REG EXP pattern ?

Hi Folks;
I need to create a reg exp pattern with these rules :
mpexprfinal      : mpexprUnit(\ OROP\ mpexprUnit)*
mpexprUnit      : mp(\ AND\ mp)*minusplusexpr
minusplusexpr :     "\ $+$\ " or "\ $-$\ "
mp : "[A-Z]{1}[0-9]{6}" (ex: I123456)
OROP : ","
Help me please!
Edited by: Moostiq on 2 mai 2011 16:52
Edited by: Moostiq on 2 mai 2011 16:52

Hi,
I don't know of any really good way to assign names to sub-patterns in a regular expression, and then use those names in bigger expressions.
You can (sort of) do the same thing in SQL, by assigning column aliases to string literals (as in def_1, below), or concatentions of literals and previouslly defined aliases (as in def_2):
WITH     def_1         AS
     SELECT     '$\+|-$'          AS minusplusexpr
     ,     '[A-Z]{1}[0-9]{6}'     AS mp
     ,     ';'               AS orop
     FROM     dual
,     def_2        AS
     SELECT def_1.*
     ,     mp || '( AND '
             || mp
             || ')*'          AS mpexprunit
     FROM    def_1
SELECT     x.*
FROM          table_x     x
CROSS JOIN     def_2     d
WHERE     REGEXP_LIKE ( x.txt
              , d.mpexprunit || '('
                             || d.orop
                       || '|'
                       || d.mpexprunit
                       || ')*'
;Take this as pseudo-code. I'm not sure it will do anything. (I can't test it until you post some sample data).
If it does run, I'm not sure it will do what you want (since you haven't explained what you want).
You may find it easier just to repeat the expressions in your code. An approach like the one above is most useful when the definitions (mp, mpexprunit, and so on) change frequently.
I hope this answers your question.
If not, post a little sample data (CREATE TABLE and INSERT statements, relevant columns only) for all tables, and also post the results you want from that data.
Explain, using specific examples, how you get those results from that data.
Always say which version of Oracle you're using. I'm not sure you'll need any features that were added after Oracle 10.1, but why take a chance?

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Reg Exp won't make a match with metacharacters?

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thanks!
package dbmanager2;
import javax.swing.*;
import java.awt.*;
import java.awt.event.*;
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* @author nkelleh
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LeWalrus wrote:
Hi All,
I'm trying to implement a filefilter using regular expressions to allow wildcard searches. Here is the code I have so far. Works fine for a literal match, but if i enter a wildcard search (eg. abc123) it will not list any files, even though if I enter the full name for a file, it matches just fine. Anyone got any ideas because I'm just comming up blank?
...Try debugging your code. A good place to start is to print the following to see where things go wrong:
public boolean accept(File dir, String name)
if (new File(dir,name).isDirectory())
    System.out.println("Ignoring: "+new File(dir,name));
    return false;
else
    Matcher matcher = pattern.matcher(dir.getName()); // shouldnt that be 'name' instead of 'dir'?
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    return matcher.matches();
}

A doubt on REG EXP

Hi friends,
Please clarify the following doubt in Reg Exp.
Table EMP has following EMP_NAMEs:
============
Anand
Bala_G
Chitra
David_C
Elango
Fathima
============
We have a set of characters as "abcdefghijklmnopqrstuvwxyz0123456789".
Now we need to find the count of EMP_NAMES whose characters (any) are not in the list of characters in the above list. In this example, the result should be 2. i.e., 'Bala_D' and 'David_C'. The query should be like:
Declare
v_string varchar2(50) := 'abcdefghijklmnopqrstuvwxyz0123456789';
v_count number(6);
Begin
select count(*)
into v_count
from emp
where regexp_like(emp_name, v_string);
dbms_output.put_line(v_count);
end;
========================
Thanks in advance!

Hi,
Welcome to the forum!
To use REGEXP_LIKE, you could say:
WHERE     REGEXP_LIKE ( emp_name
              , '[^abcdefghijklmnopqrstuvwxyz0123456789]'
              )However, it will be faster not to use regular expressions:
WHERE   LTRIM ( emp_name
           , 'abcdefghijklmnopqrstuvwxyz0123456789'
           )          IS NOT NULLEdited by: Frank Kulash on Oct 10, 2012 4:18 PM
Removed extra single-quote, after DAMorgan, below.

