How to define a regular expression using regular expressions

Hi,
I am looking for some regular expression pattern which will identify a regular expression.
Also, is it possible to know how does the compile method of Pattern class in java.util.regex package work when it is given a String containing a regex. ie. is there any mechanism to validate regular expression using regular expression pattern.
Regards,
Abhisek

I am looking for some regular expression pattern which will identify a regular
expression. Also, is it possible to know how does the compile method of
Pattern class in java.util.regex package work when it is given a String
containing a regex. ie. is there any mechanism to validate regular
expression using regular expression pattern.It is impossble to recognize an (in)valid regular expression string using a
regular expression. Google for 'pumping lemma' for a formal proof.
kind regards,
Jos

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==========
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How to get the below output using REGULAR EXPRESSIONS??
SQL> ed
Wrote file afiedt.buf
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SQL> /
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Hi,
user12240205 wrote:
... Frank, ʃʃp's method, I understand. But your method, although correct, I find it difficult to understand.
Could you explain how you did this?? What does '.*(\d{5})\D*' and '\1' mean???
Your method is better because it uses only ONE reg expression function. ʃʃp's uses 2.In Oracle 10.2 (I believe) and higher, '\d' is equivalent to '[[:digit:]]', and '\D' is equivalent to '[^[:digit:]]'. I find '\d' and '\D' easier to type, but there's nothing wrong with using '[[:digit:]]' and '[^[:digit:]]'.
'.*' means "0 or more of any character".
'\D*' means "0 or more non-digits".
The whole expression, '.*(\d{5})\D*' means:
a. 0 or more characters (any characters)
b. 5 digits
c. 0 or more non-digits.
'\1' is a Backreference . It means the sub-string that matched the pattern after the 1st '(', up to (but not including) its matching ')'. In this case, that means the sub-string that matched '\d{5}', or b. using the explanation immediately above.
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Hi All,
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:(?:(?:\r\n)?[ \t])+|\Z|(?=[\["()<>@,;:\\".\[\]]))|\[([^\[\]\r\\]|\\.)*\](?:(?
:\r\n)?[ \t])*))*\>(?:(?:\r\n)?[ \t])*)|(?:[^()<>@,;:\\".\[\] \000-\031]+(?:(?
:(?:\r\n)?[ \t])+|\Z|(?=[\["()<>@,;:\\".\[\]]))|"(?:[^\"\r\\]|\\.|(?:(?:\r\n)?
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(?:[^()<>@,;:\\".\[\] \000-\031]+(?:(?:(?:\r\n)?[ \t])+|\Z|(?=[\["()<>@,;:\\".
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]]))|"(?:[^\"\r\\]|\\.|(?:(?:\r\n)?[ \t]))*"(?:(?:\r\n)?[ \t])*)*\<(?:(?:\r\n)
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(?:[^()<>@,;:\\".\[\] \000-\031]+(?:(?:(?:\r\n)?[ \t])+|\Z|(?=[\["()<>@,;:\\".
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How to fetch substring using regular expression

Hi,
I am new to using regular expression and would like to know some basic details of how to use them in Java.
I have a String example= "http://www.google.com/foobar.html#*q*=database&aq=f&aqi=g10&fp=c9fe100d9e542c1e" and would like to get the value of "q" parameter (in bold) using regular expression in java.
For the same example, when we tried using javascript:
match = example.match("/^http:\/\/(?:(?!mail\.)[^\.]+?\.)?google\.[^\?#]+(?:.*[\?#&](?:as_q|q)=([^&]+))?/i}");
document.write('
' + match);
We are getting the output as: http://www.google.com/foobar.html#q=database,*database* where the bold text is the value of "q" parameter.
In Java we are trying to get the value of the q parameter separately or atleast resembles the output given by JavaScript. Please help me resolving this issue.
Regards
Praveen

BalusC wrote:
Regex is a cumbersome solution for fixed patterns like URL's. String#substring() in combination with String#indexOf would most likely already suffice.I usually agree, although, in this case, finding the exact parameter might be difficult without a small regex, perhaps:
"\\wq=\\s*"in conjunction with Pattern/Matcher, used similarly to an indexOf() to find the start of the parameter value.
Winston

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Hello,
I have a query,
select partition_name from dba_tab_partitions where table_owner='xxx'and num_rows <>0 and table_name = 'xxx';
P5
P6
P7
P12
P13
P14
P17
P18
P19
P20
P24
How can I use regular expression in above SQL query to get result without letter 'P', like..
5
6
7
12
13
14
17
18
19
20
24
thank you

I find answer...
select regexp_replace(partition_name,'P','')
thanks anyway

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