REGEXP_SUBSTR question

Similar Questions

• 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.

• Question about REGEXP_SUBSTR

  Dear experts.

  I would like to cut the chain TI - C, EN - C of the PBDA. JP - C, JPTYOOF-R * DPPD | TI - C, EN - C.

  SELECT REGEXP_SUBSTR (' PBDA |) JP - C, JPTYOOF-R * DPPD | TI - C, EN-it, ' [^ |] + |', 1, 3) twice;

  The statement above works well, but don't you think it would be wise to let | with exit-outside [].

  When I use the Sub statement I'm NULL.

  SELECT REGEXP_SUBSTR (' PBDA |) JP - C, JPTYOOF-R * DPPD | TI - C, EN-it, ' [^ |] + \|',1,3) twice;

  Could you please suggest me some alternatives.

  Thank you

  Rajat

  This way:

  with the data as

  (

  Select ' PBDA | JP - C, JPTYOOF-R * DPPD | TI - C, C - EN "double pass".

  )

  Select col, regexp_substr (col, ' [^ |] +' 1, 3)

  data;

  However, I would suggest using the SUBSTR function because the evaluation of the regular Expression tends to be expensive if you are dealing with huge data.

• question simple regexp_substr

  Hi all
  version 10g
  
  I want the string substr and get the point of atfer (.) string
  

  'abcd.xyz1'

  result
  
  XYZ1
  
  Thank you
  
  Published by: new learning on 10 Sep, 2010 08:12

  Hello

  Here's one way:

  REGEXP_SUBSTR ( str
             , '[^.]+$'
             )
  

  The. is not a wild-card in this regard.

  What happens if there are plusieurs.s in str?
  The expression above all will return (if any) after the last of them.

  Published by: Frank Kulash, Sep 10, 2010 11:20

  You specifically asked about REGEXP_SUBSTR.
  Regular expressions tend to be slower than the old ways, if both involve on the amout of code.
  You can get the same results faster using SUBSTR, as Iamaby suggested.
  To get the text (if any) after the last point:

  SUBSTR ( str
         , 1 + INSTR (str, '.', -1)
         )
  

• Question on connect by level 2 records

  My query begins with the records of two pairs and continues with the rank and adds two records to pair again.

  However, the unwanted extra each level (mbc 11, 12 wtc...) in the result.

  would you help me.
  

  My request;

  SELECT CUT (SUBSTR ("abc 11 12 11 22 33 kbc vbc wtc mbc mbc 17 45', INSTR kbs (" abc 11 12 11 22 33 17 mbc kbc vbc wtc mbc kbs 45 ', ' ', 1, level) + 1, "))

  INSTR ("abc 11 12 11 22 33 17 mbc kbc vbc wtc mbc kbs 45 ', ' ', 2, level + 1)-INSTR (" abc 11 12 11 22 33 17 mbc kbc vbc wtc mbc kbs 45 ', ' ', 1, level)-1)) AS PAIR

  OF THE DOUBLE

  CONNECT BY level < = LENGTH(' abc 11 mbc 12 wtc 11 vbc 22 kbc 33 mbc 17 kbs 45 ') LENGTH (REPLACE ('abc 11 12 11 22 33 17 mbc kbc vbc wtc mbc kbs 45', ' ', "))-1

  Outcome;

  ABC 11

  MBC 12

  WTC 11

  Vbc 22

  KBC 33

  MBC 17

  KBS 45

  Problem result;

  ABC 11

  11 mbc

  MBC 12

  WTC 12

  WTC 11

  11 vbc

  Vbc 22

  22 kbc

  KBC 33

  MBC 33

  MBC 17

  17 kbs

  KBS 45

  Hello

  Jimmie_M wrote:

  I thank very you much for your time. Thank you!!!

  is it possible to keep my request to get the result without regexp_substr.

  If I run regexp_substr in the table, the slow performance.

  Thank you

  You were on the right track with your first message.  The problem with this request was that you were each pair of words:

  • Word 1 and 2,
  • Word 2 and 3,
  • Word 3 and 4,
  • Word 4 and 5,

  and so on.  You don't want that every other pair:

  • Word 1 and 2,
  • Word 3 and 4,

  and so on.

  Here's a way to do it:

  WITH the data AS

  (

  SELECT ' abc 11 12 11 22 33 kbc vbc wtc mbc mbc 17 45' AS str kbs

  OF the double

  )

  got_space_pos AS

  (

  SELECT 1 * LEVEL AS n

  , INSTR (str, ' ', 1, (2 * LEVEL)-1) + 1 AS start_pos

  , INSTR (str, ' ', 1, (2 * LEVEL) + 1) AS end_pos

  FROM the data

  CONNECT BY LEVEL<= (="" length="">

  -LENGTH (REPLACE (str, ' '))

  ) / 2

  )

  SUBSTR (str, start_pos, end_pos - start_pos) SELECT as a PAIR

  OF got_space_pos

  JOIN CROSS data

  ORDER BY start_pos

  ;

  Output:

  PAIR

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

  ABC 11

  MBC 12

  WTC 11

  Vbc 22

  KBC 33

  MBC 17

  KBS 45

  Are you regular expressions of course are initially slow Paulzip query?  That's maybe something about your data you have not shown or explained.  As said by ground, you must always follow the Forum FAQ:

  Re: 2. How can I ask a question in the forums?

  so that people who want to help you can recreate the problem and test their ideas.

