• Function to remove stopwords from string

  Hello
  
  I would use a very effective feature that eliminates all the empty words of a given string (array of empty words: select distinct spw_word from ctxsys.ctx_stopwords).
  Any help will be much appreciated.
  
  Kind regards
  MR. R.

  Hello

  Seems good! I would suggest a few changes in the body:

  BEGIN
      mystr := ' ' || str || ' ' ;               -- See note 1
  
      FOR cr IN c
      LOOP
          mystr := REPLACE ( mystr
                         , ' ' || cr.x || ' '
                         , ' '                    -- See note 2
                  );
      END LOOP;                              -- See note 3
  
      mystr := TRIM ( REGEXP_REPLACE ( mystr          -- See notes 1 and 2
                                    , ' {2,}'     -- 2 or more spaces
                         , ' '     -- 1 space
                         )
               );
      RETURN  mystr;
  END f_replace_stopword;
  

  Notes:
  (1) not to clutter the code with a lot of unnecessary trimming. Before entering the loop, you must make sure that mystr begins and ends with a space, because of the problem of 'mother is in chemotherapy' as mentioned William. However, if the chain has already spaces here, is not what anyone let them hurt. The same goes for storage mystr each pass through the loop. The amount of resources, it's waste is small (if it's to lose, anyway), but I'm more concerned with having unnecessary code distract people from the real business of the function. All the extra spaces can be picked up at the end, after leaving the loop.

  (2) so-called "FUBAR" is one of the empty words that you remove, and the string contains 'A FUBAR B. If you remove just the empty word (with its spaces added), which will leave you with "AB". I guess you want 'A B', with a space between what were originally two different words. This would be particularly important if the empty word "AB" happened to appear later in the list.
  This can cause multiple consecutive spaces being left in mystr; additional spaces can be deleted once the loop is completed.

  (3) If you do not use an OPEN statement to open the cursor, then you don't need a statement CLOSE to close. Leaving a LOOP FOR cursor automatically closes the cursor.

• [JS - CS4] Help w / JS choke in double quotes in strings and regex

  Someone at - it an easy way to extract text from a string with quotes in JavaScript? I seem to have encountered some limitations in the javascript regex implementation and how a script handles a string with quotes in it.

  Here's what I'm working with: a regular as string:
  {appliedParagraphStyle: "Headline 6", changeConditionsMode:1919250519}
  I need to extract just the text "Headline 6"without the quotes

  I came across a regular expression that would get only the text without the quotes, but apparently Javascript does not support lookbacks. For example, wonderful WhatTheGrep of Jongware script can decode this regex:
  ( ? < = ») [^"]*? (?=")
  but ID can't find nothing with it using a GREP search. The same search without the hindsight find stuff:
  "[^"]*? (?=")
  But what he always finds the quote at the beginning of the string.

  I thought I could just use substrings to retrieve the text without the quotation, but apparently he has problems. See these examples from the Javascript console:

  testStr;
  Result: "Headline 6".
  testStr.length;
  Result: 1

  testStr;
  Result: Title 6
  testHL.length;
  Result: 10

  testHL;
  Result: 'title 6 '.
  testHL.length;
  Result: 1

  So if the string contains a quotation mark in it, it does not return the actual length, so I can't encode a beginning and end of the string to extract the part without the quotation.

  One of you more experienced people have a magical idea? TIA!

  Thank you, Jongware, your answer had the keys of a workaround - something I stumbled on in the review of your code.

  As new script, I had indeed crashed on the assumption that regular expressions worked the same in JS and GREP searches. ID can do lookbacks but JS can not - check out www.regular-expressions.info/refflavors.html for a comparison of the regex flavors; JS is listed as ECMA in the table.

