how to find string like "1. " with regex

[Anon]

hello,

I'm trying to parse the return from a dictionary website. I've managed to strip the HTML, script, and such using preg_replace(), but I can't figure out how to replace the entry numbers, i.e. the one and the period AND the space in "1. To go forth and program" I've tried ([1-9]+.[\s]) and ([1-9]+.\s), but that removes everything. Anyone have a solution?

Thanks muchly,
Biggus

[Anon]

Try this ...

$string = "1. To go forth and program. 2. To go forth and program.";
$string = preg_replace("/[1-9]. /i",'Replaced', $string);
echo $string;

It will produce

ReplacedTo go forth and program. ReplacedTo go forth and program.

[Anon]

Thanks, that works great! The only thing is my client will be looking up things which would have numbers (i.e. 11.37) in the definition. I changed the regexp to the following:

"/[0-9]{1,2}. /i" because chances are slim ther'll be more than 99 definitions

the to look for non-numbers, a decimal, and then one or 2 numbers:

"/[a-z]+.*[0-9]{1,2}. /i"

which works, but it drops the last word of each sentence, as you would expect, and leaves the decimals alone.

The text, after stripping the HTML etc looks like: "1. To go forth and program.2. To go forth and program." with no space after each definition. That's the kicker.

Any ideas? (Sorry for the long message)

Thanks again,
Biggus