Regular Expression ambiguous

I know the purpose of that code , but I don't understand it Sad
SQL> SELECT REGEXP_SUBSTR(
'The final test is is the implementation',
'([[:alnum:]]+)([[:space:]]+)\1') AS substr
FROM dual;
SUBST
is is
I study Regular Expression nowadays , from that link :-
http://www.oracle.com/technology/oramag/webcolumns/2003/techarticles/rischert_regexp_pt1.html
but I'm still need more good resources for ( Regular Expression white paper ) until I can understand all benefits , hence I can take Full Control of it.

just one thing to be aware of unrelated to backreferencing:
that Regex is less than perfect for the purposes of matching multiple words together:
SELECT REGEXP_SUBSTR(
'The diagram is isometric',
'([[:alnum:]]+)([[:space:]]+)\1') AS substr
FROM dual;so perhaps something like this:
SELECT REGEXP_SUBSTR(
'The diagram is isometric and and in pencil',
'([[:alnum:]]+)([[:space:]]+)\1\2') AS substr
FROM dual;using the backreference again to find the same space after the second word
but then again, this doesn't work perfectly because what if there's only one space after the second word but two after the first... and what if the duplicate word is at the end of the string without any other characters after it at all!!!!
find someone who can write a perfect regex and you'll have found someone who has too much time on their hands... :-P

Similar Messages

Regular Expression and PL/SQL help

I am using Oracle 9i, does 9i support regular expression? What functions are there?
My problem is the birth_date column in my database comes from teleform ( a scan program that reads what people wrote on paper), so the format is all jacked up.... 50% of them are 01/01/1981, 10% are 5/14/1995, 10% are 12/5/1993, 10% are 1/1/1983, 10% are 24-JUL-98. I have never really used regular expression and pl/sql, can anybody help me convert all of them to 01/01/1998?
Does Oralce 9i support regular expression? What can I do if oralce 9i does not support regular expression? Thank you very much in advance.

9i doesn't support regular expressions (at least not in the 10g regular expressions sense. There is an OWA_PATTERN_MATCH package that has some facilities for regular expressions). But it doesn't look like this is a regular expressions problem.
Instead, this is probably a case where you need to
- enumerate the format masks you want to try
- determine the order you want to try them
- write a small function that tries each format mask in succession until one matches.
Of course, there is no guarantee that you'll ever be able to convert the data to the date that the user intended because some values will be ambiguous. For example, 01/02/03 could mean Feb 1, 2003 or Jan 2, 2003 or Feb 3, 2001 depending on the person who entered the data.
Assuming you can define the order, your function would just try each format mask in turn until one generated a valid date, i.e.
BEGIN
BEGIN
    l_date := TO_DATE( p_string_value, format_mask_1 );
    RETURN l_date;
EXCEPTION
    WHEN OTHERS THEN
      NULL;
END;
BEGIN
    l_date := TO_DATE( p_string_value, format_mask_2 );
    RETURN l_date;
EXCEPTION
    WHEN OTHERS THEN
      NULL;
END;
BEGIN
    l_date := TO_DATE( p_string_value, format_mask_3 );
    RETURN l_date;
EXCEPTION
    WHEN OTHERS THEN
      NULL;
END;
BEGIN
    l_date := TO_DATE( p_string_value, format_mask_N );
    RETURN l_date;
EXCEPTION
    WHEN OTHERS THEN
      NULL;
END;
RETURN NULL;
END;Justin

[DW CC] Regular Expression and Back Reference in "Replace with": Escape

I have a problem with escaping a character in "Replace with".
Let's assume the following situation:
Content:
foobar
Search for:
(foo)(bar)
Replace by:
$1zot$2
Result:
foozotbar
Everything is fine.
But I want to insert the number "2" in between foo and bar.
When I use
Replace by: $12$2
The result is:
$12$2
It seems that the regex engine interpretes the "$12" as a whole. And because there's no back reference with the count "12" DW cannot work correctly.
The question:
How can I "escape" the "2"?
Thanks.

