REGEXP_SUBSTR output

Similar Questions

• Regexp_substr

  How to get the output below using regexp_substr

  with t as(
  select 'APP_CONTACT_INFO' s from dual union
  select 'APP_CONTACT' from dual union
  select 'APP_CONTACT_TYPE_100200' from dual
  )
  
  
  
  
  select 
  s,
  REGEXP_SUBSTR( S,'_[A-Z]{3}',1) 
  
  
  
  
  from t;
  
  
  
  
  
  
  
  
  
  CON
  CONINF
  CONTYP100
  
  

  How to repeat _ [A - Z] {3} for each underscore match and concatenate it

  Select regexp_replace ('myTWst', '. *([A-Z]+). *', '\1') twice;

  only gives you a result of 'W' because the '. '. ' * ' in the expression is "greedy".  This means that it will match both the string he can before the rest of the evaluation of the model.

  So the first '. '. "*" matches "myT", then the "([A-Z] +)" ends up with only the W to match against and the final "." "*" corresponds to the "m".

  If instead you did:

  SQL > select regexp_replace ('myTWst', '. *?) () [A-Z] +). (* $', '\1') twice;

  RE
  --
  TW

  the '? 'after the first'. '. "*" he is not greedy, so it takes what is necessary to allow the rest of the pattern to match, wherever it is just 'my' instead of 'myT '.

  the "$" at the end indicates the last '. '. "*" to match all characters up to the end of the string, but even if it is greedy in itself, it is preceded by the "([A-Z] +) ' which itself is greedy and capitals as far as possible."

  If in fact told us the "([A-Z] +)" for do not be greedy by putting a "?" after the "+" then it takes as much as necessary to then leave the rest of the pattern (in our case the greedy"." (' $* ') to match as much as possible.

  SQL > select regexp_replace ('myTWst', '. *?) () [A-Z] + ?). (* $', '\1') twice;

  R
  -
  T

  So, now the '. '. "* $" is being greedy and account as much as possible while leaving enough for non greedy "([A-Z] +?)" for a match, which is just 1 uppercase letter. "  As you can see, the result is just the first letter capitalized.

• Tokenizer output

  Hello

  I have data in the following format

  select 'A;1;B;2;C;3;D;4;E;5' from dual
  union all
  select 'U;11;V;12;W;13;X;14;' from dual
  

  I need the output voltage:

  Header 1	Header 2	Header 3	Header 4	Header 5	Heading 6
  A	1	B	2	C	3
  D	4	E	5
  U	11	V	12	W	13
  X	14

  Can you please help?

  Thank you

  Vikram R

  You will need an id column (or a generated) to determine which line is which once cut the columns upwards.

  data (id, str))

  Select 1, '; 1; B; 2; D, C; 3; 4; E; 5' of the double

  Union of all the

  Select 2, ' U; 11; V; 12; W; 13; X; 14;' of the double

  ),

  splitdata like)

  SELECT id,

  trunc ((level-1) / 6) layer.

  regexp_substr (str, ' ([^;] *) (; | $) ', 1, level, 'i', 1) divide.

  MOD (level 1, 6) + 1 hdr

  data t

  connect by level<= regexp_count(str,="" ';')="" +="" 1="">

  prior t.ID = t.id and

  prior sys_guid() is not null - is unique, so make sure the Oracle will not execute a loop on the same path

  )

  Select HEADER1-tete2, HEADER3, HEADER4, HEADER5, HEADER6

  of splitdata

  pivot)

  min (Split)

  for hdr (1 as "HEADER1", 2 as "In-tete2", 3 as "HEADER3", 4 as 'HEADER4', 5 'HEADER5", 6 as 'HEADER6')

  )

  Commander order by id, layer

  HEADER1	HEADER2	HEADER3	HEADER4	HEADER5	HEADER6
  A	1	B	2	C	3
  D	4	E	5	-	-
  U	11	V	12	W	13
  X	14	-	-	-	-

  4 selected lines.