How to make Matcher stop once a reg exp match is found

Is there a way to make the regular expression Matcher stop reading from the underlying CharSequence once it finds a match? For example, if I have a very long String and a match for some regular expression is at the beginning of the String, can I make the Matcher object stop examining the String once it finds the match? The Matcher object seems to always examine the entire CharSequence, even if a match is very near the beginning.
Thanks,
Zach Cox
[email protected]

Nope, {1}+ doesn't work either. I know it's
continuing because I created my own CharSequence
implementation that just wraps a String orwhatever,
and I print to System.out whenever Matcher callsthe
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Reg Exp - not as expected

Not often I ask questions myself, and perhaps my mind's just gone fuzzy this morning, but I'm having trouble doing a simple replace with regular expressions...
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Wrote file afiedt.buf
1* select regexp_replace('this freddies is fred fred record', 'fred', 'freddies') from dual
SQL> /
REGEXP_REPLACE('THISFREDDIESISFREDFREDRECORD'
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Wrote file afiedt.buf
1* select regexp_replace('this freddies is fred fred record', '(^| )fred( |$)', '\1freddies\2') from dual
SQL> /
REGEXP_REPLACE('THISFREDDIESISFREDFRE
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If I put an extra space inbetween the two "fred"s then it works:
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Wrote file afiedt.buf
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SQL> /
REGEXP_REPLACE('THISFREDDIESISFREDFREDRECO
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Cheers

I think this will explain ..
SQL> select regexp_replace
2         ('this freddies is fred fred fred record',
3          '(^| )(fred)($| )','\1freddies\3') str
4 from dual
5 /
STR
this freddies is freddies fred freddies record
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Reg Exp always returning false value

Hi,
Below is my code. I want to restrict the values to only alphabets and numbers and not special char.
But the below condition fails on a positive value also. Ex: ABCD, abcd, 1234. These should be acceptable.
Is the reg exp wrong?
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Imran

Try: http://stackoverflow.com/questions/9012387/regex-expression-only-numbers-and-characters

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I am trying to extract the name from the following HTML.
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My regular expression is as follows.
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Message was edited by:
VanJay011379
Sorry, I realize i was missing a ';' char. I thought the "\\)" was causing problems. Move on, nothing to see here LOL : )

why not
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Explanation of my reg exp code

Hi,
When I run my code below, the line:
Seat 4: playafly299 ($10 in chips)
results in a match, but
Seat 6: WillRyker ($5.55 in chips)
results in a miss. Can someone explain why? I read the pattern below as:
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Konigs wrote:
Thank you Sabre. When are the other brackets appropriate?The square brackets are used to define a set of characters (not a sequence) and the () brackets define a group. You have an optional group that includes the decimal point so you need () not [] .

Reg exp trouble??

Hi All,
can someone tell me what is wrong with this code. I keep getting an Illegal State Exception, indicating it hasn't made a match, but I can't find anything wrong with the syntax. Yes, java.util.regex.*, is included at the start of the class.
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        System.out.println(myMatch.start());
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LeWalrus wrote:
Hi All,
can someone tell me what is wrong with this code. I keep getting an Illegal State Exception, indicating it hasn't made a match, but I can't find anything wrong with the syntax. Yes, java.util.regex.*, is included at the start of the class.
private void testReg()
Pattern myPattern = Pattern.compile("dog");
System.out.println(myPattern.pattern());
Matcher myMatch = myPattern.matcher("dogdoggydogilicious");
System.out.println(myMatch.start());
}Any help would be appreciated. Thanks.Hi, you have to invoke e.g. myPattern.matches(); before you can use start!
start
public int start()
    Returns the start index of the previous match.
    Specified by:
        start in interface MatchResult
    Returns:
        The index of the first character matched
    Throws:
        IllegalStateException - If no match has yet been attempted, or if the previous match operation failedSo you should use one of these methods before invoking start.
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      The lookingAt method attempts to match the input sequence, starting at the beginning, against the pattern.
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http://java.sun.com/javase/6/docs/api/java/util/regex/package-summary.html
http://java.sun.com/javase/6/docs/api/java/util/regex/Pattern.html
http://java.sun.com/javase/6/docs/api/java/util/regex/Matcher.html
Greetings Michael
Edited by: Urmech on Aug 13, 2008 7:04 AM