Dreamweaver uses JavaScript for its Find and Replace with regex.
I've just checked the Regular Expressions Cookbook by Jan Goyvaerts and Steven Levithan (O'Reilly). It deals with exactly this sort of situation where capturing groups can be ambiguous. This is what it says:
"Java and JavaScript try to be clever with $10 [and above]. If a capturing group with the two-digit number exists in your regular expresssion, both digits are used for the capturing group. If fewer capturing groups exist, only the first digit is used to reference the group, leaving the second as a literal. Thus $23 is the 23rd capturing group, if it exists. Otherwise, it is the second capturing group followed by a literal 3."
In other words, there is no way to escape the 2 in the Replace field. It would appear there's a bug in Dreamweaver's use of capturing groups.

Handling Regular expression ambiguitie s

Hi All,
I am attemting to make my own regular expression class. (Just for fun) :)
I have successfully converted the regular expression into a DFA ( deterministic finite automaton ). However, as i have noticed, there is still ambiguity in transition when it comes to certain classes of regular expressions such as a+[^0-9]+w. this means one or more a's followed by one or more symbols that are NOT numbers, followed by a w. Now if we have a string "aaaw", the first a is matched quite nicely, but should the second 'a' be treated as an a that is to be matched with "a+" or "[^0-9]" ??
Is there any algorithm that resloves such an ambiguity?
Any help would be much appreciated. I am stuck with this issue and don't seem to have any leads at all :)

Hi All,
My problem is now solved. I have refined the suggestions to come up with the following algorithm ( excuse the errors, and ineffeciencies, if any )
push all DFA states into a queue
while !(dfaStateQueue.empty())
       currentDFAState = dfaStateQueue.pop()
       push all transitions of currentDFAState to transitionQueue.
       store all transitions into transitionArray.
        while ( !transitionQueue.empty() )
               compareWithTransition = transitionQueue.pop();
               for(i = 1 to length(transitionArray) )
                      compareWithTransition = transitionArray;
Set intersec = set_intersection(compareWithTransition.getChars, compareToTransition.getChars);
if( !intersect.empty() )
//characters in compareWith but not in compareTo
Set minus1 = set_difference(compareWithTransition, compareToTransition)
//characters in compareTo but not in compareWith.
Set minus2 = set_difference(compareToTransition, compareWithTransition)
//compareWithTransition is subset of compareToTransition
if(minus1.empty())
replace characters of compareTo with characters in set minus2
replace characters of a closure transition of compareTo
with characters in set minus2
compareWithTransition.nextDfa.addAllTransitions
(compareToTransition.nextDfa.getTransitions())
continue
//compareToTransition is subset of compareWithTransition
else if(minus2.empty())
replace characters of compareWithTransition with characters in set minus1
replace characters of a closure transition of compareWithTransition
with characters in set minus2
compareToTransition.nextDfa.addAllTransitions
(compareWithTransition.nextDfa.getTransitions())
continue;
end if
create new DFA state D;
currentDFAState.addTransition(intersect, D);
push D onto queue.
D.addAllTransitions(compareWithTransition.nextDfa.getTransitions());
D.addAllTransitions(compareToTransition.nextDfa.getTransitions());
replace characters of compareWithTransition with characters in set minus1
replace characters of compareToTransition with characters in set minus2
end if
end for
end while
end while
I hope this helps everyone at large. Now there is no need for backtracking at all and DFAs are now truly "DFAs" :)
I dont know if other tools employ this technique or not. Please ignore the glaring ineffeciencies. I have give the algorithm the way i coded it ( Never considered myself as effecient programmer in the first place) HAHAHA
Chears,
AUTOMATON

Logical AND in Java Regular Expressions

I'm trying to implement logical AND using Java Regular Expressions.
I couldn't figure out how to do it after reading Java docs and textbooks. I can do something like "abc.*def", which means that I'm looking for strings which have "abc", then anything, then "def", but it is not "pure" logical AND - I will not find "def.*abc" this way.
Any ideas, how to do it ?
Baken

First off, looks like you're really talking about an "OR", not an "AND" - you want it to match abc.*def OR def.*abc right? If you tried to match abc.*def AND def.*abc nothing would ever match that, as no string can begin with both "abc" and "def", just like no numeric value can be both 2 and 5.
Anyway, maybe regex isn't the right tool for this job. Can you not simply programmatically match it yourself using String methods? You want it to match if the string "starts with" abc and "ends with" def, or vice-versa. Just write some simple code.