  In case you or anyone who wants to know, my problem concerns styles conversion a Word doc that may or may not have styles, the script expects. ID hesitate looking for a style that is not part of the document, so I had to intercept it. Rather than extracting the text from the style of the chain and comparing it to a list of styles for the document, I made a regexp to the list of style and use to match the string, avoiding having to make quotes in there at all.

  var app.activeDocument = docRef;
  var myStyles, s;
  for (var s = docRef.paragraphStyles.length - 1; s > = 2; s-) {}
  get the list styles, but not the default styles in brackets, because they're going to screw up a regular expression pattern
  myStyles = docRef.paragraphStyles [s] .name + "|" + myStyles;
  }

  myRE = new RegExp (myStyles);

  Str = "{findWhat:------"do", appliedParagraphStyle:------"Topic 1\", appliedFont:------"Wingdings\", fontStyle:------"Bold\"}";
  f = str.match (Myrie);
  If (f)

  Alert ("search for:" + f + "work" "");
  on the other
  Alert ("do not attempt a search... ») ;

  Thanks again for your kind response, it was just the additional perspective that I needed!

• Parse qualified double quote delimited strings

  Hi all
  
  I have a few strings of data that I get I want to analyze. My problem is that I am not able to understand how I can analyze the fields that are double quote qualified ("19 999")?
  
  I played with the regex_substr, but I can't quite make what I want.
  
  I want to analyze the myString column in my example below in four areas:
  Product
  OnOrder
  OnHand
  TotalSold
  
  Here is an example... I've been playing with it for a bit, and I can't get it.
  

  WITH parseString AS (Select 
                          '"Grado, Headphones",123,2222,"19,999" ' myString
                       from dual
                       UNION ALL
                       Select 
                          'Audio Slave,222,3333,444' myString
                       from dual
                       
                                  )
  Select 
      myString
  from parseString

  Oracle Database 10 g Enterprise Edition release 10.2.0.3.0 - 64bi
  PL/SQL version 10.2.0.3.0 - Production
  CORE Production 10.2.0.3.0
  AMT for Linux: release 10.2.0.3.0 - Production
  NLSRTL Version 10.2.0.3.0 - Production
  
  
  Any suggestions greatly appreciated.
  
  S
  
  Published by: ScarpacciOne on April 22, 2010 15:24

  Hello

  As Centinul said, it would help a lot if you have published your desired results and explains how get you from the sample data.
  Here's something that can help, according to your specific needs:

  WITH     got_item_cnt     AS
  (
       SELECT     myString
       ,     1 + LENGTH (myString)
              - LENGTH (REPLACE (myString, ','))     AS item_cnt
       FROM     parsestring
  )
  ,     cntr     AS
  (
       SELECT     LEVEL     AS n
       FROM     dual
       CONNECT BY     LEVEL <= ( SELECT  MAX (item_cnt)
                         FROM    got_item_cnt
                       )
  )
  ,     comma_separated_values     AS
  (
       SELECT     i.myString
       ,     c.n
       ,     REGEXP_SUBSTR ( i.myString
                       , '[^,]+'
                       , 1
                       , c.n
                       )          AS value
       FROM     got_item_cnt     i
       JOIN     cntr          c     ON     c.n     <= i.item_cnt
  )
  ,     got_quote_cnt     AS
  (
       SELECT     myString
       ,     n
       ,     REPLACE (value, '"')     AS value
       ,     NVL ( SUM ( LENGTH (value)
                   - LENGTH (REPLACE (value, '"'))
                   ) OVER ( PARTITION BY  myString
                         ORDER BY      n
                         ROWS BETWEEN  UNBOUNDED PRECEDING
                           AND      1        PRECEDING
                          )
                , 0
                )          AS quote_cnt
       FROM     comma_separated_values
  )
  SELECT     myString
  ,     n
  ,     LTRIM ( SYS_CONNECT_BY_PATH (value, ',')
             , ','
             )          AS token
  FROM     got_quote_cnt
  WHERE     CONNECT_BY_ISLEAF     = 1
  START WITH     MOD (quote_cnt, 2)     = 0
  CONNECT BY     MOD (quote_cnt, 2)     = 1
       AND     myString     = PRIOR myString
       AND     n          = PRIOR n + 1
  ORDER BY     myString,     n
  ;
  

  Output of your sample data:

