Doubt with Regular Expression

Hi for All,
how can I get the expected result using regexp_substr or regexp_replace?
with tab as
Select 'TEST( XX_XXXXXX, 12 ) ; AAAA' txt from dual union all
Select 'TEST( AAAAAAAAA , 67 ); 1234' txt from dual union all
Select 'TEST( 92233 ,   47   ); 5234' txt from dual union all
Select 'TEST( AAAAAAAAA , AA ); 897'   txt from dual union all
Select 'TEST( CCCCC 25 );'             txt from dual union all
Select 'TEST CCCCC, 45 );'             txt from dual union all
Select 'TEST( EDCCCC, 45 ;)'           txt from dual union all
Select 'TEST(BBBBBBBBB,12);'           txt from dual
Select regexp_substr(txt, '[^.*,. *][[:digit:]][^\).*$]') from tab
REGEXP_SUBSTR(TXT,'[^.*,.*][[:
12
67
922
897
25
45
45
8 rows selected
Mandatory Criteria
"TEST" And "(" And "," And ")" And ";"
expected result
12
67
47
Null
Null
Null
Null
12Regards,

Something like:
with tab as
Select 'TEST( XX_XXXXXX, 12 ) ; AAAA' txt from dual union all
Select 'TEST( AAAAAAAAA , 67 ); 1234' txt from dual union all
Select 'TEST( 92233 ,   47   ); 5234' txt from dual union all
Select 'TEST( AAAAAAAAA , AA ); 897'   txt from dual union all
Select 'TEST( CCCCC 25 );'             txt from dual union all
Select 'TEST CCCCC, 45 );'             txt from dual union all
Select 'TEST( EDCCCC, 45 ;)'           txt from dual union all
Select 'TEST(BBBBBBBBB,12);'           txt from dual
select txt,
        case
          when regexp_like(txt,'^TEST$.+, *\d+ *$ *;.*$') then regexp_replace(txt,'(^TEST$.+, *)(\d+)( *$ *;.*$)','\2')
        end expected_result
from tab
TXT                          EXPECTED_RESULT
TEST( XX_XXXXXX, 12 ) ; AAAA 12
TEST( AAAAAAAAA , 67 ); 1234 67
TEST( 92233 ,   47   ); 5234 47
TEST( AAAAAAAAA , AA ); 897
TEST( CCCCC 25 );
TEST CCCCC, 45 );
TEST( EDCCCC, 45 ;)
TEST(BBBBBBBBB,12);          12
8 rows selected.
SQL> SY.

Similar Messages

Doubt with Regular Expression (REGEXP_REPLACE)

Hi for All,
how can I get the expected result using regexp_replace?
With Tab As( Select 1 Id, ' .XX ( A => B' TXT From dual Union All
             Select 1 Id, ' , C => 783 );'     TXT From dual )
SELECT REGEXP_REPLACE(Upper(TXT || Lead(TXT)Over(Partition BY id Order BY rownum))
                     ,'(^.*XX *$.+,.+)(\d+)( *$ *;.*$)','\2') AA
FROM tab
AA
3
, C => 783 );
Expected result
AA
783
, C => 783 );Regards,

Hi,
This gets the right results, perhaps for the right reason:
SELECT REGEXP_REPLACE ( Upper(TXT || Lead (TXT) Over (Partition BY id Order BY rownum))
                       , '(^.*XX *$.+,\D+)(\d+)( *$ *;.*$)','\2'
                 )      AS AA
FROM tab
;That is, change the 2nd '.+' to '\D+'.
Consider this regular expression:
SELECT     REGEXP_REPLACE ( 'C => 783'
                 , '(.+)(\d+)'
                 , '\2'
                 )     AS aa
FROM     dual
;Clearly, the 1st argument matches the pattern in the 2nd, but how? All of the interpretations below are equally valid:
\1          \2
C => 78          3
C => 7          83
C =>           783When there is a choice like this, it seems that \1 is made as big as possible.

