Changeparticular the characters of a string using regular expressions...

Similar Questions

• How to capture multiple line String using regular expressions?

  Hello

  I have a simple program like this:

  

  What I want to accomplish is to capture everything between > start and > to end with a single regular expression matching node. It seems that the definition of multiples? true or False does not help.

  I'm using LabVIEW 2012.

  If it is impossible to capture using a single node, that's fine. But I want to assure you that I can make full use of this node without combining several others.

  Thank you!

  > start([\w\s]*) > end

  A point matches any character except line break characters.  You have two of them.

  

• Operation of string using regular expressions

  SQL> with raw_data as
    2  (
    3      select 1 id ,'A11P75 P76 R01       U01 U12 U40 U42 V53 VL VR A11' str from dual union all
    4      select 2,'A11P75 P76 R01       U01 U12 U40 U42 V53 VL VR' str from dual union all
    5       select 3,'A11P75 P76 R01       U01 U12 U40 U42 V53 VL  A11 VR' str from dual union all
    6      select 4,'A11P75 P76 R01       U01 U12 U40 U42 V53 A11 VL VR' str from dual union all
    7      select 5,'P75 P76 R01  A11     U01 U12 U40 A11 U42 V53 VL VR ' str from dual
    8  )
    9  select id
   10            ,str
   11            ,case when instr(replace(substr(str,instr(str,' ',-1,3)),'A11','#'),'#')>1
   12                      then replace(str,' A11','')
   13                      else str
   14             end Expected
   15    from raw_data
   16    /
  
  
          ID STR                                                 EXPECTED
  ---------- --------------------------------------------------- ---------------------------------------------------
           1 A11P75 P76 R01       U01 U12 U40 U42 V53 VL VR A11  A11P75 P76 R01       U01 U12 U40 U42 V53 VL VR
           2 A11P75 P76 R01       U01 U12 U40 U42 V53 VL VR      A11P75 P76 R01       U01 U12 U40 U42 V53 VL VR
           3 A11P75 P76 R01       U01 U12 U40 U42 V53 VL  A11 VR A11P75 P76 R01       U01 U12 U40 U42 V53 VL  VR
           4 A11P75 P76 R01       U01 U12 U40 U42 V53 A11 VL VR  A11P75 P76 R01       U01 U12 U40 U42 V53 VL VR
           5 P75 P76 R01  A11     U01 U12 U40 A11 U42 V53 VL VR  P75 P76 R01  A11     U01 U12 U40 A11 U42 V53 VL VR
  
  
  SQL>
  
  
  

  Logic of waiting: there will be a WORD like here "A11" who is looking within the chain

  If this word found in the last three words separated by space, delete this letter as a guide / appearing in the column PROVIDED

  First of all, your solution will not delete words A11 but instead of words that begin with A11:

  with raw_data as)

  Select the 1 id, str "A11P75 P76 R01 U01 A11MIDDLE U40 U42 V53 VL VR A11FAKE' of the double

  )

  SELECT id,

  Str,

  -case when instr (replace (substr (str, instr (str,' ', - 1, 3)), "A11", "#"), "#") > 1

  then replace (str, 'A11', ")

  of another str

  End expected

  of raw_data

  /

  ID STR                                                          EXPECTED
  ---------- ------------------------------------------------------------ -----------------------------------------------------
  1 A11P75 P76 R01, U01 A11MIDDLE U40-U42 V53 A11FAKE A11P75 P76 R01 U01MIDDLE U40-U42 V53 VRFAKE VL VL RV

  SQL >

  In any case, something like:

  with raw_data as)

  Select the 1 id, str "A11P75 P76 R01 U01 U12 U40 U42 V53 VL VR A11' Union double all the

  Select 2, str "A11P75 P76 R01 U01 U12 U40 U42 V53 VL VR' Union double all the

  Select 3, str "A11P75 P76 R01 U01 U12 U40 U42 V53 VL A11 VR' Union double all the

  Select option 4, str "A11P75 P76 R01 U01 U12 U40 U42 V53 A11 VL VR' Union double all the

  Select 5, str "P75 P76 R01 A11 U01 U12 U40 A11 U42 V53 VL VR" of double

  )

  SELECT id,

  Str,

  case

  Where regexp_like (str,'(^ |) (\s+))A11(\s+\w+){2}$) ')

  then regexp_replace (str '(^ |) (\s+))A11((\s+\w+){2}$)','\3) ')