  MYSTRING                                N TOKEN
  -------------------------------------- -- --------------------
  "Grado, Headphones",123,2222,"19,999"   2 Grado, Headphones
  "Grado, Headphones",123,2222,"19,999"   3 123
  "Grado, Headphones",123,2222,"19,999"   4 2222
  "Grado, Headphones",123,2222,"19,999"   6 19,999
  Audio Slave,222,3333,444                1 Audio Slave
  Audio Slave,222,3333,444                2 222
  Audio Slave,222,3333,444                3 3333
  Audio Slave,222,3333,444                4 444
  

  The basic strategy is to divide myString to all the commas, then re - unite consecutive elements that are in double - quotes.
  You never said what version of Oracle are you uisng.
  This solution works in Oracle 10.1. REGEXP_COUNT, introduced in Oracle 11.1, would make it a bit simpler.

  Here is each subquery:
  got_item_count finds the number of elements by commas in each myString
  CNTR generates integers 1, 2, 3,... until the maximum number of elements by commas on line any
  comma_separated_values myString splits its different elements, ignoring the quotes at the moment. I assumed myString is unique. If this isn't the case, you need some other unique identifier for each row of parsestring.
  * quote_cnt had "notes how quotes took place in myString, each element not included.  If (and only if) this number is odd, then the element is really a continuation of a previous item, in other words, the comma that separates this point from the previous one was within quotation marks and therefore should not have divided myString.  I assumed that the double-quote character always indicates the grouping; It is never to be taken literally, and therefore the final results will never contain quotes.
  The main query string aggregation on consecutive points, bringing together all the elements that are weird quote_cnts with the last item previous had a same quote_cnt.

  Published by: Frank Kulash, 22 April 2010 23:33

• AppleScript - quotes from removal of cells

  I need to remove the quotes of cells in number.

  Currently, I have a script that imports data from Yahoo Finance in numbers in a range of cells (for example b2:j8).

  Unfortunately, much of this data as of the dates & times that are imported are surrounded by double quotes ".» It is therefore impossible to manipulate the data in these cells (for example dates are imported in the format mm/dd/yy and I wish I could change it to the jj/mm/aa but because the format was imported as 'dd/mm/yy' column cannot be formatted into a date format).

  Also other cells that can contain text have quotes. Fortunately the data that contains numbers have no quotes.

  I've looked everywhere and can't find a script that would be able to remove the quotes of cells.

  Help would be really appreciated. Thank you.

  The script below will remove the first and last characters for the values in the cells 3 numbers.

  To use:

  1. Copy-paste in the Script Editor (in Applications > utilities)
  2. Select the cells of numbers with dates
  3. With the selected cells, lick the button 'run '.

  Because this changes the contents of the cells, best to test first on a copy of your data.

  BTW, you don't specify what version of numbers that you use.  If you use numbers 2, some changes will be necessary.

  And have you seen this thread on the use of API finance from Yahoo to retrieve data on stocks and place it in numbers?

  SG

  say application "Numbers".

  say front document to tell the worksheet active

  say ( class is worn) fromfirst table whose selection range

  Repeat with ch in (get selection range) of cells

  c the value of the value to the (c value as text)'s text 2 thru -2

  end Repeat

  end say

  end say

  end say

• Scan from String - Remove comma

  Hello LabVIEW community.  Here's an easy one that made me a bit puzzled.  Then of course, it's not easy for me.  I am using string analysis for city, State, Zip from a string.  I can build the string that I have just described by commas using the Format string in.  But you can't do the opposite and get an error 85.  Yes, I have read the white paper on this and am not entirely sure how to apply their solution.  Yes, I could put bandaids on the code and remove the comma "manually", but I was wondering if the community was able to remove the commas of the chain by simply using the Scan of the chain.

  

  

  Analysis of the chain is "gourmet" you've discovered.  %s will basically take anything that's possible, including commas, so the first %s will simply take the whole string.  You must put the brakes on it a little.  For simple as cases this (no comma allowed inside the different parts) just replace %s, %s, %s % [^,], % [^,], % [^,] which means corresponds to something that isn't a comma followed by a comma.

• delete a random amount of extra spaces from string

  Hello I'm a string input that sperates a table of data with an amount random spaces.