• How to remove the REGEXP_SUBSTR space

  Choose the length (TRIM (REGEXP_SUBSTR ('V, 370498, 658561477, 20150917, 4.37

  (((' [^,] +' 1, 5))) of the double

  output

  ---

  9

  Select TRIM (REGEXP_SUBSTR ('V, 370498, 658561477, 20150917, 4.37

  ((' [^,] +' 1, 5)) of the double

  output

  -----

  20150917

  
  

  last value actual length 8 only

  can help me please for these

  Thank you

  Thanks for the help

  to replace suggested used

  fix

  Choose the length (regexp_substr (regexp_replace ('V, 370498, 658561477, 20150917, 4.37

  (' [[: space:]] *', ")," [^,] (+' 1, 5)) solution, "

  length (rtrim (regexp_substr ('V, 370498, 658561477, 20150917, 4.37

  ', '[^,]+', 1, 5)

  10) has contributed to

  Double;

  Hi odie_63

  I replied to your previous one (csv to xml using ult_file pl/sql conversion ), it is also very good

  Thank you...

• How to deal with NULL values in regexp_substr?

  DECLARE
     TYPE t_rec IS TABLE OF VARCHAR2(200);
  
  /* case 1 */
     l_input    varchar2(100) := '28/09/2009 00:00:00|AFRISB|FC|773|7|17|512|76|60|U|';
  /* case 2 */
  --   l_input    VARCHAR2(100) := '06/10/2009 00:00:00|PDIERR|PS|999|0|1|0|0||U|';
     l_bars     PLS_INTEGER;
     l_fields   PLS_INTEGER;
     l_rec      t_rec := t_rec();
  BEGIN
     l_bars     := regexp_count(l_input, '\|', 1, 'i');
     l_fields   := l_bars + 1;
  
     FOR fld IN 1 .. l_fields
     LOOP
        l_rec.EXTEND();
  
        l_rec(fld)   := REGEXP_SUBSTR(l_input, '[^|]+', 1, fld);
        dbms_output.put_line('Fld ' || fld || ': ' || l_rec(fld));
     END LOOP;
  END;
  

  The above code is that I break a string of pipe delimited into individual elements.

  Looking at the case 1 in the code, the output displays each field correctly. However in case 2, the 9th field is null (nothing between the two tubes) so the regexp_substr 19 online is assigning the 'U' in the box 10 as a field 9.

  What should I do to get the line 19 to correctly assign a NULL value to field 9?

  Information about the version of the database are:

  BANNER

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

  12 c Oracle database Release 12.1.0.1.0 - 64 bit Production
  PL/SQL Release 12.1.0.1.0 - Production
  CORE Production 12.1.0.1.0
  AMT for 64-bit Windows: Version 12.1.0.1.0 - Production

  NLSRTL Version 12.1.0.1.0 - Production

  Thank you

  Steve

  Hello

  If you want to use regexes, so instead of

  REGEXP_SUBSTR (l_input, "[^ |]") +', 1, fld);

  use

  REGEXP_SUBSTR (l_input

  , '([^|] *)(\|| $)'

  -P

  fld

  NULL

  -P

  );

  . The 6th argument to REGEXP_SUBSTR is a new feature of Oracle 11.1.  It's kind of a back reference. Normally, REGEXP_SUBSTR returns the substring that matches the pattern in the 2nd argument.  If the 6th argument is 1, however, REGEXP_SUBSTR returns only the part of the pattern between the 1st on the left '(' et sa correspondance)."

• REGEXP_SUBSTR I need any word val, including space, or null

  Hi all

  I need to separate the last value in the field, but my query spaces not considered between the delimiters, please help thanks...