  Where regexp_like (str,'(^ |) (\s+))\w+\s+A11\s+\w+$) ')

  then regexp_replace (str,'((^ |))) (\s+))\w+\s+)A11\s+(\w+)$','\1\4) ')

  Where regexp_like (str,'(^ |) (\s+))\w+\s+\w+\s+a11$) ')

  then regexp_replace (str,'((^ |))) (\s+))\w+\s+\w+)\s+a11$','\1) ')

  of another str

  End expected

  of raw_data

  /

  ID EXPECTED STR
  ---------- --------------------------------------------------- ---------------------------------------------------
  1 A11P75 P76 R01, U01 U12 U40-U42 V53 VR A11 A11P75 P76 R01, U01 U12 U40-U42 V53 RV VL VL
  A11P75 2 P76 R01, U01 U12 U40-U42 V53 A11P75 P76 R01, U01 U12 U40-U42 V53 RV VL VL RV
  3 A11P75 P76 R01, U01 U12 U40-U42 V53 VL A11 A11P75 P76 R01, U01 U12 U40-U42 V53 VL VR VR
  A11P75 4 P76 R01, U01 U12 U40-U42 V53 A11 VL A11P75 P76 R01, U01 U12 U40-U42 V53 VL VR VR
  5 P75 P76 R01 A11 U01 U12 U40 A11 U42 V53 RV VL R01 A11 U01 U12 U40-U42 V53 VL VR A11 P76 P75

  SQL >

  SY.

• Manipulating strings using regular expressions

  Hello guys,.

  I stuck in a situation in which I want to extract specific data from a column of the table.

  Here are the values for a particular column in which I want to ignore the values as well as support in the support and who are as .pdf, .doc.

  TRIS (dibenzylideneacetone) dipalladium (0) 451CDHA.pdf

  AM57001A (ASRM549CDH). DOC

  Identity sulfate AM23021A (project)

  PG-1183. E.2 (0.25 mg CTF)

  AS149656A (DEV APPL AERO WHT PROVENTIL HFA)

  Report on the stability (ISR) annex.2 semi-solid form (internal information)

  TSE(batch#USLF000332)-242CDH, Lancaster synthesis.pdf

  Additive TR3018520A 1 (REF 3018520)

  AM10311A Particle size airjet by sieving (constant sieving) (project)

  ASE00099B additive (REF E000099) mesh 90

  Palladium AM37101_312-99 (Z11c) by DCP.doc

  PS21001A_1H - NMR.doc (PN 332-00)

  AM68311A (Q - a CP 33021.02) attachment

  Attachment AM68202-1 has (BioReliance No. 02.102006)

  I want below output to above the values for the column

  Trisdipalladium451CDHA

  AM57001A

  Identity AM23021A sulfate

  PG-1183. E.2

  ...

  ..

  .

  Thanks in advance

  Like this?

  SQL> with t  2  as  3  (  4  select 'Tris(dibenzylideneacetone)dipalladium (0) 451CDHA.pdf' str from dual  5  union all  6  select 'AM57001A(ASRM549CDH).DOC' str from dual  7  union all  8  select 'AM23021A Identity of sulfate (draft)' str from dual  9  union all 10  select 'PG-1183.E.2 (0.25 mg FCT)' str from dual 11  union all 12  select 'AS149656A (DEV AERO APPL HFA WHT PROVENTIL)' str from dual 13  union all 14  select 'Stability report (RSR) Annex2 semi-solid form (internal information)' str from dual 15  union all 16  select 'TSE(Batch#USLF000332)-242CDH, Lancaster synthesis.pdf' str from dual 17  union all 18  select 'TR3018520A Addendum 1 (PN 3018520)' str from dual 19  union all 20  select 'AM10311A Particle size air-jet sieving (constant sieving) (draft)' str from dual 21  union all 22  select 'ASE00099B Addendum (PN E000099) 90 mesh' str from dual 23  union all 24  select 'AM37101_312-99 (Z11c) Palladium by DCP.doc' str from dual 25  union all 26  select 'PS21001A_1H-NMR.doc (PN 332-00)' str from dual 27  union all 28  select 'AM68311A (Q-One CP 33021.02) Attachment' str from dual 29  union all 30  select 'AM68202-1A (BioReliance no. 02.102006) Attachment' str from dual 31  ) 32  select str 33      , regexp_replace(str, '(\([^)]+\))|(\..{3})') str_new 34    from t;
  