Grouping & Back-references with regular expressions on Replace Text window

I really appreciate the inclusion of the Regular Expressions in the search & replace feature. One thing I am missing is back-references in the replacement expression. For instance, in the unix tools vi or sed, I might do something like this:
s/$firstPart$ $secondPart$ $oldThirdPart$/\2 \1 newThirdPart/g
which would allow me to switch the places of firstPart and secondPart, and totally replace thirdPart. If grouping and back-references are already present in the Replace Text window, how does one correctly invoke them?

duplicate of Grouping & Back-references with regular expressions on Replace Text window

Assistance with Regular Expression and Tcl

Assistance with Regular Expression and Tcl
Hello Everyone,
I recently began learning Tcl to develop scripts for automating network switch deployments.
In my script, I want to name the device with a location and the last three octets of the base mac address.
I can get the Base MAC address by :
show version | include Base
Base ethernet MAC Address       : 00:00:00:DB:CE:00
And I can get the last three octets of the MAC address using the following regular expression.
([0-9a-f]{2}[:-]){2}([0-9a-f]{2}$)
But I have not been able to figure out how to call the regular expression in the tcl script.
I have checked several resources but have not been able to figure it out. Suggestions?
Ultimately, I want to set the last three octets to a variable (something like below) and then call the variable when I name the switch.
set mac [exec "sh version | i Base"] (include the regular expression)
ios_config "hostname location$mac"
Thanks for any assistance in advance.
Chris

This worked for me.
Switch_1(tcl)#set result [exec show ver | inc Base]
Base ethernet MAC Address       : 00:1B:D4:F8:B1:80
Switch_1(tcl)#regexp {([0-9A-F:]{8}\r)} $result -> mac
1
Switch_1(tcl)#puts $mac
F8:B1:80
Switch_1(tcl)#ios_config "hostname location$mac"
%Warning! Hostname should contain at least one alphabet or '-' or '_' character
locationF8:B1:80(tcl)#

How to search with regular expression

I make pdx files so that I can search text quickly. But Acrobat doesn't provide a way to search with regular expression. I'm wondering if there is a way that I don't know to search for regular expression in Acrobat Pro 9?

First, Acrobat must "mount" the PDX.
As "Find" does not use the cataloged index, use Shift+Ctrl+F to open the advanced search dialog.
It may be helpful to first enter Acrobat Preferences and for the Search category tick "Always use advanced search options".
Back to the Search dialog - use the drop down menu for "Look In" to pick "Select Index" then, if no PDXs show, click the Add button.
In the Open Index File dialog, browse to the location of the desired PDX and select it.
OK out and use "Return results containing" to pick a "Match ..." requirement or Boolean.
To become familiar with query syntax, for Acrobat, it is good to review Acrobat Help.
http://help.adobe.com/en_US/Acrobat/9.0/Professional/WS58a04a822e3e50102bd615109794195ff-7 c4b.w.html
Be well...

Problem with Regular Expression

Hi There!!
I have a problem with regular expression. I want to validate that one word and second word are same. For that I have written a regex
Pattern p=Pattern.compile("([a-z][a-zA-Z]*)\\s\1");
Matcher m=p.matcher("nikhil nikhil");
boolean t=m.matches();
if (t)
          System.out.println("There is a match");
     else
          System.out.println("There is no match");
The result I am getting is always "There is no match
Your timely help will be much appreciated.
Regards

Ram wrote:
ErasP wrote:
You are missing a backward slash in the regex
Pattern p = Pattern.compile("([a-z][a-zA-Z]*)\\s\\1");
But this will fail in this case.
Matcher m = p.matcher("Nikhil Nikhil");It is the reason for that *[a-z]*.The OP had [a-z][a-zA-Z]* in his code, so presumably he know what that means and wants that String not to match.

Get the string between li tags, with regular expression

I have a unordered list, and I want to store all the strings between the li tags (<li>.?</li>)in an array:
<ul>
<li>This is String One</li>
<li>This is String Two</li>
<li>This is String Three</li>
</ul>
This is what have so far:
<li>(.*?)</li>
but it is not correct, I only want the string without the li tags.
Thanks.

No one?
Anoyone here experienced with Regular Expression?