  SELECT RTRIM (REGEXP_SUBSTR ('C | 00013010101 |)) C00009 | 009402 | 00000100.000 | 1. AZUSDINE 500 MG TABLET | AZULFIDINE | 00016010101 | 1. 19950118 | 20150330 | 00000000 | 00000000 | 0 | F | 9. 1. 1. 0 | 1. 1. AB | 00000000 | 1. 0 | 000 | 0 | 0000001 | 00,000 | 00,000 | 00,000 | 0000012 | 0000012 | AB | TAB | 00000001.000 | 19900930 | 19900930 | 00000000 | 1. 19950701 | 2. 0 | BOTTLE | 0 | 20080925 | 20110928 | 0 | 0 | 0 | AB | 00000100.000 | 0 | 00 | AB | 2. 2. 9. 3. 1. I have | AZULFIDINE 500 MG TABLET '

  (' [a-zA-Z0-9] +--------|?', 1, 72),' | ') Double val

  output

  VAL

  -----

  AZULFIDINE

  Mandatory

  PUT out, VAL: AZULFIDINE 500 MG TABLET

  Thank you

  Olivier

  If the number of fields is fixed, if you always want (in this case) the 69th field, you could do that (I replaced your channel hardcoded with a column called "input_string" for clarity):

  SELECT RTRIM (REGEXP_SUBSTR (input_string |'|)) ',' [^ \ |] *\|',1,69),'|') Val from your_table

  If you want just the entire last section, you can do it without using regular expressions at all:

  SELECT SUBSTR (input_string, INSTR(input_string,'|',-1) + 1) val from your_table

• REGEXP_SUBSTR simple query

  Hi all

  Select RTRIM (REGEXP_SUBSTR('675809532|11787148|0','[a-zA-Z0-9]+\|'),'| ') double cardid

  Output

  --------

  675809532

  form above query I had the first value 675809532 how can I second valueor third value only pls help for these

  I need to pass the value to the variable individally

  Thank you...

  By adding the occurece to the RegExp as:

  Select RTRIM (REGEXP_SUBSTR ("675809532 |")) 11787148 | 0','[a-zA-Z0-9] + \| ?', 1,2),'| ') the double cardid

  and the regular expression would be easier that way:

  Select REGEXP_SUBSTR ('675809532 | 11787148 | 0', ' [^ \ |] +' 1, 3) double cardid

  HTH

• using regexp_substr

  Hello

  I need to extract the first five characters digital of the specified column. I'm able to reach the requirement using substr and definition of functions.

  I learned that the use of regular expressions we can achieve in a simple way. Please find below the test data.

  with t as (select '123456suri' double empid

  Union all select 'suri436789' of the double

  Union all select '521521suri' of the double

  Union select all 'suri785643abc' of the double)

  select * from t ;

  Desired output:

  EmpID

  ---------

  12345

  52152

  Please help me how to extract the results required using REGEXP_SUBSTR.

  I tried with query below, but not able to restrict the result to only five digits.

  Select double regexp_substr('3456789suri123','[0-9]+',1,1)

  Thank you

  Suri ;-)

  Post edited by: Suri - added query

  Good first try

  \d is a shortcut for [0-9]

  What happens when you try to \d\d?

  How about \d\d\d?

  You should be able to guess the rest of there

  You can also use {n} means "repeat exactly n times.

• Questioning REGEXP_SUBSTR, regular expressions

  With the expression below, I expect to come back the first 4 words and do not understand why it is not doing.

  db > select regexp_substr('This is a sample text string', '([[:alpha:]]+){3,4}') from dual;
  REGE
  ----
  This
  

  This works as expected.

  db > select regexp_substr('This is a sample text string', '([[:alpha:]]+ ){3,4}') from dual;
  REGEXP_SUBSTR('TH
  -----------------
  This is a sample
  

  Hello

  Solomon Yakobson says:

  Frank Kulash wrote:

  If you are looking for 3 or 4 instances of 1 or more letters, without anything in addition to the letters between them.  Any group of 3 or more consecutive letters matches.  For example,.

  1. "Th" as the 1st appearance of 1 or more letters,
  2. 'i' as the 2nd appearance of 1 or more letters, and
  3. of ' as the 3rd appearance of 1 or more letters

  is a substring that matches the pattern you described.

  In fact T is the 1st appearance of 1 or several letters, h is the second, I ranked third and s are defined. Otherwise:

  SQL > select regexp_substr ('the following is an example of text string', ' ([[: alpha:]] +) {3,4}') of the double
  2.