  STR                                                                    STR_NEW---------------------------------------------------------------------- --------------------------------------Tris(dibenzylideneacetone)dipalladium (0) 451CDHA.pdf                  Trisdipalladium  451CDHAAM57001A(ASRM549CDH).DOC                                              AM57001AAM23021A Identity of sulfate (draft)                                  AM23021A Identity of sulfatePG-1183.E.2 (0.25 mg FCT)                                              PG-1183AS149656A (DEV AERO APPL HFA WHT PROVENTIL)                            AS149656AStability report (RSR) Annex2 semi-solid form (internal information)  Stability report  Annex2 semi-solid formTSE(Batch#USLF000332)-242CDH, Lancaster synthesis.pdf                  TSE-242CDH, Lancaster synthesisTR3018520A Addendum 1 (PN 3018520)                                    TR3018520A Addendum 1AM10311A Particle size air-jet sieving (constant sieving) (draft)      AM10311A Particle size air-jet sievingASE00099B Addendum (PN E000099) 90 mesh                                ASE00099B Addendum  90 meshAM37101_312-99 (Z11c) Palladium by DCP.doc                            AM37101_312-99  Palladium by DCPPS21001A_1H-NMR.doc (PN 332-00)                                        PS21001A_1H-NMRAM68311A (Q-One CP 33021.02) Attachment                                AM68311A  AttachmentAM68202-1A (BioReliance no. 02.102006) Attachment                      AM68202-1A  Attachment
  
  14 rows selected.
  

• Chips with 3 delimiter characters using regular expressions

  Hello world

  I have a function that is able to mark the input in a collection string using regular expressions.

  In case the input string is a character such as the comma or semicolon delimiter,

  We can just get the result we want like the example below.

  SQL> select * from v$version;
  
  
  BANNER
  --------------------------------------------------------------------------------
  Oracle Database 11g Enterprise Edition Release 11.2.0.1.0 - 64bit Production
  PL/SQL Release 11.2.0.1.0 - Production
  CORE    11.2.0.1.0      Production
  TNS for 64-bit Windows: Version 11.2.0.1.0 - Production
  NLSRTL Version 11.2.0.1.0 - Production
  

  SQL> with tab1 as (
    2      select 'abc,dfg,h,,1234' as col1 from dual
    3  )
    4  select regexp_substr(col1, '([^,]*)(,|$)', 1, level, 'i', 1) as result
    5  from tab1
    6  connect by level <= regexp_count(col1, ',')+1;
  
  
  RESULT
  ---------------
  abc
  dfg
  h
  
  1234
  

  But in the case where the channel of entry has 2 types of delimiter and each delimiter consists of 3 characters as below

  I wonder if it is possible to get the result as below.

  The input string: 01| ^ | ABCD| ^ | 111| * | 02| ^ | efgh| ^ | 222

  Separators: | * | is divided into lines, | ^ | is divided into columns

  Expected result:

  col1 col2 col3

  Row1 - > 01 abcd 111

  row2-> efgh 02 222

  Simply put, take a next

  The input string: 01| ^ | ABCD |^| 111 |*| 02 |^| efgh |^| 222

  Separator: | * |

  Result:

  01. ^ | ABCD | ^ | 111

  02. ^ | efgh | ^ | 222

  How can I achieve this using regular expressions?

  Kind regards

  Euntaek

  You need to know the number of the column from the outset:

  with tab1 as)

  Select ' 01 | ^ | ABCD | ^ | 111. * | 02. ^ | efgh | ^ | 222' as double col1

  )

  Select rownum,

  regexp_substr (regexp_substr (col1 '(.*?) ((\|\*\|)| $) ', 1, level, null, 1),'(.*?) ((\|\^\|)| $) ', 1, 1, null, 1) col1,.

  regexp_substr (regexp_substr (col1 '(.*?) ((\|\*\|)| $) ', 1, level, null, 1),'(.*?) ((\|\^\|)| $) ', 1, 2, null, 1) col2.

  regexp_substr (regexp_substr (col1 '(.*?) ((\|\*\|)| $) ', 1, level, null, 1),'(.*?) ((\|\^\|)| $) ', 1, 3, null, 1) col3

  of tab1

  connect by level<= regexp_count(col1,'\|\*\|')="" +="">

  /

  ROWNUM COL1 COL2 COL3

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

  1 01 abcd 111

  2 efgh 02 222

  SQL >

  SY.