Help in Regular expression

Hello..
I wanted to write a regular expression to match the foll string..
 
 NEW ORLEANS, Louisiana (CNN) 
-- Two years after Hurricane Katrina devastated coastal areas of Louisiana and Mississippi, residents say much of America has forgotten their plight.
 
I tried doing..
Matcher matcher= Pattern.compile(" ([^<^>]+?)", Pattern.CASE_INSENSITIVE).matcher(story);
Its not working...
is there any other soln?

Theres probably a better way to do this but here's a way that works.
import java.util.regex.*;
public class RegexTester{
public static void main(String[] args){
 String text =
 " " +
 " NEW ORLEANS, Louisiana (CNN) " +
 "-- Two years after Hurricane Katrina devastated coastal areas of Louisiana and Mississippi," +
 "residents say much of America has forgotten their plight." +
 " ";
 String regex = ">((?:\\s*[\\S&&[^<>]]+\\s*)*?)<";
 Pattern p = Pattern.compile(regex);
 Matcher m = p.matcher(text);
 while(m.find()){
 System.out.println("Match: '" + m.group(1) + "'");
}

Help in regular expression matching

I have three expressions like
1) [(y2009)(y2011)]
2) [(y2008M5)(y2011M3)] or [(y2009M5)(y2010M12)]
3) [(y2009M1d20)(y2011M12d31)]
i want regular expression pattern for the above three expressions
I am using :
REGEXP_LIKE(timedomainexpression, '???[:digit:]{4}*[:digit:]{1,2}???[:digit:]{4}*[:digit:]{1,2}??', 'i');
but its giving results for all above expressions while i want different expression for each.
i hav used * after [:digit:]{4}, when i am using ? or . then its giving no results. Please help in this situation ASAP.
Thanks

I dont get your question Can you post your desired output? and also give some sample data.
Please consider the following when you post a question.
1. New features keep coming in every oracle version so please provide Your Oracle DB Version to get the best possible answer.
You can use the following query and do a copy past of the output.
select * from v$version 2. This forum has a very good Search Feature. Please use that before posting your question. Because for most of the questions
that are asked the answer is already there.
3. We dont know your DB structure or How your Data is. So you need to let us know. The best way would be to give some sample data like this.
I have the following table called sales
with sales
as
 select 1 sales_id, 1 prod_id, 1001 inv_num, 120 qty from dual
 union all
 select 2 sales_id, 1 prod_id, 1002 inv_num, 25 qty from dual
select *
from sales 4. Rather than telling what you want in words its more easier when you give your expected output.
For example in the above sales table, I want to know the total quantity and number of invoice for each product.
The output should look like this
Prod_id sum_qty count_inv
1 145 2 5. When ever you get an error message post the entire error message. With the Error Number, The message and the Line number.
6. Next thing is a very important thing to remember. Please post only well formatted code. Unformatted code is very hard to read.
Your code format gets lost when you post it in the Oracle Forum. So in order to preserve it you need to
use the {noformat}{noformat} tags.
The usage of the tag is like this.
<place your code here>\
7. If you are posting a *Performance Related Question*. Please read
 {thread:id=501834} and {thread:id=863295}.
 Following those guide will be very helpful.
8. Please keep in mind that this is a public forum. Here No question is URGENT.
 So use of words like *URGENT* or *ASAP* (As Soon As Possible) are considered to be rude.

Regular expression alphabets

Hi
I want to retrieve the data if the data contains a character or a space or '-' thru select query .
Please help me in writing the combination of 3 with regular expression.
Thanks!!