  Is it possible to replace all the sequences of space by space?

  for example

  CPU PID MEM

  15.689 12.2 6610

  21.8. 156-6611

  (if the spaces are random)

  I have ried replaces the double spaces with single spaces a total of 8 times in the list, but for some reason which is also (somewhat) deleting simple spaes as well and im not sure why.

  You can do this with a single search and replace String.  You must enable regular Expressions for this function (with the right button of the function, choose Regular Expression).  The regular Expression '+' (space followed by a sign +) will match one or more spaces - son from that in the search string.  Son of a single "" ((espace) space) in the replacement string, true Set "replace all", then run the function and it will replace all multiple spaces (from the current line) by a space.  If you want to manage several lines in the string, enable this option as well.

  Bob Schor

• Remove the unused characters from string

  Hi, I try to remove unused characters from a string, but I can't find a function for this.

  The string like: 0000 0000 0000 0000 0000 0000 0130 3130 3030 3030 3004 0000 0000 0000 0000 0000 0000 0000, I need the characters from 01 to 04.

  I need some advice.

  If the characters between the SOH and EOT had a NULL value, then Match Regular Expression can be a choice:

  Find \x01, find something else THAN \x04, then find \x04

  

• Read from String

  Hi all

  The question I have is probably very trivial, but I apparently I'm stuck with it. The task that I am doing is to transfer a 'string data' that I get from reading via RS232 port of a controller of movement (relative coordinates of the position of the engine), in a table, so I can use it following the curve of the path. The chain is as follows:

  TPM: * TPM + 1274.36, + 0, + 0, + 56.149

  and I want two of these four numbers in the data file, something like this + 1274.36, + 56.149, by rank

  When I put 'write LVM' express VI and wired it "read rs232" to "signal in VI" me gives error that the data do not match. I therefore feel that the string must be changed somehow before it can be written in the data file.

  Can someone help me with this problem please?

  Thanks in advance

  It's my VI:

  You will need to use a function such as match on the palette of the string pattern.  Would be the model to look for the TPM secure: * party TPM assuming that is common on all lines of incoming data.  Then write the part after string matching in a text file.

  Since you just want to write the data to a text file, do not use the file function measure Express.  That really does work well if you already have a data type of 'Signal' as those coming from other Express VI.  Just use functions open close text, write files, in the range of file IO instead.  Search for Finder example for some examples.

  

• Question about the transition from string values to the Partition clause in a merge statement

  Hi all
  
  I use the code to update the data of specific secondary partition using oracle merge statements below.
  
  I'm getting the name of the secondary partition and pass this string to the secondary partition clause.
  
  The Merge statement is a failure, indicating that the specified secondary partition does not exist. But the partition under do exists for the table.
  
  We use a server Oracle 11 GR 2.
  
  Here is the code I use to fill in the data.
  
  declare
  ln_min_batchkey PLS_INTEGER;
  ln_max_batchkey PLS_INTEGER;
  lv_partition_name VARCHAR2 (32767).
  lv_subpartition_name VARCHAR2 (32767).
  Start
  
  FOR m1 IN (SELECT (year_val + 1) AS year_val, year_val AS orig_year_val)
  FROM (SELECT DISTINCT
  To_char (batch_create_dt, 'YYYY') year_val
  OF stores_comm_mob_sub_temp
  ORDER BY 1)
  ORDER BY year_val)
  LOOP
  lv_partition_name: =.
  () scmsa_handset_mobility_data_build.fn_get_partition_name
  nom_table_p = > 'STORES_COMM_MOB_SUB_INFO ',.
  p_search_string = > m1.year_val);
  
  FOR m2
  IN (SELECT DISTINCT
  'M' || To_char (batch_create_dt, 'MM') AS month_val
  OF stores_comm_mob_sub_temp
  WHERE TO_CHAR (batch_create_dt, 'YYYY') = m1.orig_year_val)
  LOOP
  lv_subpartition_name: =.
  () scmsa_handset_mobility_data_build.fn_get_subpartition_name
  nom_table_p = > 'STORES_COMM_MOB_SUB_INFO ',.
  p_partition_name = > lv_partition_name,
  p_search_string = > m2.month_val);
  