• To delete the characters only at the beginning of a string using regexp_replace

  I want to be able to delete the characters in a string only until the first numeric value.  If it takes place after the digital, I want to keep them.

  "for example"AB1234' becomes '1234', 'AB1234C' becomes ' 1234 C.

  It might be also spaces and other types of characters at the beginning.  I tried this:

  Select

  regexp_replace ('ABC1234', ' [^ [: digit:]]').

  regexp_replace ('AB 1234',' [^ [: digit:]]').

  regexp_replace ('AB 01234',' [^ [: digit:]]').

  regexp_replace ('AB1234C', ' [^ [: digit:]]').

  regexp_replace ('^ 1234', ' [^ [: digit:]]').

  regexp_replace ('* #1234Z ',' [^ [: digit:]]')

  OF THE DOUBLE

  For all except "AB 01234' returned '1234' so it retains the digital.    For "AB 01234' it returns correctly '01234'

  The bad are those with a character after digital, for example "AB1234C".

  I know it I'm missing something here, can someone please help with this?

  Thank you

  SQL> with t
    2  as
    3  (
    4  select 'ABC1234' str from dual
    5  union all
    6  select 'AB 1234' str from dual
    7  union all
    8  select 'AB 01234' str from dual
    9  union all
   10  select 'AB1234C' str from dual
   11  union all
   12  select '^1234' str from dual
   13  union all
   14  select '*#1234Z' str from dual
   15  )
   16  select str
   17       , regexp_replace(str, '^[^[:digit:]]+') str_new
   18    from t;
  
  STR      STR_NEW
  -------- --------
  ABC1234  1234
  AB 1234  1234
  AB 01234 01234
  AB1234C  1234C
  ^1234    1234
  *#1234Z  1234Z
  
  6 rows selected.
  

• Find the words (wild cards) using regular expressions

  I'm testing to see if the words are present for revision 1 of a drawing of the cartridge.

  The script search the digit 1 followed by a date, a title, and 4 sets of initials.

  The number 1 is static, (date, title and original are the cards that they are different for each design).

  I use regular expressions to match the words.

  The regular expression highlighted in blue is the number 1 and the date.

  Him remains highlighted in orange does not match the title and initials.

  If anyone can help with the regular expression that is most appreciated.

  Once I got that work will add the form fields for the initials, noting only the console at this point for the tests.

  numWords = this.getPageNumWords (0);

  number of words on the page

  loop through the words on the page

  for (var j = 0; j < numWords-1; j ++)

  {/ / get the pair of words to test}

  ckWords = this.getPageNthWord (0, j) + ' ' + this.getPageNthWord (0, j + 1); test words

  Check if 1 26.05.16 THE STRENGTHENING REVISED MM SB AE GM word string is present

  If (ckWords.match(/ ^ 1\s [0-9] {1,2}.)) [0-9] {1,2}. [0-9] {2} \s\w+(\s+\w+){1,7}/))

  {

  Console.println (ckWords);

  }

  }

  You can use something like this:

  ckWords = this.getPageNthWord (0, j) + ' ' + this.getPageNthWord (j, 0, + 1) + ' ' + this.getPageNthWord (0, + 2 j) + ' ' + this.getPageNthWord (0, j + 3) + ' ' + this.getPageNthWord (0, j + 4) + ' ' + this.getPageNthWord (0, j + 5) + ' ' + this.getPageNthWord (0, j + 6);

  If (! ckWords.match (/ ^ 1\s\d {1,2} \.\d {1,2} \.\d {2} \s\w+ (?:-s + \w +) {1.8} \s([A-Z]{2})\s([A-Z]{2})-s ([A - Z] {2}) \s([A-Z]{2})$ /)) ckWords + ckWords + "" + this.getPageNthWord (0, j + 7);

  If (! ckWords.match (/ ^ 1\s\d {1,2} \.\d {1,2} \.\d {2} \s\w+ (?:-s + \w +) {1.8} \s([A-Z]{2})\s([A-Z]{2})-s ([A - Z] {2}) \s([A-Z]{2})$ /)) ckWords + ckWords + "" + this.getPageNthWord (0, j + 8);

  If (ckWords.match (/ ^ 1\s\d {1,2} \.\d {1,2} {2} \s\w+ \.\d (?:-s + \w +) 1.8 ([A - Z] {2}) \s([A-Z]{2})\s([A-Z]{2})\s {} \s([A-Z]{2})$ /))

  {

  ...