VT wrote:
Hi,
Try this
SELECT *
FROM <TABLE> WHERE REGEXP_LIKE(<COLUMN>, '[a-z -][A-Z -]');cheers
VTThat won't work as it's expecting at least two characters with the first having to be a-z (lower case) or space or "-" followed by A-Z (upper case) or space or "-".
The correct way is either:
[a-zA-Z -]or
[[:alpha:] -]using the alpha set is often preferable as it can work differently with different character sets/languages rather than restricting to just the a-zA-Z ranges.
Generating a reference for your own database characterset/language can be useful...
SQL> select level-1 as asc_code, decode(chr(level-1), regexp_substr(chr(level-1), '[[:print:]]'), CHR(level-1)) as chr,
2 decode(chr(level-1), regexp_substr(chr(level-1), '[[:graph:]]'), 1) is_graph,
3 decode(chr(level-1), regexp_substr(chr(level-1), '[[:blank:]]'), 1) is_blank,
4 decode(chr(level-1), regexp_substr(chr(level-1), '[[:alnum:]]'), 1) is_alnum,
5 decode(chr(level-1), regexp_substr(chr(level-1), '[[:alpha:]]'), 1) is_alpha,
6 decode(chr(level-1), regexp_substr(chr(level-1), '[[:digit:]]'), 1) is_digit,
7 decode(chr(level-1), regexp_substr(chr(level-1), '[[:cntrl:]]'), 1) is_cntrl,
8 decode(chr(level-1), regexp_substr(chr(level-1), '[[:lower:]]'), 1) is_lower,
9 decode(chr(level-1), regexp_substr(chr(level-1), '[[:upper:]]'), 1) is_upper,
10 decode(chr(level-1), regexp_substr(chr(level-1), '[[:print:]]'), 1) is_print,
11 decode(chr(level-1), regexp_substr(chr(level-1), '[[:punct:]]'), 1) is_punct,
12 decode(chr(level-1), regexp_substr(chr(level-1), '[[:space:]]'), 1) is_space,
13 decode(chr(level-1), regexp_substr(chr(level-1), '[[:xdigit:]]'), 1) is_xdigit
14 from dual
15 connect by level <= 256
16 /
ASC_CODE C IS_GRAPH IS_BLANK IS_ALNUM IS_ALPHA IS_DIGIT IS_CNTRL IS_LOWER IS_UPPER IS_PRINT IS_PUNCT IS_SPACE IS_XDIGIT
 0 1
 1 1
 2 1
 3 1
 4 1
 5 1
 6 1
 7 1
 8 1
 9 1 1
 10 1 1
 11 1 1
 12 1 1
 13 1 1
 14 1
 15 1
 16 1
 17 1
 18 1
 19 1
 20 1
 21 1
 22 1
 23 1
 24 1
 25 1
 26 1
 27 1
 28 1
 29 1
 30 1
 31 1
 32 1 1 1
 33 ! 1 1 1
 34 " 1 1 1
 35 # 1 1 1
 36 $ 1 1 1
 37 % 1 1 1
 38 & 1 1 1
 39 ' 1 1 1
 40 ( 1 1 1
 41 ) 1 1 1
 42 * 1 1 1
 43 + 1 1 1
 44 , 1 1 1
 45 - 1 1 1
 46 . 1 1 1
 47 / 1 1 1
 48 0 1 1 1 1 1
 49 1 1 1 1 1 1
 50 2 1 1 1 1 1
 51 3 1 1 1 1 1
 52 4 1 1 1 1 1
 53 5 1 1 1 1 1
 54 6 1 1 1 1 1
 55 7 1 1 1 1 1
 56 8 1 1 1 1 1
 57 9 1 1 1 1 1
 58 : 1 1 1
 59 ; 1 1 1
 60 < 1 1 1
 61 = 1 1 1
 62 > 1 1 1
 63 ? 1 1 1
 64 @ 1 1 1
 65 A 1 1 1 1 1 1
 66 B 1 1 1 1 1 1
 67 C 1 1 1 1 1 1
 68 D 1 1 1 1 1 1
 69 E 1 1 1 1 1 1
 70 F 1 1 1 1 1 1
 71 G 1 1 1 1 1
 72 H 1 1 1 1 1
 73 I 1 1 1 1 1
 74 J 1 1 1 1 1
 75 K 1 1 1 1 1
 76 L 1 1 1 1 1
 77 M 1 1 1 1 1
 78 N 1 1 1 1 1
 79 O 1 1 1 1 1
 80 P 1 1 1 1 1
 81 Q 1 1 1 1 1
 82 R 1 1 1 1 1
 83 S 1 1 1 1 1
 84 T 1 1 1 1 1
 85 U 1 1 1 1 1
 86 V 1 1 1 1 1
 87 W 1 1 1 1 1
 88 X 1 1 1 1 1
 89 Y 1 1 1 1 1
 90 Z 1 1 1 1 1
 91 [ 1 1 1
 92 \ 1 1 1
 93 ] 1 1 1
 94 ^ 1 1 1
 95 _ 1 1 1
 96 ` 1 1 1
 97 a 1 1 1 1 1 1
 98 b 1 1 1 1 1 1
 99 c 1 1 1 1 1 1
 100 d 1 1 1 1 1 1
 101 e 1 1 1 1 1 1
 102 f 1 1 1 1 1 1
 103 g 1 1 1 1 1
 104 h 1 1 1 1 1
 105 i 1 1 1 1 1
 106 j 1 1 1 1 1
 107 k 1 1 1 1 1
 108 l 1 1 1 1 1
 109 m 1 1 1 1 1
 110 n 1 1 1 1 1
 111 o 1 1 1 1 1
 112 p 1 1 1 1 1
 113 q 1 1 1 1 1
 114 r 1 1 1 1 1
 115 s 1 1 1 1 1
 116 t 1 1 1 1 1
 117 u 1 1 1 1 1
 118 v 1 1 1 1 1
 119 w 1 1 1 1 1
 120 x 1 1 1 1 1
 121 y 1 1 1 1 1
 122 z 1 1 1 1 1
 123 { 1 1 1
 124 | 1 1 1
 125 } 1 1 1
 126 ~ 1 1 1
 127 1
 128 Ç 1 1 1
etc.
{code}