  REG
  ---
  The

  SQL >

  returns no rows.

  Not that it is important for the OP's question, but I think the way I presented it is more precise.

  How can tell us if 'This' match of the 4 occurrences (all 1 character) or 3 occurrences (1 with 2 characters and 2 with 1 character each)?

  In more general terms, when we say something like

  (expr +) {x, y}

  is '+' more greedy than "{x, y}.

  It seems to me that '+' is dishonest in this case.  Consider this query:

  WITH sample_data AS

  (

  SELECT "This is an example of text string" dual UNION ALL STR

  SELECT 'the following is an example of text string' FROM dual UNION ALL

  SELECT 'Fubar' double UNION ALL

  SELECT 'chat' FROM dual

  )

  SELECT str

  REGEXP_REPLACE (str

  , ' ([[: alpha:]] {1,2})([[:alpha:]]+) ([[: alpha:]] +)([[:alpha:]]+)?'

  , "\1=1;\2=2;\3=3;\4=4".

  ) And matches

  OF sample_data

  ;

  which produces this output:

  STR MATCHES

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

  This is an example of text string Th = 1; i = 2; s = 3; = 4 is a samp = 1; l =

  2; e = 3; 4 = = you 1; x = 2; t = 3; = 4 = stri

  1; n = 2; g = 3; = 4

  The following is an example of text string T = 1; h = 2; e = 3; = 4 followi = 1; n = 2; g

  = 3 ; = 4 is a samp = 1; l = 2; e = 3; = 4 t

  e = 1; x = 2; t = 3; = stri 4 = 1; n = 2; g = 3;

  = 4

  Fubar Fub = 1; a = 2; r = 3; = 4

  cat                                  c=1;a=2;t=3;=4

  In this case, '+' appears to be labels as "{3.4}.

  I don't have not seen this documented anywhere, but it seems it's because + gets to go first in the model.  In other words, there is nothing to inherently rapacity on ' + '.  If we change the first '+' above "{1,9}", like this

  WITH sample_data AS

  (

  SELECT "This is an example of text string" dual UNION ALL STR

  SELECT 'the following is an example of text string' FROM dual UNION ALL

  SELECT 'Fubar' double UNION ALL

  SELECT 'chat' FROM dual

  )

  SELECT str

  REGEXP_REPLACE (str

  , ' ([[: alpha:]]{1,9}) ([[: alpha:]] +) ([[: alpha:]] +)([[:alpha:]]+)?'

  , "\1=1;\2=2;\3=3;\4=4".

  ) And matches

  OF sample_data

  ;

  the result remains the same.

  Of course, is using the first query I posted above

  (expr) (expr) (expr) (expr)?

  It is possible that

  (expr) {3,4}

  which is what OP posted, is handled differently.

• XML and SQL output

  Hi all

  Please find test data and results as XML and SQL output.

  Version : Oracle Database 11 g Enterprise Edition Release 11.1.0.7.0 - 64 bit Production

  create table emp_xml (empno number, emp_name varchar2 (100), by e-mail to varchar2 (100));

  insert into emp_xml values (100, 'DAVID',' [email protected]');

  insert into emp_xml values (101, 'ROBERT',' [email protected]');

  insert into emp_xml values (102, 'DANIEL',' [email protected]');

  CREATE table emp_dept_xml (empno number, dept_list varchar2 (4000));

  insert into emp_dept_xml values (' 100,'10, 20, 30, ");

  insert into emp_dept_xml values (' 101,'40, 20, 10, 30, 50, ");