• String format using regular expressions

  Input string output format...
  
  SELECT q'<select ab_c "ABC", efg "EFG" from dual>' str FROM DUAL
  
  Output:
  
  STR                                  
  -------------------------------------
  select ab_c "ABC", efg "EFG" from dual
  
  
  Required output format using regular expression...
  
  
  STR                                  
  -------------------------------------
  select 'ab_c' "ABC", 'efg' "EFG" from dual

  Regular expressions have many limitations as tools of analysis, and you specify the rules you want. This expression puts quotation marks around a non-empty string before a quoted string:

  SELECT regexp_replace(q'',
                        '([^" ]+)( +"[^ ]*")' , '''\1''\2' ) str FROM DUAL;
  
  STR
  ------------------------------------------
  select 'ab_c' "ABC", 'efg' "EFG" from dual
  {code}
  
  It is not robust - a missing " will confuse it, and you should be using bind variables anyway.                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                                
  		    
              
             
   
  		     
  How to extract the value of a tag XML using regular Expressions
   
  			 
  We get a response XML from a WEB SERVICE.
  I convert it to VARCHAR2.
  Now, I want to get the real answer that is inside the tag of the response.
  I tried this:
  DECLARE
  V_1 VARCHAR2 (30000): = ' < soap: Body > < ns:ProcessArgusFeedsResponse xmlns:ns = "urn: PegaRULES:SOAP:ArgusToPegaFeeds:Services" > ' |
  '< response > good < / answer >< / ns:ProcessArgusFeedsResponse > < / soap: Body > ';
  v_response VARCHAR2 (100);
  BEGIN
  DBMS_OUTPUT. PUT_LINE (V_1);
  v_response: = REGEXP_SUBSTR (v_1, ' / < >(.+?) < \/Response > answer /');
  dbms_output.put_line (v_response);
  END;
  It does not work.
  Any help would be greatly appreicated.
  			 
  Hello
  user12240205 wrote:
  We get a response XML from a WEB SERVICE.
  I convert it to VARCHAR2.
  Why?  XML has its own native ways of analysis; Why not use them?
  If you use regular expressions, then REGEXP_REPLACE, as shown above, is a good option in Oracle 10, but starting in Oracle 11.1, you can use REGEXP_SUBSTR like this:
  REGEXP_SUBSTR (v_1
  , '(.+?)'
  1
  1
  NULL
  1
  )
  The 6th argument is like a backreference; "He tells REGEXP_SUBSTR did not return to the entire organization, but only the part inside the 1st left '(' et correspondant à sa droite).
  The '?' to make it non-greedy is necessary only if v_1 can contain more than one response.
  Now, I want to get the real answer that is inside the tag of the response.
  I tried this:
  DECLARE
  V_1 VARCHAR2 (30000): = "" |
  
  'Good';
  v_response VARCHAR2 (100);
  BEGIN
  DBMS_OUTPUT. PUT_LINE (V_1);
  v_response: = REGEXP_SUBSTR (v_1, ' /(. +?)) <\ esponse="">/');
  
  dbms_output.put_line (v_response);
  END;
  It does not work.
  
  That's because it's looking for a slash ("/") before the '' tag and another after the ' tag.
  The backslash ("\") is not necessary here, but it is not nothing wrong.
  		   
   
              
             
   
  		     
  One for the era: how to get this output using REGULAR EXPRESSIONS?
   
  			 How to get the bottom of output using REGULAR EXPRESSIONS?
  SQL> ed
  Wrote file afiedt.buf
  
    1* CREATE TABLE cus___addresses    (full_address                   VARCHAR2(200 BYTE))
  SQL> /
  
  Table created.
  
  SQL> PROMPT Address Format is: House #/Housename,  street,  City, Zip Code, COUNTRY
  House #/Housename,  street,  City, Zip Code, COUNTRY
  SQL> INSERT INTO cus___addresses VALUES('1, 3rd street, Lansing, MI 49001, USA');
  
  1 row created.
  
  SQL> INSERT INTO cus___addresses VALUES('3B, fifth street, Clinton, OK 74103, USA');
  
  1 row created.
  