Help in query using regular expression

HI,
I need a help to get the below output using regular expression query. Please help me.
SELECT REGEXP_SUBSTR ('PWRPKG(P/W+P/L+CC)', '[^+]+', 1, lvl) val, lvl
FROM DUAL,(SELECT LEVEL lvl FROM DUAL
CONNECT BY LEVEL <=(SELECT MAX ( LENGTH ('PWRPKG(P/W+P/L+CC)') - LENGTH (REPLACE ('PWRPKG(P/W+P/L+CC)','+',NULL))+ 1) FROM DUAL));
I need the output as
correct result:
==============
val lvl
P/W 1
P/L 2
CC 3
But i tried the above it is not coming the above result. Please help me where i did a mistake.
Thanks in advance

Frank gave you a solution in your other thread. You could simplify it if you are on 11g:
SQL> select * from table_x
2 /
TXT
TECHPKG(INTELLI CC+FRT SONAR)
PWRPKG(P/W+P/L+CC)
select txt,
 regexp_substr(
 txt,
 '(.*\()*([^+)]+)',
 1,
 column_value,
 null,
 2
 ) element,
 column_value element_number
from table_x,
 table(
 cast(
 multiset(
 select level
 from dual
 connect by level <= regexp_count(txt,'\+') + 1
 as sys.OdciNumberList
order by rowid,
 column_value
TXT ELEMENT ELEMENT_NUMBER
TECHPKG(INTELLI CC+FRT SONAR) INTELLI CC 1
TECHPKG(INTELLI CC+FRT SONAR) FRT SONAR 2
PWRPKG(P/W+P/L+CC) P/W 1
PWRPKG(P/W+P/L+CC) P/L 2
PWRPKG(P/W+P/L+CC) CC 3
SQL> SY.