  DBMS_OUTPUT. Put_line (' lv_subpartition_name = > ' | lv_subpartition_name |' and lv_partition_name = > ' | lv_partition_name);
  
  IF lv_subpartition_name IS NULL
  THEN
  DBMS_OUTPUT. Put_line ("to the INTERIOR of FI = > ' |") M2.month_val);
  INSERT INTO STORES_COMM_MOB_SUB_INFO (T1)
  T1.ntlogin,
  T1.first_name,
  T1.last_name,
  T1.job_title,
  T1.store_id,
  T1.batch_create_dt)
  SELECT t2.ntlogin,
  T2.first_name,
  T2.last_name,
  T2.job_title,
  T2.store_id,
  T2.batch_create_dt
  OF stores_comm_mob_sub_temp t2
  WHERE TO_CHAR (batch_create_dt, 'YYYY') = m1.orig_year_val
  AND'M '. To_char (batch_create_dt, 'MM') =
  M2.month_val;
  ELSIF lv_subpartition_name IS NOT NULL
  THEN
  DBMS_OUTPUT. Put_line (' INSIDE ELSIF = > ' | m2.month_val);
  MERGE (SELECT *)
  OF stores_comm_mob_sub_info
  SUBPARTITION (lv_subpartition_name)) T1
  USING (SELECT *)
  OF stores_comm_mob_sub_temp
  WHERE TO_CHAR (batch_create_dt, 'YYYY') =
  M1.orig_year_val
  AND'M '. To_char (batch_create_dt, 'MM') =
  M2.month_val) T2
  WE (T1.store_id = T2.store_id
  AND T1.ntlogin = T2.ntlogin)
  WHEN MATCHED
  THEN
  GAME UPDATE
  T1.postpaid_totalqty =
  (NVL (t1.postpaid_totalqty, 0))
  (+ NVL (t2.postpaid_totalqty, 0));
  T1.sales_transaction_dt =
  LARGEST)
  NVL (t1.sales_transaction_dt,
  T2.sales_transaction_dt),
  NVL (t2.sales_transaction_dt,
  T1.sales_transaction_dt)),
  T1.batch_create_dt =
  LARGEST)
  NVL (t1.batch_create_dt, t2.batch_create_dt),
  NVL (t2.batch_create_dt, t1.batch_create_dt))
  WHEN NOT MATCHED
  THEN
  INSERT (t1.ntlogin,
  T1.first_name,
  T1.last_name,
  T1.job_title,
  T1.store_id,
  T1.batch_create_dt)
  VALUES (t2.ntlogin,
  T2.first_name,
  T2.last_name,
  T2.job_title,
  T2.store_id,
  T2.batch_create_dt);
  END IF;
  END LOOP;
  END LOOP;
  
  COMMIT;
  
  end;
  /
  
  
  
  Really appreciate your input here.
  
  Thank you
  MK.

  Hello

  You can use "immediate execution" what works.

  Thank you

• Issue of conversion from string to DateTime to subtract the duration of 1 day in the XSLT file

  Hello
  
  I need to subtract the duration 1 day after the date of entry which is in the format "yyyy-mm - ddT00:00:00"(data type = String)
  I tried to use the available function: subtract-dayTimeDuration-from - dateTime () but this function expects the entry in the datetime data type and so on use of this I get Null as the result of the transformation.
  
  I used to convert the input string, dateTime() but it doesn't seem to work because I get the error "Invalid Xpath expression (null).
  
  Please help identify any other way I can convert the string to dateTime format and then subtract the duration. Or another way to subtract the length of the day 1 of the string itself.
  
  Thanks for the help!

  Hello

  It should work with the date in the format that you have... Look at the code, there may be a typing error... In addition, in BPEL, you must use the xp20 functions...

  Take a look at this...
  https://blogs.Oracle.com/Reynolds/entry/whats_the_time_mr_bpel

  I hope this helps...

  See you soon,.
  Vlad