  In addition, the sample data (CREATE TABLE and INSERT statements), be sure to upgrade your version of Oracle.  A recursive (available in Oracle 11.2 and upwards) WITH clause might help in this problem. or maybe CROSS APPLY (Oracle 12.1) would be useful.

  Simplify the prblem as much as possible; for example, you generally need not show all columns in your actual tables, only those involved in the problem.  But make sure that you validate the example data illustrate the real problem.

• regexp_substr: retrieve a block of text between delimiters

  Hello

  regexp_substr is very powerful, but not so easy to understand. I'll try to do my best, but I need your help.

  My question is this: I have a multiline text, stored in a database field. The text is structured as a "windows.ini" file: there are sections in square brackets and the lines of text in the section.

  I want to retrieve all of the text in a given section, and I think it's possible using regexp_substr.

  Here is the text of the field (in bold the part I want back):

  ---

  [ARTICLE 1]

  AAA = 123

  BBB = 456

  CCC = 789

  [ARTICLE 2]

  DDD = 987

  EEE = 654

  ---

  In other words, is there a way to retrieve the text between [ARTICLE 1] and [ARTICLE 2]?

  Thank you in advance.

  Concerning

  Select

  regexp_substr (txt, ' [de] [^] +', 1, level * 2) s

  t

  connect by level<= regexp_count(txt,="">

• regexp_substr: a regular expression for the separate comma operator of witn of string literals IN

  The following regular expression separates simple values separated by commas (SELECT regexp_substr (: pCsv,'[^,] +', 1, level) FROM DUAL CONNECT BY regexp_substr (: pCsv, ' [^,] +', 1, level) IS NOT NULL); Exampple: 300100033146068, 300100033146071, 300100033146079 returns 300100033146068 300100033146071 300100033146079

  This works very well if we use the regex with SQL IN operator select * from mytable where t.mycolumn IN (SELECT regexp_substr (: pCsv,'[^,] +', 1, level) FROM DUAL CONNECT BY regexp_substr (: pCsv, ' [^,] +', 1, level) IS NOT NULL);

  But this query does not work if the comma-separated value is a single literal string 'one', 'two', 'three '.

  Thanks for your reply. my request was mainly on regexp_substr.  Need to request was simple: any table with a column of varchar type can be used. Next time I'll give you an example.

  All ways working answer for my question is is SELECT regexp_substr (: pCsv,'[^, "] +', 1, level) FROM DUAL CONNECT BY regexp_substr (: pCsv, ' [^,"] +', 1, level) IS NOT NULL

• REGEXP_SUBSTR, CONNECT BY slowing down with value

  I have the following sql code

  SELECT distinct q.broker_ID, q.user_ID, q.MGC_BREACH_ALERT, REGEXP_SUBSTR (nvl(q.CGC_ALERT_THRESHOLDS,0), "[^ |]") +', 1, LEVEL) value

  Of

  (select u.broker_ID, t.user_ID, t.CGC_ALERT_THRESHOLDS, ATSD_USE_TRADERS t.MGC_BREACH_ALERT t, ATS_USERS u where u.user_ID = t.user_ID and t.sponsored_user = 1) q

  CONNECT TO  (REGEXP_SUBSTR (nvl(q.CGC_ALERT_THRESHOLDS,0), "[^ |]")) +', 1, LEVEL) IS NOT NULL)

  order by 1

  When all the values in the table has CGC_ALERT_THRESHOLDS as two pipes separate values (for example 34 | 44) the query runs in less than 1 second.

  
  

  When the CGC_ALERT_THRESHOLDS are the three values of separate pipes (for example 34: 44 | 54) then the request takes longer than 30 seconds to complete.

  The table has only 150 lines.

  
  

  Can you please explain where I have gone wrong / or any other suggestions for improvements to this request?

  Hello

  Whenever you have a question, please post a small example of data (CREATE TABLE and only relevant columns, INSERT statements) for all of the tables involved and the accurate results you want from this data, so that people who want to help you can recreate the problem and test their ideas.

  Explain, using specific examples, how you get these results from these data.

  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?