Query help in regular expression

Hi all,
SELECT * FROM emp11
WHERE INSTR(ENAME,'A',1,2) >0;
Please let me know the equivalent query using regular expressions.
i have tried this after going through oracle regular expressions documentation.
SELECT * FROM emp11
WHERE regexp_LIKE(ename,'A{2}')
Any help in this regard would be highly appreciated .
Thanks,
P Prakash

please go here
Introduction to regular expressions ...
Thanks,
P Prakash

Urgent!!! Problem in regular expression for matching braces

Hi,
For the example below, can I write a regular expression to store getting key, value pairs.
example: ((abc def) (ghi jkl) (a ((b c) (d e))) (mno pqr) (a ((abc def))))
in the above example
abc is key & def is value
ghi is key & jkl is value
a is key & ((b c) (d e)) is value
and so on.
can anybody pls help me in resolving this problem using regular expressions...
Thanks in advance

"((key1 value1) (key2 value2) (key3 ((key4 value4)
(key5 value5))) (key6 value6) (key7 ((key8 value8)
(key9 value9))))"
I want to write a regular expression in java to parse
the above string and store the result in hash table
as below
key1 value1
key2 value2
key3 ((key4 value4) (key5 value5))
key4 value4
key5 value5
key6 value6
key7 ((key8 value8) (key9 value9))
key8 value8
key9 value9
please let me know, if it is not possible with
regular expressions the effective way of solving itYes, it is possible with a recursive regular expression.
Unfortunately Java does not provide a recursive regular expression construct.
$_ = "((key1 value1) (key2 value2) (key3 ((key4 value4) (key5 value5))) (key6 value6) (key7 ((key8 value8) (key9 value9))))";
my $paren;
   $paren = qr/
           [^()]+ # Not parens
         |
           (??{ $paren }) # Another balanced group (not interpolated yet)
    /x;
my $r = qr/^(.*?)$(\w+?) (\w+?|(??{$paren}))$\s*(.*?)$/;
while ($_) {
     match()
# operates on $_
sub match {
     my @v;
     @v = m/$r/;
     if (defined $v[3]) {
          $_ = $v[2];
          while (/\(/) {
               match();
          print "\"",$v[1],"\" \"",$v[2],"\"";
          $_ = $v[0].$v[3];
     else { $_ = ""; }
C:\usr\schodtt\src\java\forum\n00b\regex>perl recurse.pl
"key1" "value1"
"key2" "value2"
"key4" "value4"
"key5" "value5"
"key3" "((key4 value4) (key5 value5))"
"key6" "value6"
"key8" "value8"
"key9" "value9"
"key7" "((key8 value8) (key9 value9))"
C:\usr\schodtt\src\java\forum\n00b\regex>

Bracket in Regular Expression constant?

I am a bit puzzled by the behavior I am experiencing in LV 2011. I hope to get some light from experts out there.
I am trying to parse a messy ASCII header file and after having split it into individual lines (strings), I use the "Match Regular Expression" function to remove some of the info before the substantial information.
Some of the strings include square brackets ([, ]), which are special characters for the function, therefore, as documented in the help, one needs to precede them with a backslash.
Example:
I want to parse the following line:
#PR [PR_DEV,I,2]
One way (which I am using because of considerations related to the rest of the header) is the the following:
Note that the first string constant is using "Code Display" whereas the second one is using "Normal Display".
Why did I not put a backslash in front of the bracket in the first string, you may ask? Well, I did, but it disappeared after I typed the other characters. And reverting to "Normal Display" did not restore it.
Of course, the first version does not parse the input string correctly, whereas the second one does it fine.
In other words, the custom display string (which is convenient for cryptic codes such as \s* or to distinguish between space and tab...or simply ENTER tabs!) seems to mess up with the \[ combo (likewise with the \] one).
It is not a huge deal. I can use the "Normal Display" mode, but I tend to think that this qualifies as a hidden "feature". And again, it is still a pain in the ... when dealing with special characters such as tabs, etc...
Solved!
Go to Solution.

I think that [ is a special character which needs to be preceded by a backslash, but it is not one of the defined backslash characters (like \s). So, you need to put in two \\ to get one \ while in '\' Codes Display.
You can put in any character by using \xx where the xx is a hex character using only upper case letters for A..F. I converted the strings to byte arrays and tried to see what made the arrays match and the Match work.
Lynn

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:
http://forum.java.sun.com -
So, the URL is always preceeded by:

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....
(http:\/\/[a-zA-Z0-9.]+) [-]
i dont know the java methods to call...just the reg exp...
Sanjay Acharya

Litte help with regular expression?