Search with regular expression like this "[^\(]+mytype"

I tried regular expression which find me line which has "mytype" but does not precede with "("
I wrote it like this: [^\(]+mytype
It works perfectly with python reg exp. But it does not work in MSVC 2013 search(ctrl+shift+f). Do you know why? Is it bug?
Is there a way to find variable definition with search function through whole project

Hi Vit,
According to your description, you mean that when you use the
ctrl+shift+f shortcut key to search which line has "mytype" word in the whole project from VS2013 IDE like the following screen shot, but it does not work
in VS2013 IDE, am I right?
If yes, could you please provide me a screen shot about how you search the
mytype?
Generally, when we used the
ctrl+shift+f shortcut key to search, it works fine in VS IDE. Therefore, to check if you install some third-party Add-ins impact your VS IDE, I suggest you can search
the mytype in VS safe mode check if you get same issue.
Reference:
http://msdn.microsoft.com/en-us/library/ms241278.aspx
(1)If you did not get same issue, I doubt that you may be install some Add-ins impact the VS IDE, so I suggest
you can try to disable or un-install them and then check this issue again.
(2)If you still get same issue, I suggest you can try to reset the VS setting file or you can try to repair your
VS check this issue again.
In addition, I suggest you can try to use Ctrl + F shortcut key to search the mytype word.
If I misunderstand your issue, please provide me more detail message about your issue.
Best Regards,
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Need help with regular expression

I'm trying to use the java.util.regex package to extract URLs from html files.
The URLs that I am interested in extracting from the HTML look like the following:
<font color="#008000">http://forum.java.sun.com -
So, the URL is always preceeded by:
<font color="#008000">
and then followed by a space character and then a hyphen character. I want to be able to put all these URLs in a Vector object. This doesn't seem like it should be too difficult but for some reason I can't get anywhere with it. Any help would be greatly appreciated. Thanks!

hi gupta am not sure of the java syntax but i can tell u about the regular expression...try this....
<font color="#008000">(http:\/\/[a-zA-Z0-9.]+) [-]
i dont know the java methods to call...just the reg exp...
Sanjay Acharya

Help with Regular Expression for field validation

I'm fairly new to using regular expressions and using Acrobat. This is probably a simple question, but I've been unable to figure it out.
I have a text field on a PDF that I would like to be 9 characters in length. The first 2 characters can only be alphanumeric, the last 7 characters can only be numeric.
At first I was using the following, which allows all the characters to be alphanumeric:
var re = /^[A-Za-z0-9 :\\_]$/;
if (event.change.length >0) {
if (event.willCommit == false) {
    if (!re.test(event.change)) {
        event.rc = false
That works fine, but it's not quite what I needed. With some assistance I changed it (see below) to fit what I was looking for. However, this didn't work; it prevents anything from being entered in the field:
var re = /^[A-Za-z0-9]{2}\d{7}$/;
if (event.change.length >0) {
if (event.willCommit == false) {
    if (!re.test(event.change)) {
        event.rc = false
Any help would be greatly appreciated.
Thanks...

Here's a function you can call form the field's custom Format script. It should be placed in a document-level JavaScript:
function custom_ks1() {
    // Define non-commited regular expression
    var re = /^[A-Za-z0-9]{0,2}([0-9]{0,7})?$/;
    // Get all of the characters the user has entered
    var value = AFMergeChange(event);
    // Allow field to be cleared
    if(!value) return;
    if (event.willCommit) {
        // Define commited regular expression
        var re = /^[A-Za-z0-9]{2}[0-9]{7}$/;
        if (!re.test(value)) { // If final value doesn't match, alert user
            app.alert("Your error message goes here.");
            // event.rc = false
    } else { // not commited
        // Only allow characters that match the regular expression
        event.rc = re.test(value);
Call it like this:
// Custom Keystroke script
custom1_ks();

Help with regular expression needed

Hi,
Perhaps someone here can help me with my regular expression I'm trying to build in my Java code.
The regular expression that I'm looking to build consists of any non-whitespace character up until it finds one or two <>= symbols and then any character thereafter. So both these Strings would match the expression:
City 1==London
Age>=18
The regular expression that I'm using is as follows:
(\\S+)([><=]){1,2}(.+)However, group 1 always retrieves the first <>= symbol as in "City 1=". How can I make the <>= part greedy so that it retrieves both operator symbols?
Thanks.

