More efficient way to extract number from string

Hello guys,
I am using this Regexp to extract numbers from a string, and I doubt that there is a more efficient way to get this done:
SELECT regexp_replace (regexp_replace ( REGEXp_REPLACE (' !@#$%^&*()_+= '' + 00 SDFKA 324 000 8702 234 | " ' , '[[:punct:]]',''), '[[:space:]]',''), '[[:alpha:]]','') FROM dual
{code}
Is there a more efficient way to get this done ?
Regards,
Fateh

Or, with less writing, using Perl syntax \D (non-digit):
SELECT regexp_replace(' !@#$%^&*()_+= '' + 00 SDFKA 324 000 8702 234 | " ','\D')
FROM dual
REGEXP_REPLACE(
003240008702234
SQL>
{code}
SY.

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you can do this in the Results section with a computed column using the native string functions. to break it down i have split it up a bit.
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Need a way to extract substring from string in oracle

Hi all,
I have one requirement related to extracting string from a paramater.
suppose the string may like this in various format
string:= 'This my string <Rid//problem/123456>'
or
string:= '<Rid//problem/123456> This my string'
or
string:= ' This is <Rid//problem/123456> my string'
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Thanks n regards
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Hi,
What version of Oracle ?
How do you delimit the string to extract ?
- always between the last / and before the >
- the last string mande of number ?
- the first string made of number ?
Here are 3 possible answers :SQL> with s as (
2 select 'This my string <Rid//problem/123456>' s from dual
3 union all select '<Rid//problem/123456> This my string' from dual
4 union all select ' This is <Rid//problem/123456> my string' from dual
5 )
6 select s.s,
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8 regexp_replace(s,'^[^0-9]*(.*?)[^0-9]*$','\1') r2,
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10 from s ;
S                                        R1       R2       R3
This my string <Rid//problem/123456>     123456   123456   123456
<Rid//problem/123456> This my string     123456   123456   123456
This is <Rid//problem/123456> my string 123456   123456   123456If you're on 10g or more...
Of course it also works in PL/SQL :SQL> l
1 declare
2 r1 varchar2(50) := 'This my string <Rid//problem/123456>';
3 r2 varchar2(50) := '<Rid//problem/123456> This my string';
4 r3 varchar2(50) := ' This is <Rid//problem/123456> my string';
5 r1b varchar2(50);
6 r2b varchar2(50);
7 r3b varchar2(50);
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10 r2b := regexp_replace(r2,'^[^0-9]*(.*?)[^0-9]*$','\1') ;
11 r3b := regexp_replace(r3,'^.*/(.*?)>.*$','\1') ;
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14 dbms_output.put_line('3 : '||r3||' -> '||r3b);
15* end;
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1 : This my string <Rid//problem/123456> -> 123456
2 : <Rid//problem/123456> This my string -> 123456
3 : This is <Rid//problem/123456> my string -> 123456
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I know this is a simple formula, but I can't quite get it.

Are the two alpha and remaining numeric characters always in the same position within the string?
Are the two alpha characters always going to be the same?
Is the length of the numeric value following the two alpha characters always going to be the same?
The more no answers you have to the above the tougher it will be!
Get back with the above answers and possibly copy and paste some examples of the ENTIRE strings from which you will be extracting the values and let us know which are the values you'll be pulling from within the larger string.
"I need to extract the number, if it exists, from an alphanumeric text string. I am told that the first two characters will be alpha and the remaining character(s) will be numeric, if the numeric exists at all.
I know this is a simple formula, but I can't quite get it."

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Hi, all
If I want to compare a requirement number and extract a number form a continuous data, . for example,
the string like this ( ******** 123 ******
*********456 ******
*********789*******)
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What if no element is greater?
What if there are multiple solutions?
What if the string does not contain any numbers?
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LabVIEW Champion . Do more with less code and in less time .
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May be I don't know how to use the following:
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===
String pattern = "###.###";
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DecimalFormat myFormatter = new DecimalFormat(pattern);
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Hi
You made the best choice. JRegex is closer to java.util.regex.* and fine works with excellent performance in any JVM version so you can use examples/tutorials from both JRegex and Java.Sun.Com to learn more about regular expression.
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How to use JRegex? no secrets!
import jregex.*;
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String patternString ="^([+-]?(?:\\d+(?:\\.\\d*)?|\\.\\d+)(?:[eE][+-]?\\d+)?)\s+(.*)$";
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// below is identical in both packages
Matcher m = pattern.matcher(anyTargetString);
if (m.matches()) {
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String scratch = m.group(2);
// your stuff goes here
..Remember to put JRegex jarfile in classpath!
Regards and Success.

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is there a way to extract data from one of e.g. 7 rows from the multicolumn listbox?

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