Regex to capture everything before first optional string

Question

I want to capture a pattern upto but not including the first instance of an optional other pattern with preg_match, eg:

ABCDEFGwTW$%                         | capture ABCD
@Q%HG@H%hg afdgwsa g   weg#D DEFG    | capture @Q%HG@H%hg afdgwsa g   weg#D D
@Q%HDEFG@H%hg afdgwsa g   weg#D DEFG | capture @Q%HD

So in the above case anything before the first instance of the string EFG is captured. also, if the EFG string is not present then I want to capture the whole string.

I would have thought that the following would work, but no such luck:

$pattern = '/(.*)(?:EFG)?/';
preg_match($pattern, 'Q$TQ@#%GEFGw35hqb', $matches);
print_r($matches);
//should give: 'Q$TQ@#%G'

Answer 1

You can use

'/(.*?)(?=EFG|$)/'

Answer 2

Try this: (.*?)(?:EFG|$)

This will match any character (as few as possible) until it finds EFG.