Make the first group, the non-spaces, reluctant:
"(\\S+?)([<>=]{1,2})(.+)"

Help with regular expression to find a pattern in clob

can someone help me writing a regular expression to query a clob that containts xml type data?
query to find multiple occurrences of a variable string (i.e <EMPID-XX> - XX can be any number). If <EMPID-01> appears twice in the clob i want the result as EMPID-01,2 and if EMPID-02 appears 4 times i want the result as EMPID-02,4.

with
ofx_clob as
(select q'~
<EMPID>1
< UNQID>123456
< TIMESTAMP>...
< ADDRINFO>
< TITLE>^@~*
< FIRST>ABCD
< MI>
< LAST>EFGH
< ADDR1>ADDR1
< ADDR2>^@~*
< CITY>CITY
<EMPID>2
< UNQID>123457
< TIMESTAMP>...
< ADDRINFO>
< TITLE>^@~*
< FIRST>ABCD
< MI>
< LAST>EFGH
< ADDR1>ADDR1
< ADDR2>^@~*
< CITY>CITY
<EMPID>1
< UNQID>123458
< TIMESTAMP>...
< ADDRINFO>
< TITLE>^@~*
< FIRST>ABCD
< MI>
< LAST>EFGH
< ADDR1>ADDR1
< ADDR2>^@~*
< CITY>CITY
~' ofx from dual
select '<EMPID>' || to_char(ids) || '(' || to_char(count(*)) || ')' multi_empid
from (select replace(regexp_substr(ofx,'<EMPID>\d*',1,level),'<EMPID>') ids
          from ofx_clob
        connect by level <= regexp_count(ofx,'<EMPID>')
group by ids having count(*) > 1
MULTI_EMPID
<EMPID>1(2)
with
ofx_clob as
(select q'~
<EMPID>1
< UNQID>123456
< TIMESTAMP>...
< ADDRINFO>
< TITLE>^@~*
< FIRST>ABCD
< MI>
< LAST>EFGH
< ADDR1>ADDR1
< ADDR2>^@~*
< CITY>CITY
<EMPID>2
< UNQID>123457
< TIMESTAMP>...
< ADDRINFO>
< TITLE>^@~*
< FIRST>ABCD
< MI>
< LAST>EFGH
< ADDR1>ADDR1
< ADDR2>^@~*
< CITY>CITY
<EMPID>1
< UNQID>123456
< TIMESTAMP>...
< ADDRINFO>
< TITLE>^@~*
< FIRST>ABCD
< MI>
< LAST>EFGH
< ADDR1>ADDR1
< ADDR2>^@~*
< CITY>CITY
<EMPID>2
< UNQID>123456
< TIMESTAMP>...
< ADDRINFO>
< TITLE>^@~*
< FIRST>ABCD
< MI>
< LAST>EFGH
< ADDR1>ADDR1
< ADDR2>^@~*
< CITY>CITY
<EMPID>1
< UNQID>123458
< TIMESTAMP>...
< ADDRINFO>
< TITLE>^@~*
< FIRST>ABCD
< MI>
< LAST>EFGH
< ADDR1>ADDR1
< ADDR2>^@~*
< CITY>CITY
~' ofx from dual
select '<EMPID>' || listagg(to_char(ids) || '(' || to_char(count(*)) || ')',',') within group (order by ids) multi_empid
from (select replace(regexp_substr(ofx,'<EMPID>\d*',1,level),'<EMPID>') ids
          from ofx_clob
        connect by level <= regexp_count(ofx,'<EMPID>')
group by ids having count(*) > 1
MULTI_EMPID
<EMPID>1(3),2(2)
Regards
Etbin
Message was edited by: Etbin
used listagg to report more than one multiple <EMPID>

Help With Regular Expression In Apex Validation

Apex 3.2
There is a validation type of regular expression in apex, but I have never used regular expression before,
so a little help is appreciated.
I need to validate a field. It is only allowed to contain alpha characers, numbers, spaces and the - (dash) character.
I have tried several times to get this working
eg
[[:alpha:]]*[[:digit:]]*[[:space:]]*[-]*
^[[:alpha:][:digit:][:space:]-]+?
and others, but just can't to get the syntax correct.
Can someone help me with this please
Gus

