Regular Expressions

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jack.smith.sam@gmail.com - 07 Oct 2006 13:54 GMT

Hi All,

I have a string which can be in these two formats : "word" or "word"
( may or may not be present)
I used the following regular expression to extract word (word is
similar to email ID and may containt letter,-,_):

[]*([\\w-_]+)

but for some words like "best' it outputs "est" and removes the "b". I
think [] means < or b or >.
Can you help me solving this issue(preferably without using additional
parentheses)?
Thanks a lot.

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Pielmeier Markus - 07 Oct 2006 14:03 GMT

Hi Jack,

On Sat, 07 Oct 2006 05:54:17 -0700, jack.smith.sam wrote:

> I have a string which can be in these two formats : "word" or "word"
> ( may or may not be present)
> I used the following regular expression to extract word (word is
> similar to email ID and may containt letter,-,_):
>
> []*([\\w-_]+)

Try this: "(?:)?([\\w-_]+)"

> but for some words like "best' it outputs "est" and removes the "b". I
> think [] means < or b or >.

The problem is, that the [] parenthesis treats the characters not as
string, but as individual characters which can appear individially.

Greetings,
Markus

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Martin Gerner - 07 Nov 2006 22:55 GMT

jack.smith.sam@gmail.com wrote in news:1160225657.635244.130570
@m7g2000cwm.googlegroups.com:

> I have a string which can be in these two formats : "word" or "word"
> ( may or may not be present)
[quoted text clipped - 7 lines]
> Can you help me solving this issue(preferably without using additional
> parentheses)?

Another, perhaps simpler, alternative is of course to solve it with indexOf
and substring. It takes a bit more space in your code, though.

String extractedString = null;
if (myString.indexOf("") == 0)
extractedString = myString.substring(3);
else
extractedString = myString;

Signature

Martin Gerner

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Lew - 12 Nov 2006 16:24 GMT

> Another, perhaps simpler, alternative is of course to solve it with indexOf
> and substring. It takes a bit more space in your code, though.
[quoted text clipped - 4 lines]
> else
> extractedString = myString;

Why the "throwaway" initialization of extractedString = null?

- Lew

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Lars Enderin - 12 Nov 2006 17:23 GMT

Lew skrev:

>> Another, perhaps simpler, alternative is of course to solve it with
>> indexOf and substring. It takes a bit more space in your code, though.
[quoted text clipped - 5 lines]
>
> Why the "throwaway" initialization of extractedString = null?

I prefer

String extractedString = (myString.indexOf("") == 0 ?
myString.substring(3) : myString);

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Regular Expressions

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