Greetings all,
I have a simple regular expression "(\\w+)\\s(\\w+)\\s(.+)"
Which I want to match against the strings like "Acetobacter pasteurianus LMD22.1"
But this always fails whenever there is a dot (.) character like "LMD22.1" in above string.
How to solve this ?
Thanks in advance.

Shouldn't that be Acinetobacter?
edit: nope, I'm wrong, you're right.
Edited by: Encephalopathic on Apr 7, 2009 7:34 PM

Help regarding regular expression

HI All ,
Please see the following string
String s = "IF ((NOT NUM4 IS ALPHABETIC ) AND NUM3 IS ALPHABETIC-UPPER AND (NUM5 IS GREATER OR EQUAL TO 3) AND (NUM5 IS NOT GREATER THAN 3) AND (NUM3 GREATER THAN 46) AND (NUM5 GREATER THAN NUM3) OR NUM3 LESS THAN 78) .";
My problem is: i want to capture the part of this line which contains "ALPHABETIC ,ALPHABETIC-UPPER for ex :NOT NUM4 IS ALPHABETIC , NUM3 IS ALPHABETIC-UPPER.from that I have to capture the word num4 , num3 which are in these phrases only ;from the whole string whereever it exists along with the phrase,Can any one help me out by suggesting something.num4 and num3 are variable names

I suspect you're right, Sabre, but I can't resist...
import java.util.regex.*;
* A rewriter does a global substitution in the strings passed to its
* 'rewrite' method. It uses the pattern supplied to its constructor, and is
* like 'String.replaceAll' except for the fact that its replacement strings
* are generated by invoking a method you write, rather than from another
* string. This class is supposed to be equivalent to Ruby's 'gsub' when given
* a block. This is the nicest syntax I've managed to come up with in Java so
* far. It's not too bad, and might actually be preferable if you want to do
* the same rewriting to a number of strings in the same method or class. See
* the example 'main' for a sample of how to use this class.
* @author Elliott Hughes
public abstract class Rewriter
private Pattern pattern;
private Matcher matcher;
   * Constructs a rewriter using the given regular expression; the syntax is
   * the same as for 'Pattern.compile'.
public Rewriter(String regularExpression)
    this.pattern = Pattern.compile(regularExpression);
   * Returns the input subsequence captured by the given group during the
   * previous match operation.
public String group(int i)
    return matcher.group(i);
   * Overridden to compute a replacement for each match. Use the method
   * 'group' to access the captured groups.
public abstract String replacement();
   * Returns the result of rewriting 'original' by invoking the method
   * 'replacement' for each match of the regular expression supplied to the
   * constructor.
public String rewrite(CharSequence original)
    this.matcher = pattern.matcher(original);
    StringBuffer result = new StringBuffer(original.length());
    while (matcher.find())
      matcher.appendReplacement(result, "");
      result.append(replacement());
    matcher.appendTail(result);
    return result.toString();
public static void main(String[] args)
    String s = "IF ((NOT NUM4 IS ALPHABETIC ) " +
                "AND NUM3 IS ALPHABETIC-UPPER " +
                "AND (NUM5 IS GREATER OR EQUAL TO 3) " +
                "AND (NUM5 IS NOT GREATER THAN 3) " +
                "AND (NUM3 GREATER THAN 46) " +
                "AND NUM645 IS ALPHABETIC " +
                "AND (NUM5 GREATER THAN NUM3) " +
                "OR NUM3 LESS THAN 78 " +
                "AND NUM34 IS ALPHABETIC-UPPER " +
                "AND NUM92 IS ALPHABETIC-LOWER " +
                "AND NUM0987 IS ALPHABETIC-LOWER) .";
    String result =
      new Rewriter("(NUM\\d+) +IS +(ALPHABETIC(?:-(?:UPPER|LOWER))?)")
        public String replacement()
          String type = group(2);
          if (type.endsWith("UPPER"))
            return "Character.isUpper(" + group(1) + ")";
          else if (type.endsWith("LOWER"))
            return "Character.isLower(" + group(1) + ")";
          else
            return "Character.isLetter(" + group(1) + ")";
      }.rewrite(s);
    System.out.println(result);
}

Regular Expression ambiguous

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