  insert into emp_dept_xml values (102, ' 10'),

  Expected in SQL output

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

  EMPNO EMP_NAME EMAIL DEPTNO

  DAVID 100 [email protected] 10

  DAVID 100 [email protected] 20

  DAVID 100 [email protected] 30

  ROBERT 101 [email protected] 10

  ROBERT 101 [email protected] 20

  ROBERT 101 [email protected] 30

  ROBERT 101 [email protected] 40

  ROBERT 101 [email protected] 50

  DANIEL 102 [email protected] 10

  Results in XML

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

  < USERS_DETAILS >

  < USER >

  < EMPNO > 100 < / EMPNO >

  DAVID < NAME > < / NAME >

  < EMAIL > [email protected] < / EMAIL >

  < ASSIGNED_DEPT >

  < DEPT > 10 < / DEPT >

  < DEPT > 20 < / DEPT >

  < DEPT > 30 < / DEPT >

  < / ASSIGNED_DEPT >

  < / USER >

  < USER >

  101/EMPNO < EMPNO > >

  ROBERT < NAME > < / NAME >

  < EMAIL > [email protected] < / EMAIL >

  < ASSIGNED_DEPT >

  < DEPT > 10 < / DEPT >

  < DEPT > 20 < / DEPT >

  < DEPT > 30 < / DEPT >

  < DEPT > 40 < / DEPT >

  < DEPT > 50 < / DEPT >

  < / ASSIGNED_DEPT >

  < / USER >

  < USER >

  < > 102 EMPNO < / EMPNO >

  DANIEL < NAME > < / NAME >

  < EMAIL > [email protected] < / EMAIL >

  < ASSIGNED_DEPT >

  < DEPT > 10 < / DEPT >

  < / ASSIGNED_DEPT >

  < / USER >

  < / USERS_DETAILS >

  Thank you

  Rambeau

  SQL>   select e.empno
    2         , e.emp_name
    3         , e.email
    4         , regexp_substr(d.dept_list, '[^,]+', 1, level) deptno
    5      from emp_xml e
    6      join emp_dept_xml d
    7        on e.empno = d.empno
    8   connect
    9        by level <= length(d.dept_list) - length(replace(dept_list, ','))
   10       and prior e.empno = e.empno
   11       and prior dbms_random.value() is not null;
  
       EMPNO EMP_NAME             EMAIL                DEPTNO
  ---------- -------------------- -------------------- --------------------
         100 DAVID                [email protected]      10
         100 DAVID                [email protected]      20
         100 DAVID                [email protected]      30
         101 ROBERT               [email protected]     40
         101 ROBERT               [email protected]     20
         101 ROBERT               [email protected]     10
         101 ROBERT               [email protected]     30
         101 ROBERT               [email protected]     50
         102 DANIEL               [email protected]     10
  
  9 rows selected.
  
  SQL> select xmlelement
    2         (
    3              "USERS_DETAILS"
    4            , xmlagg(xmlelement
    5              (
    6                  "USER"
    7                , xmlelement("EMPNO", empno)
    8                , xmlelement("NAME", emp_name)
    9                , xmlelement("ASSIGNED_DEPT", xmlagg(xmlelement("DEPT", deptno)))
   10              ))
   11         ).extract('*') xml_output
   12    from (
   13            select e.empno
   14                 , e.emp_name
   15                 , e.email
   16                 , regexp_substr(d.dept_list, '[^,]+', 1, level) deptno
   17              from emp_xml e
   18              join emp_dept_xml d
   19                on e.empno = d.empno
   20           connect
   21                by level <= length(d.dept_list) - length(replace(dept_list, ','))
   22               and prior e.empno = e.empno
   23               and prior dbms_random.value() is not null
   24         )
   25   group
   26      by empno
   27       , emp_name;
  
  XML_OUTPUT
  --------------------------------------------------------------------------------
  
    
      100
      DAVID
      
        10
        20
        30
      
    
    
      101
      ROBERT
      
        40
        50
        30
        20
        10
      
    
    
      102
      DANIEL
      
        10
      
    
  
  
  SQL>
  

• regexp_substr within pl/sql

  Table:

  REFERRING_PAGE	SEARCH_TERM
  ESTENCHERES.FR	4RUNER; HIGHLANDER; COROLLA; CAMRY;
  HONDA.COM	AGREEMENT; CIVIC; PILOT
  FORD.COM	EXPLORER; ESCAPE SUV; MUSTANG; TAURUS

  Outputs of the screen:

  REFERRING_PAGE	Search_term	WORDS
  ESTENCHERES.FR	4RUNNER; HIGHLANDER; COROLLA; CAMRY;	4RUNNER;
  		HIGHLANDER;
  		COROLLA;
  		CAMRY;

  HONDAAGREEMENT; CIVIC; PILOTAGREEMENT;CIVIC;PILOTFORD.COMEXPLORER;EXPLORERESCAPE SUV;MUSTANG;TAURUS

  Dear OTN forum;

  How can I build this function RUN IMMEDIATELY REGEXP_SUBSTR in PL/SQL. Thanks in advance.