Reg Exp and non capturing groups

Hello,
I got problems with the non capturing groups in java.
I want to extract the 2nd ocurenc of a given patern from a string.
This ist the string =" 12.12.2004 [Logging] [Success] ### More text [Perhaps more brackets here] some text"
Now I want to extract the 2nd occurecy of the brackets with an unknow text pattern.
I tried this pattern = "(?:\[\w+\] )\[w+\]"
I thought that the non-capturing pattern helps me to find the first pair of [] brackets. the capturing group should give me want i want.
But I still get the result: [Logging] [Success]"
Even if I try just to use only the non cpaturing patern "(?:\[\w+\])" I still get an output from Matcher.group() which I didnt expect here!
What I am doing wrong?
Thanx for any help

Sorry, sabre150 I just missed it to answer you.
I have placed it into eclipse in a scrapbookpage into the following code:
final String regexp = "[^\\]]*\\[[^\\]]*\\]\\s*(.*)";
final String zeile = " 12.12.2004 [Logging] [Success] ### More text [Perhaps more brackets here] some text";
java.util.regex.Pattern pattern = java.util.regex.Pattern.compile(regexp);
java.util.regex.Matcher matcher = pattern.matcher("");
matcher.reset(zeile);
System.out.println("Found: "+matcher.find());
System.out.println(matcher.group());Output was:
Found: true
12.12.2004 [Logging] [Success] ### More text [Perhaps more brackets here] some text
and not what I wanted
"[Success] ### More text [Perhaps ...."
I dont know what I am doing wrong.
I thought I could solve it with the non-capturing group, but they doesnt seem to work.
My question to the non-capturing group is:
I have this pattern [code]"(?:.*)".
This pattern will probably match the String "Hello Mary!"
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Reg exp help

Hi
I want to extract the first token of filename (and not the directory path) from a string like this:
D:\MIRACLE\UPLOADS\1361\HG-ORA02#xaldb#alertlog_contents#22022010111941.mirdf.loadfailed_240210095258_62
HG-ORA02#xaldb#alertlog_contents#22022010111941.mirdf.loadfailed_240210095258_62
\HG-ORA02#xaldb#alertlog_contents#22022010111941.mirdf.loadfailed_240210095258_62
/HG-ORA02#xaldb#alertlog_contents#22022010111941.mirdf.loadfailed_240210095258_62
the tokens would bed HG-ORA02 in all instances
Could you help me creating a reg expression to get the token?
best regards
Mette

Hi,
Whenever you have a problem, it helps if you post your sample data in a form people can use.
CREATE TABLE and INSERT statements are great; so is:
CREATE TABLE     table_x
AS          select 'D:\MIRACLE\UPLOADS\1361\HG-ORA02#xaldb#alertlog_contents#22022010111941.mirdf.loadfailed_240210095258_62'
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UNION ALL     SELECT 'HG-ORA02#xaldb#alertlog_contents#22022010111941.mirdf.loadfailed_240210095258_62'
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UNION ALL     SELECT '\HG-ORA02#xaldb#alertlog_contents#22022010111941.mirdf.loadfailed_240210095258_62'
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UNION ALL     SELECT '/HG-ORA02#xaldb#alertlog_contents#22022010111941.mirdf.loadfailed_240210095258_62'
               AS filename, 4 as x_id FROM dual
;It also helps if you explain how you get the results you want from that data.
Assuming that
(a) the filename is divided, first, by '/' or '\' characters, and
(b) each of those parts may be divided into tokens by '.' or '#' characters, and that
(c) we want the first (b) part of the last (a) part:
SELECT     REGEXP_SUBSTR ( REGEXP_SUBSTR ( filename
                          , '[^/\]*$'
                , '[^.#]+'
                ) AS token_1
FROM     table_x;Here, the inner REGEXP_SUBSTR finds the last part delimited by / or \. You can add other delimiters by putting them in the square brakets, anywhere after ^.
The outer REGEXP_SUBSTR operates on the results of the inner one. It returns the first part delimited by . or #. Again, you can add other delimiters.