Example:
SQL> ed
Wrote file afiedt.buf
1 with t as (select 'This is some example text' as txt from dual union all
2             select 'And this is the 2nd one with numbers' from dual union all
3             select 'And this allows double-barrelled words with hyphens' from dual union all
4             select 'But this one shouldn''t be allowed!' from dual
5            )
6 --
7 select *
8 from t
9* where regexp_like(txt, '^[[:alnum:] -]*$')
SQL> /
TXT
This is some example text
And this is the 2nd one with numbers
And this allows double-barrelled words with hyphens

Help with Regular Expressions and regexp_replace

Oh great Oracle Guru can I can gets some help
I need to clean up the phone numbers that have been entered in Oracle eBusiness per_phones table. Some of the phone numbers have dashes, some have spaces and some have char. I would just like to take all the digits out and then re-format the number.
Ex.
914-123-1234 .. output (914) 123-1234
9141231234 ..again (914) 123-1234
914 123 1234 .. (914) 123-1234
myphone ... just null
(914)-123-1234.. (914) 123-1234
I really tried to understand the regular expressions statments, but for some reason I just can't understand it.

Hi,
Welcome to the forum!
I would create a user-defined function for this. I expect there will be a lot of exceptions to the regular rules (for example, strings that do not contain exactly 10 digits, such as '1-800-987-6543') that can be handled, but would require lots of nested fucntions and othwer complicted code if you had to do it in a single statement.
If you really want to do it with a regular expression:
SELECT     phone_txt
,     REGEXP_REPLACE ( phone_txt
                 , '^\D*'          || -- 0 or more non-digits at the beginning of the string
                       '(\d\d\d)'     || -- \1 = 3 consecutive digits
                '\D*'          || -- 0 or more non-digits
                       '(\d\d\d)'     || -- \2 = 3 consecutive digits
                '\D*'          || -- 0 or more non-digits
                       '(\d\d\d)'     || -- \3 = 4 consecutive digits
                '\D*$'             -- 0 or more non-digits at the end of the string
                 , '(\1) \2-\3'
                 )          AS new_phone_txt
FROM    table_x
;

Help with Regular Expressions

I have some sample code for string editing using regular expressions but I'm a little confused as to it's behavior. Here's what I have:
public class WordFixer
    protected final String PUNC_MATCH = "[\\d\\p{Punct}]+";
    protected final String PUNC_PREFIX = "^" + PUNC_MATCH;
    protected final String PUNC_SUFFIX = PUNC_MATCH + "$";
    public String fixPrefix (String w)
        return w.replaceFirst(PUNC_PREFIX, "");
    public String fixSuffix (String w)
        return w.replaceFirst(PUNC_SUFFIX, "");
    }This replaces all leading and trailing punctuation with "" and it works. However, changing the replaceFirst's to just replace doesn't work. I don't understand why that is. Doesn't the ^ mean in the front, and doesn't the $ mean in the back, so shouldn't this work by just changing replaceFirst to replace?

JFactor2004 wrote:
I have some sample code for string editing using regular expressions but I'm a little confused as to it's behavior. Here's what I have:
public class WordFixer
protected final String PUNC_MATCH = "[\\d\\p{Punct}]+";
protected final String PUNC_PREFIX = "^" + PUNC_MATCH;
protected final String PUNC_SUFFIX = PUNC_MATCH + "$";
public String fixPrefix (String w)
return w.replaceFirst(PUNC_PREFIX, "");
public String fixSuffix (String w)
return w.replaceFirst(PUNC_SUFFIX, "");
}This replaces all leading and trailing punctuation with "" and it works. However, changing the replaceFirst's to just replace doesn't work. I don't understand why that is. The first parameter of the replaceFirst(...) is a regex-String. And the ^, $ and [...] stuff all have a special meaning in the regex language. The replace(...) method takes plain-Strings as a parameter, so the String "^\[\\d\\p{Punct}\]+" is interpreted as just that, without any special meaning.
JFactor2004 wrote:
Doesn't the ^ mean in the front, and doesn't the $ mean in the back, Yes, ^ means the start of the String (sometimes the start of a line) and $ means the end of the String (sometimes the end of a line).
JFactor2004 wrote:
so shouldn't this work by just changing replaceFirst to replace?Nope, see the explanation above.

Doubt with Regular Expression

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