  SET SERVEROUTPUT SIZE 100000;

  DECLARE

  v_referring_page dealers.referring_page%type;

  v_search_term dealers.search_term%type;

  MyCursor CURSOR IS

  SELECT referring_page

  search_term

  AT SELECT RETAILERS

  WHERE search_term IS NOT NULL

  AND TRUNC (log_date) = TO_DATE('30-JAN-14','DD-MON-YY');

  iRec myCursor % ROWTYPE;

  BEGIN

  OPEN myCursor;

  LOOP

  FETCH myCursor INTO iRec;

  OUTPUT WHEN myCursor % NOTFOUND;

  DBMS_OUTPUT. Put_line (iRec.referring_page |) »  '|| iRec.search_term);

  DBMS_OUTPUT. PUT_LINE(' ');

  DBMS_OUTPUT. Put_line (EXECUTE IMMEDIATE ' SELECT REGEXP_SUBSTR (iRec.search_term, "[^;]")) + ", 1, level) FROM DUAL

  CONNECT BY REGEXP_SUBSTR (iRec.search_term, "[^;]") (+ 1, level) IS NOT NULL ");"

  END LOOP;

  CLOSE MyCursor;

  END;

  Hello

  2745978 wrote:

  Table:

  REFERRING_PAGE	SEARCH_TERM
  ESTENCHERES.FR	4RUNER; HIGHLANDER; COROLLA; CAMRY;
  HONDA.COM	AGREEMENT; CIVIC; PILOT
  FORD.COM	EXPLORER; ESCAPE SUV; MUSTANG; TAURUS

  Outputs of the screen:

  REFERRING_PAGE	Search_term	WORDS
  ESTENCHERES.FR	4RUNNER; HIGHLANDER; COROLLA; CAMRY;	4RUNNER;
  		HIGHLANDER;
  		COROLLA;
  		CAMRY;
  HONDA	AGREEMENT; CIVIC; PILOT	AGREEMENT;
  		CIVIC;
  		PILOT
  FORD.COM	EXPLORER;	EXPLORER
  		ESCAPE SUV;
  		MUSTANG;
  		TAURUS

  Dear OTN forum;

  How can I build this function RUN IMMEDIATELY REGEXP_SUBSTR in PL/SQL. Thanks in advance.

  SET SERVEROUTPUT SIZE 100000;

  DECLARE

  v_referring_page dealers.referring_page%type;

  v_search_term dealers.search_term%type;

  MyCursor CURSOR IS

  SELECT referring_page

  search_term

  AT SELECT RETAILERS

  WHERE search_term IS NOT NULL

  AND TRUNC (log_date) = TO_DATE('30-JAN-14','DD-MON-YY');

  iRec myCursor % ROWTYPE;

  BEGIN

  OPEN myCursor;

  LOOP

  FETCH myCursor INTO iRec;

  OUTPUT WHEN myCursor % NOTFOUND;

  DBMS_OUTPUT. Put_line (iRec.referring_page |) »  '|| iRec.search_term);

  DBMS_OUTPUT. PUT_LINE(' ');

  DBMS_OUTPUT. Put_line (EXECUTE IMMEDIATE ' SELECT REGEXP_SUBSTR (iRec.search_term, "[^;]")) + ", 1, level) FROM DUAL

  CONNECT BY REGEXP_SUBSTR (iRec.search_term, "[^;]") (+ 1, level) IS NOT NULL ");"

  END LOOP;

  CLOSE MyCursor;

  END;

  REGEXP_SUBSTR, like almost all other single-row functions, can be invoked directly in PL/SQL; you need not EXECUTE IMMEDIATE or double table.