  Jones wrote:

  I have the following sql code

  SELECT distinct q.broker_ID, q.user_ID, q.MGC_BREACH_ALERT, REGEXP_SUBSTR (nvl(q.CGC_ALERT_THRESHOLDS,0), "[^ |]") +', 1, LEVEL) value

  Of

  (select u.broker_ID, t.user_ID, t.CGC_ALERT_THRESHOLDS, ATSD_USE_TRADERS t.MGC_BREACH_ALERT t, ATS_USERS u where u.user_ID = t.user_ID and t.sponsored_user = 1) q

  CONNECT TO  (REGEXP_SUBSTR (nvl(q.CGC_ALERT_THRESHOLDS,0), "[^ |]")) +', 1, LEVEL) IS NOT NULL)

  order by 1

  When all the values in the table CGC_ALERT_THRESHOLDS as two pipes separate values (for example 34: 44) the query runs in less than 1 second.

  When the CGC_ALERT_THRESHOLDS are three pipes separated values (for example 34: 44 | 54) then the query takes more than 30 seconds to complete.

  The table has only 150 lines.

  Can you please explain where I have gone wrong / or any other suggestions for improvements to this request?

  Whenever you are tempted to use SELECT DISTINCT, think very carefully about why you need it.  It is often a way to hide an error in the query.

  In this case, the error is using CONNECT BY on a multiple row table, without conditions PRIOR involving.  If you have 150 rows in the table, this means that you will

  150 lines level = 1,

  150 * 150 = 22500 ranks LEVEL = 2,

  150 * 150 * 150 = 3375000 rows = 3 LEVEL.

  ...

  Soon, which will be a large number of lines.

  You must change the CONNECT BY clause into something like:

  CONNECT BY REGEXP_SUBSTR (NVL (q.cgc_alert_thresholds, '0'))

  , '[^|] +'

  1

  LEVEL

  ) IS NOT NULL

  AND PRIOR SYS_GUID () IS NOT NULL

  AND PREREQUISITE uk = uk - uk is a unique key

  I can't be sure without your sample data.

• Simple question: how to divide the string into multiple lines concatenated?

  Hi people,
  
  Maybe it's an easy question.
  
  How to split a string that is concatenated into multiple lines by using the SQL query?
  
  ENTRY:

  select 'AAA,BBB,CC,DDDD' as data from dual

  Delimiter = ', '.
  
  Expected results:

  data
  ------------
  AAA
  BBB
  CCC
  DDDD

  I'm looking for something nice to feature "an opposite to «sys_connect_by_path»»
  
  Thank you
  Tomas

  with t as (select 'aaaa,,bbbb,cccc,dddd,eeee,ffff' as txt from dual)
  -- end of sample data
  select REGEXP_SUBSTR (txt, '[^,]+', 1, level)
  from t
  connect by level <= length(regexp_replace(txt,'[^,]*'))+1
  
  REGEXP_SUBSTR(TXT,'[^,]+',1,LE
  ------------------------------
  aaaa
  bbbb
  cccc
  dddd
  eeee
  ffff
  

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  Hello

  I have a couple of "faces" in the issue of people who are coming in white, but acknowledged same 'face' of many times.  Is anyway to update it for photo comes actually?  At a few faces, I don't know that facial recognition found since it is coming from white.

  Hi JohnP007,

  Congratulations on your iPhone 7 more running iOS 10! I understand that some of your faces in the album of people pull up as a draft and you want to refresh. You can try to use the steps below to fix the faces on the thumbnails in albums.
  

  Difficulty faces and names mixed-up

  If you notice that there is a photo of someone in a collection that is poorly identified, you can remove it.

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  2. Type Show done face to emphasize his face in every photo.
  3. Press on each photo that is not the person.
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  You can hide the people or groups that you don't want in your album of people.

  1. Open the album people and press Select.
  2. Touch the people you don't want to see.
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  If you want to see the people that you have hidden, press on show hidden people.

  People in the Photos on your iPhone, iPad or iPod touch

  This should be corrected without delay faces. Please use the Apple Support communities to post your question. Good day.

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  If you forgot the answers to your questions of security of Apple ID - Apple Support

• I can't reset the security questions. We received notice as below:

  Hello world

  I can't reset the security questions. We received notice as below:

  "Cannot reset Security Questions."

  We have insufficient information to reset your security questions. "

  Please help me as soon as possible! Thank you very much.

  Hello

  You will need to contact the Apple Support.

  The information is available here:

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  (I'm afraid that no one here can solve the problem for you - this is a user-based community).

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  Now I'm trying to 2012 Macbook upgrade my Pro partner (on 10.11.6).  I got a Crucial SSD MX300 to replace his HARD drive.  Crucial comes with (or recommend) Acronis software.  And a lot of the messages of the forum recommend Carbon Copy Clone.

  My question is if I can use the same method for the cloning of the HD as my Macbook Air (just restore disk of Mac OS X utilities)?  This time, I'm upgrading HARD drive and I don't know if something is different.  At the same time, if I got lucky the first time, I don't not ruin Macbook Pro my spouse this time.

  Thank you.

  Yes, you can use disk utility to clone your MBP wives, but unlike CCC, it will not clone the recovery and Partition.

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  Same question perhaps for music. Some of my songs have been bought before the latest Apple codecs.

  Any help is appreciated!

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  Do exactly what you did to make this post, but your question in there instead.