  SQL> INSERT INTO cus___addresses VALUES('Rose Villa, Stanton Grove, Murray, TN 37183, USA');
  
  1 row created.
  
  SQL> SELECT * FROM cus___addresses;
  
  FULL_ADDRESS
  ----------------------------------------------------------------------------------------------------
  1, 3rd street, Lansing, MI 49001, USA
  3B, fifth street, Clinton, OK 74103, USA
  Rose Villa, Stanton Grove, Murray, TN 37183, USA
  
  SQL> The REG EXP query shouLd output the ZIP codes: i.e. 49001, 74103, 37183 in 3 rows.
  Published by: user12240205 on June 18, 2012 03:19			 
  /* Formatted on 2012/06/18 17:25 (Formatter Plus v4.8.8) */
  SELECT REGEXP_SUBSTR ((REGEXP_SUBSTR (full_address, '[^,]+', 1, 4)), '[[:digit:]]+') RESULT
    FROM cus___addresses
  		   
   
              
             
   
  		     
  Using Regular Expressions in the Code of edge
   
  			 
  Hello.
  I am quite new to the Edge Code but find it quite interesting to use.
  The find/replace feature is pretty but I got a little confused on how to use regular expression matching.
  I tried to clean up the coordinates of a file Adobe Edge animate full of 120.17px (heavy and less accurate)
  So basically you're looking \d+\.\d+px
  First of all, she says "use /re/ for regex" even though I know the Code is made for developers who * courses * know that it took me a little time to understand I should just put my regex between slashes.
  /\d+\.\d+PX/
  then the time comes to replace them. ALT-cmd-F and it says "replace".
  Puzzled again.
  Will I type my full sentence there? Maybe a «...» "written up somewhere around would avoid this question because, there, of course, a second field to come.
  And there, I got stuck.
  I can not find how to get the replacement of the model.
  tried to \1 $1 \1/ $1 / nothing seems to work... any hint of welcome.
  			 
  Franck,
  You are right that it does not work. I open a topic. Here is the link if you want to follow: https://github.com/adobe/brackets/issues/1861
  FYI, I know exactly how you want to replace the search results, but here are a few tips:
  You must set the text that you want to retrieve by using parentheses. Thus, for example, if you want the integer part of the number of the result, then your regexp would be: / (\d+)\.\d+px/
  Then you must specify the first result using $1, so (when it's fixed), you can use something like: $1px
  Thank you
  Randy
  		   
   
              
             
   
  		     
  A special character validation using regular expressions in ADF
   
  			 
  Hi guys,.
  I want to put the validation of a special as character (,.') ((en) &, -) using regular expressions.
  I asked the posting as [a-zA-Z0-9'(.),--/ &] but it does not work properly.
  Special characters should be like:
  Comma,
  Hyfan-
  Dot.
  Open and close braces (and).
  Ampercent &
  Apastrophy '
  Space ""
  Please help if anyone has idea.
  And I also tried to put under expression as...
  [a-zA-Z] + (\\s* [0-9] * [a-zA-Z] *-* & * \\(*\\) *, *'*. *) * [a-zA-Z0-9] + but we need the validation if we put special characters between the charater as "ab," chain "& (bc).
  his does not work if I put a special character at the beginning and the end of the string in the ADF
  			 
  Thanks Timo...
  its working fine...
  		   
   
              
             
   
  		     
  Validation using regular expressions or oracle fn
   
  			 
  Hello
  I want to apply validation to a column. I'm trying to see if I can use the regexp_replace function to find a boss and see if I can replace it to a better format. Can someone help me with these rules?
  1. only the characters a - z, A - Z, 0-9, ', - is allowed. If there are no other characters, then these characters must be removed.
  2. name should not begin with a symbol.
  3. If there is any symbol permitted in the name, (for example the hyphen) then the hyphen should be preceded and followed by a space.
  I would like to know if the above can be obtained by regular expressions or any other function will be useful to achieve the same.
  
