regular expressions

safra

Hi,

I need to remove part of a string. The part starts with a space and then any (unknown) contents between brackets.

For example:

$example = "This is an example (contents)";

should become:

$example = "This is an example";

How can this be done?

Thanks,
Raoul

stolzyboy

http://www.php.net/manual/en/function.ereg-replace.php

safra

I realize I have to use something like ereg_replace. The problem is how to set the regular expression to match the space and then brackets with contents in between?

I tried things like:
\s(^>)
\s(.)
\s([^>.]*)

Can't get it to work.

laserlight

This seems to work for me:

$example = ereg_replace("(.*)", "", $example);

safra

Thanks laserlight!

it does remove the brackets and the contents, the only thing now is to also remove the whitespace before the bracket.

I tried:

" (.*)"

just a space at the beginning of the expression, and it works. But I wonder if this is correct or if there is a more proper way?

laserlight

I suspect that this might be better:

$example = ereg_replace(" ?(.*)", "", $example);

which would match and remove if there's 1 space or no space before the opening parenthesis.

Then again you might use
$example = ereg_replace(" (.)", "", $example);

which should match and remove if there's 0 or more spaces before the opening parenthesis.

safra

thanks again Laserlight!