Reg Exp help needed (trouble with line breaks)

Hello,
I am attempting to parse a line of text with regular expressions. I am having difficulty with the line breaks.
I want to divide the following line by the first '\n' char.
For instance, I wan the output to be as follows:
found 1: one
found 2: ddd
dddHowever, I get the following with my attached code:
found 1: one
found 2: dddHowever, the second line break seems to be the end of the second group.
How do I get the any character '.' to read past any subsequent line breaks?
import java.util.regex.*;
public class Test
     String rawSequenceArg = ">one\nddd\nddd";
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Reg exp in form

Hello,
I need some help on reg expressions.
I want to allow users to type a url on my html page. The way I currently handle this situations is this:
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However, if the user types in something like this:
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How can I make sure that only those instances get replaced where only URL is typed which is not part of other html. Should I add blank spaces before and after the url?
Thank you.

Will there be any links in the source document that are already in the processed format? That is, http://www.somewhere.org/If so, you'll need to match the whole element so don't accidentally match a URL within it. Then you can deal with the situation you brought up, where a start tag has a URL attribute. For that, just match any complete start tag and plug it back in. That should take care of any URL's that are contained within larger structures, leaving only "naked" URL's that need to be processed.
A simple replaceAll() won't work here, because you're doing different things with the text depending on what was matched. The Matcher class provides lower-level operations that let you do that kind of thing, but Elliott Hughes has written a nice little utility called Rewriter that does most of the work for you. Here's how you might use it:     String regex = "(?is)" // CASE-INSENSITIVE and DOTALL modes
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        String found = group(0);
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Strange behavior when searching a phrase using reg exp and dynamic sql

Hi,
I have a strange issue while using dynamic sql for an apex page. I have a requirement to search a string in the database column which is entered by user on a page item. The search process should search the whole phrase only.
I have a query generated dynamically in the back end and use it in a cursor in the stored procedure
SELECT t.group_cn , t.group_desc, t.group_type, t.partner_organization_id, t.partner_organization
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   IF p_search_process NOT in ('PARTNER') THEN
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      OPEN v_cursor FOR v_sql USING p_search_id;
   END IF;
   LOOP
      FETCH v_cursor INTo v_obj.group_cn, v_obj.group_desc, v_obj.group_type, v_obj.partner_organization_id,
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      v_search_array(v_search_array.last) := v_obj;
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However, if the search string contains any special character, it does not return any thing. This strange issue happens only if I call the procedure from the apex page and the search string contains a special character. (please note that the procedure works fine even from apex if the string does not have a special character). When I debugged this, found that, the cursor does not fetch any rows (it is supposed to fetch two rows) for unknown reason. When I run the query separately, it returns the two rows (in which the column group_desc contains the search string "HR Selection & Assignment") as desired. Also, when I test the procedure in the back end (PLSQL developer), it works fine.
Any idea, what is causing this strange behaviour?
Advance thanks.
Regards,
Natarajan

i don't see anything about a dataProvider. you're assigning a source for a scrollpane. scrollpane's don't have a dataProvider property.
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Using reg exp's in filters

We have followed the recommendation to convert the cache database from Derby to MSSQL.
There seems to be an un-documented disadvantage in this:
regular expressions are no longer supported in filters !
Can anyone confirm this ?
We are running TES 6.1.0 with MSSQL 2008 database.

you can try a combination of instr() and substr() function. the example below is using an instr() function to determine the position of the last comma that is used in the string. the substr() function takes cares of the getting the last word that is used in the string with the help of the instr() function.
SQL> select substr(a.str,instr(a.str,',',-1)+1) last_string
2 from (select 'abcd,def,hijk,lmno,pqrs' str from dual) a ;
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