  Thank you very much
  Cherkaoui
  			 
  HI, Charan,
  Whenever you have a problem, please post a small example data (CREATE TABLE and only relevant columns, INSERT statements) of all the tables involved, so that people who want to help you can recreate the problem and test their ideas.
  Also post the exact results you want from this data, as well as an explanation of how you get these results from these data, with specific examples.  Post in this case, maybe 5 or 10 strings, and what you would like to see (a string corrected, or a 'yes' / 'No' flag saying if this string follows rules.)
  Always say what version of Oracle you are using (for example, 11.2.0.2.0).
  See the FAQ forum: Re: 2. How can I ask a question on the forums?
  2784427 wrote:
  Hello
  I want to apply validation to a column. I'm trying to see if I can use the regexp_replace function to find a boss and see if I can replace it to a better format. Can someone help me with these rules?
  1. only the characters a - z, A - Z, 0-9, ', - is allowed. If there are no other characters, then these characters must be removed.
  2. name should not begin with a symbol.
  3. If there is any symbol permitted in the name, (for example the hyphen) then the hyphen should be preceded and followed by a space.
  I would like to know if the above can be obtained by regular expressions or any other function will be useful to achieve the same.
  Thank you very much
  Cherkaoui
  Sorry, we don't know what you want to do.
  1 seems to mean you want to change the given string so that it follows the rules.  (What exactly are special characters?  Are there only 2 special characters: single quotation mark and hyphen?)
  2 seems to tell us that you want to check if it meets the rules
  3. can be a
  I'll assume that you want to modify the string so that the returned value corresponds to rules.  This means that if the given string violates, 2 symbols will be removed from the beginning of the string, and if it violates 3, then spaces will be added.
  I don't think that there is a way to do this with a regular expression.
  You can make each with a separate REGEXP_REPLACE corrections, and make all in the same query by one nesting inside each other, like this:
  SELECT REGEXP_REPLACE (REGEXP_REPLACE (REGEXP_REPLACE (: str)))
  , "[^ a-zA-Z0-9 cm-]"
  )
  , '^[''-]+'
  )
  , ' *([''-]) *'
  , ' \1 '
  ) AS new_str
  OF the double
  ;
  of course, I can't test it, until post you some sample data and the exact results you want from these data.
  Here, the inner REGEXP_REPLACE rule 1 applies.  It deletes all characters except a to z small letters, capital letters A to Z, numbers, spaces, single quotes and hyphens.
  Rule 2 applies the REGEXP_REPLACE middle.  It removes the single quotes and hyphens that appear at the beginning of the string.
  
  The external REGEXP_REPLACE rule 3.  It guarantees that each single quotation mark or a hyphen is preceded and followed by a space.  If there is already a space (or several spaces) before and/or after the special symbol, they are replaced by a single space.  If this isn't what you want, it can be fixed, but it will make the expression even messier.
  You might consider to write a user-defined function to normalize strings.  In a function procedure, you can apply the rules one after the other. you need not nesting complicated like that.  Of course, a user-defined function will be slower, but maybe it's not a problem.
  Still, the above query converts the string so that the production rules.
  If you just want to check whether a string conforms to the rules, you can compare the original string to the output of the above expression.  If they are identical, then the string is in accordance with the rules.
  		   
   
              
             
   
  		     
  Using regular expressions to solve sys_refcursor of a record
   
  			 Regarding my Question about sys_refcursor with record type of thread, I thought it can be solved differently. It is:
  
  
  I have a string like ' 8:1706, 1194, 1817 ~ 1:1217, 1613, 1215, 1250'
  
  
  I need to do a few things using regular expressions and get something like
  
  
  select * from <table> where
  c1 in (8,1)
  and c2 in (1706,1194,1817,1217,1613,1215,1250);
  Is it possible by using regular expressions in a single select statement?			 
  Hello
  Game 6' - 8 "wrote:
  Your understanding is quite correct. But unfortunately it doesn't Frank.
  SQL> SELECT COUNT (*)
  2    FROM (SELECT sp.*
  3            FROM spml sp, spml_assignment spag
  4           WHERE sp.spml_id = spag.spml_id
  5             AND spag.class_of_svc_id = 8
  6             AND spag.service_type_id IN (1706, 1194, 1817)
  7             AND spag.carrier_id = 4445
  8             AND NVL (spag.haulage_type_id, -1) = NVL (NULL, -1)
  9             AND spag.effdate = TO_DATE ('01/01/2000', 'mm/dd/yyyy')
  10             AND spag.unit_id = 5
  11             AND sales_org_id = 1
  12          UNION ALL
  13          SELECT sp.*
  14            FROM spml sp, spml_assignment spag
  15           WHERE sp.spml_id = spag.spml_id
  16             AND spag.class_of_svc_id = 1
  17             AND spag.service_type_id IN (1217, 1613, 1215, 1250)
  18             AND spag.carrier_id = 4445
  19             AND NVL (spag.haulage_type_id, -1) = NVL (NULL, -1)
  20             AND spag.effdate = TO_DATE ('01/01/2000', 'mm/dd/yyyy')
  21             AND spag.unit_id = 5
  22             AND sales_org_id = 1);
  