  You can do something like this:

  item_cnt: = REGEXP_COUNT (iRec.search_term

  , '[^;] +'

  );

  IF item_cnt > 0

  THEN

  FOR j IN 1... item_cnt

  LOOP

  dbms_output.put_line (REGEXP_SUBSTR (iRec.search_term

  , '[^;] +'

  1

  j

  )

  );

  END LOOP;

  END IF;

• REGEXP_SUBSTR question

  Good day to you all:

  I have a column in a table that contains the names of files as such:

  EXAMPLE1.05232014.BUSINESS.PDF

  EXAMPLE2.05182014.SUPPORT.DOCX

  EXAMPLE3.02252014.OPERATIONS.GIF

  How to use the REGEXP_SUBSTR function to extract the file extension?  My desired to above data output would be:

  . PDF

  . DOCX

  . GIF

  Please, if there is a way easier to extract file extension share, as I have mucho problemos with REGEXP_SUBSTR command.

  SQL DEVELOPER VERSION Version 3.2.20.09 - Oracle 10 g environment

  Thank you very much for your help!

  I appreciate the expertise here...

  Hello

  With the help of regular expressions:

  REGEXP_SUBSTR (file_name

  , '\. [^.] *$'

  )

  Without them (probably faster):

  SUBSTR (file_name

  , INSTR (filename, ".", - 1).

  )

  Want to sure if the filename does not contain any '.'s at all?  The above expression REGEXP_SUBSTR returns NULL and the expression of SUBSTR returns the file_name, but either can be replaced.

• Split the cell enclosed in several lines using REGEXP_SUBSTR - Performance?

  Hello

  I have a table that has about 20,000 lines. There is a column called key word which has values like below:

  File_id	Keyword
  1	SMITH; ALLEN; WARD; JONES; BRADY
  2	S & P500; TOPIX
  3	SMALL; LARGE; MEDIUM

  I want to display the data like this:

  File_id	KEYWORD
  1	SMITH
  1	ALLEN
  1	WARD
  1	BRADY
  2	S & P500
  2	TOPIX
  etc.

  I use this application and it works:

  SELECT STG. FILE_ID, REGEXP_SUBSTR (STG. KEYWORD ' [^;] +', 1, LEVEL) AS A KEYWORD OF STG STG_TABLE

  CONNECT REGEXP_SUBSTR (STG. KEYWORD ' [^;] +', 1, LEVEL) IS NOT NULL

  But its so slow its unusable.

  Is there a faster way to return this output?

  Other info:

  KEY Word is varchar2 (4000), but rarely more than 100 bytes are used

  Oracle 11g 2

  Thank you!

  Wrong approach, there are too many lines that are generated

  Apart from replacing regexp with substr/instr, try the first slot

  SELECT

  STG. FILE_ID

  REGEXP_SUBSTR (STG. KEYWORD ' [^;] +', 1, LEVEL) AS KEYWORD

  OF STG STG_TABLE

  CONNECT

  level<= regexp_count(stg.keyword,';')="" +="">

  and prior STG. FILE_ID = STG. FILE_ID

  and prior sys_guid() is not null

• Problem with REGEXP_SUBSTR related query.

  I'm having a problem with this query:

  SELECT *.

  FROM (SELECT REGEXP_SUBSTR ('{SUMMER} |)) {} POINT OF SALE}',

  '({)([A-Z]+ *[A-Z]*)(})',

  1,

  LEVEL,

  'I,

  2)

  Val

  OF THE DOUBLE

  CONNECT BY LEVEL < =.

  REGEXP_COUNT ('{SUMMER} |) {} POINT OF SALE} ',' |') + 1)

  WHERE val IS NOT NULL

  I need the output in 2 rows in this format:

  VAL

  ====

  SUMMER

  POINT OF SALE

  But I'm not able to get the "POINT OF SALE" in the output due to space, or perhaps another reason. Anyone can correct my query?

