Regular expression to replace <

Question

I have the data below,Is any way to write common regular expression to get desired output in oracle using regexp_replace function.

<Tier><grade><><sdlc><17,10><>  : result should be 17.10
<><sdlc><16,909312> :   16.909312
<><sdlc><11396,87> :11396.87
<20121217>        :20121217
<UNIT><6086>  : 6086
<Tier1><><sdlc><0,47> :0.47

Thanks Reimeus,But i want it using regular expression,that is the limitation i have it. — user1726550, Dec 18 '12 at 22:34
Is your data XML ? if so, better read this- http://stackoverflow.com/questions/1732348/regex-match-open-tags-except-xhtml-self-contained-tags/1732454#1732454 — A.B.Cade, Dec 18 '12 at 22:43
@shellter His question is quite clear. `<><17,10><>` convert to `17.10` — Smit, Dec 18 '12 at 22:49
@shellter It happens cuz we just took brief look and miss details. Happens to best of us. ;-} — Smit, Dec 18 '12 at 22:54

score 2 · Answer 1 · answered Dec 18 '12 at 22:35

2

You could you a simple regex like below:

Steps:

Replace , witha .
use regex to find a match

Code:

String s="<Tier><grade><><sdlc><17,10><>";
            s = s.replace(',', '.');
            Pattern p = Pattern.compile("\\d+\\.*\\d+");
            Matcher m = p.matcher(s);
            if(m.find()){
            System.out.println(m.group());// 17.10
        }

answered Dec 18 '12 at 22:35

PermGenError

45,977
8
87
106

thanks,How it works for other fields :<><16,909312> : 16.909312 <><11396,87> :11396.87 <20121217> :20121217 <6086> : 6086 <><0,47> :0.47 – user1726550 Dec 18 '12 at 23:09
apply the same regex for other aswell. – PermGenError Dec 18 '12 at 23:09
how to use this in oracle regexp_replace function? – user1726550 Dec 18 '12 at 23:13
hmm, i am not quite sure how to do this in oracle as i am not familiar with oracle's regex engine, i suggest you to update thequestion as how to parse this string using oracle regex, you'd get more help. – PermGenError Dec 18 '12 at 23:16

Regular expression to replace <

1 Answers1