  COUNT(*)
  ----------
  88
  
  SQL> SELECT COUNT (*)
  2    FROM spml sp, spml_assignment spag
  3   WHERE sp.spml_id = spag.spml_id
  4     AND spag.carrier_id = 4445
  5     AND NVL (spag.haulage_type_id, -1) = NVL (NULL, -1)
  6     AND spag.effdate = TO_DATE ('01/01/2000', 'mm/dd/yyyy')
  7     AND spag.unit_id = 5
  8     AND sales_org_id = 1
  9     AND REGEXP_LIKE ('8:1706,1194,1817~1:1217,1613,1215,1250',
  10                      '(^|~)' || spag.class_of_svc_id || ':'
  11                     )
  12     AND REGEXP_LIKE ('8:1706,1194,1817~1:1217,1613,1215,1250',
  13                      '(:|,)' || spag.service_type_id || '(,|$)'
  14                     );
  
  COUNT(*)
  ----------
  140
  
  SQL> 
  
  Published by: release 6' - 8 "August 11, 2009 20:04
  Serving what you ordered!
  Originally, you said that you are looking for something that produces the same result as
  where   c1 in (8, 1)
  and      c2 in (1706, 1194, 1817, 1217, 1613, 1215, 1250)
  
  in other words, the c1s could be coupled with any of the c2s.
  Now, it seems that what you want is
  where     (     c1 = 8
       and     c2 IN (1706, 1194, 1817)
       )
  or     (     c1 = 1
       and     c2 IN (1217, 1613, 1215, 1250)
       )
  
  in other words, c1 = 8 and c2 = 1250 is not good; is not c1 = 1 and c2 = 1706.
  In this case, try
  WHERE     REGEXP_LIKE ( s
                , '(^|~)' || c1
                       || ':([0-9]+,)*'
                       || c2
                       || '(,|~|$)'
                )
  		   
   
              
             
   
  		     
  ADF Email of Validation using regular expressions
   
  			 Hello
  
  
  Wanted to add search Email Validation VO.
  
  
  It works if I put
  <af:validateRegExp pattern="[A-Z0-9._%+-]+@[A-Z0-9.-]+\.[A-Z]{2,4}" 
                                           messageDetailNoMatch="The value {1} is not a valid email address:"/>
  However, this requires identification of email to be entered in capital letters.
  
  
  I tried with below the option does not work.
  
  
                 <af:inputText value="#{bindings.xxEmail.inputValue}" label="Email"
                                        required="#{bindings.xxEmail.hints.mandatory}"
                                        columns="#{bindings.xxEmail.hints.displayWidth}"
                                        maximumLength="#{bindings.xxEmail.hints.precision}"
                                        shortDesc="#{bindings.xxEmail.hints.tooltip}" id="it5">
                              <f:validator binding="#{bindings.xxEmail.validator}"/>
                              <f:validateLength minimum="6"/>              
                              <af:validateRegExp pattern="[A-Za-z0-9._%+-]+@[A-Za-z0-9.-]+\.[A-Za-z]{2,4}" 
                                           messageDetailNoMatch="The value {1} is not a valid email address:"/>
                          </af:inputText>
  I got over information of
  
  ADF Email of Validation using regular expressions
  
  The user enter email id without @.
  
  
  
  
  
  
  Please suggest this model to reach.
  
  
  
  
  Thank you
  JIT
  
  
  Published by: appsjit on January 25, 2013 19:08			 
  Hello.
  My English is not very good.
  I use below format and it works
  
  "^ [_A-Za-z0 - 9-] + (\.). [_A-Za-z0-9-]+)*@[A-Za-z0-9][A-Za-z0-9-]+ (-.)] ([A Za-z0-9] +) * (-.) [A-Za - z] {2,}) $»
  messageDetailNoMatch = "the value {1} is not a valid email address" / >
  Habib
  Published by: Habib Eslami on January 26, 2013 01:22
  		   
   
             
            
           
  		
   
            
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