  So, you want something like that?

  SQL > ed
  A written file afiedt.buf

  1. SELECT *.
  2 (SELECT REGEXP_SUBSTR ('{SUMMER} |)) {} POS} ',' [^ {|}] + ", LEVEL, 1, 'i') val"
  3 FROM TWO
  4. CONNECT BY LEVEL<>
  5 REGEXP_COUNT ('{SUMMER} |) {} POINT OF SALE} ',' [^ |] +'))
  6 * val WHERE IS NOT NULL
  SQL > /.

  VAL
  ------------------------
  SUMMER
  POINT OF SALE

  SQL >

• SQL: REGEXP_SUBSTR CSV with string delimiter

  Hello

  I would like to create a function to split a CSV file with string using REGEXP delimiter.

  Because the REGEXP function does not deal with NULL values, in the first place, I use a REGEXP_REPLACE:

  SELECT REGEXP_REPLACE (' ' ' ';) "2011-12-30 15:33:15 '; "" 116 "; » « ; » 1 » ; "" ANSWER "; "" SIP/SDX20016-b6ec9c10 '; (""', '("") + ',' 'NULL' ') FROM dual;

  Result:

  'NULL '; "2011-12-30 15:33:15 '; "" 116 "; "' NULL '; » 1 » ; "" ANSWER "; "" SIP/SDX20016-b6ec9c10 '; "' NULL '.

  If I can deal with the empty columns.

  Then I apply this REGEXP:

  	SELECT REGEXP_SUBSTR (' 'NULL';) "2011-12-30 15:33:15 '; "" 116 "; "' NULL '; » 1 » ; "" ANSWER "; "" SIP/SDX20016-b6ec9c10 '; ("' NULL ' ',' [^(";")] +', 1, level)-LOOSE COLLECTION ltab_SplittedTelList
  	OF the double
  	CONNECT REGEXP_SUBSTR (' 'NULL';) "2011-12-30 15:33:15 '; "" 116 "; "' NULL '; » 1 » ; "" ANSWER "; "" SIP/SDX20016-b6ec9c10 '; » NULL » ', ' [^ « ; »] +', 1, level) IS NOT NULL;

  Result:

  NULL VALUE

  2011-12-30 15:33:15

  116

  NULL VALUE

  1

  REPLIED

  SIP/SDX20016-b6ec9c10

  NULL VALUE

  Very good, but if I put a delimiter column ';' in the middle of a string, for example in the middle of the date, I had my date cut in two.

  I can't find the correct model for my REGEXP.

  Jump, it is possible to do it with REGEXP.

  Thanks in advance,

  Greg.

  Hi, Greg.

  Here's a way to do it in pure SQL:

  WITH got_str AS

  (

  SELECT "' ';'" 2011 12-30; 15:33:15 '; 116. " » « ; » 1 » ; "" ANSWER "; "" SIP/SDX20016-b6ec9c10 '; ' ' ' ' ' AS str

  OF the double

  )

  SELECT THE LEVEL

  , REGEXP_SUBSTR (' ";' |") Str | ';"'

  , '";" ((([^"]*"[^;]) *[^"])*)'

  1

  LEVEL

  NULL

  1

  ) AS sub_str

  OF got_str

  CONNECT BY LEVEL<= 1="" +="" regexp_count="" (="">

  , '";"'

  )

  ;

  Note that I added one; after 2011-12-30.

  Output:

  LEVEL SUB_STR

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

  1

  2 on 12-30-2011; 15:33:15

  3 116

  4

  5 1

  6 ANSWERED

  7 SIP/SDX20016-b6ec9c10

  8

  As always, the solution depends on your version.  It works in Oracle 11.2.

  Your message, it seems that you do this in PL/SQL.  The above query will work in PL/SQL, of course, but it might be simpler to use the code of procedure.  Parse the string from left to right, followed the quotation marks and semicolons.  If you find a semicolon, and the total number of quotes spent so far is the same, then you have found the beginning of a new secondary channel.