[RESOLVED] preg_replace without replacing qualifiers

[bookbuyer2000]

How would I use preg_replace to replace everything between two qualifiers without replacing the qualifiers themselves? In the following example, I'm wanting to replace everything between the "x" and the "p" but what happens is that the "x" and the "p" also get replaced:

preg_replace('/(x).*?(p)/',......,'example')

will output: e......le
instead of: ex......ple

Thank you 🙂

[bradgrafelman]

The x and the p are part of the expression you're matching, so either a) include the matched parts in the replacement (e.g. "$1...$2") or, better yet, don't even match them in the first place.

In your example, I would do a positive lookbehind and lookahead to locate the 'x' and the 'p' (respectively) without including them in the 'matched' part of the expression.

[bookbuyer2000]

Thank you for the super quick reply 🙂
I didn't understand your first suggestion (I'm new to regex), so I tried your second suggestion which produces the desired result:
preg_replace('/(?<=a).*(?=p)/','......','example'); // prints exa......le

I am curious about your first suggestion? What would be the line of code to implement the $1 or $2 ??

Thank you. 🙂

[bradgrafelman]

bookbuyer2000;10960276 wrote:

I am curious about your first suggestion? What would be the line of code to implement the $1 or $2 ??

Since the 'x' is in the first matched sub-group, '$1' in the replacement string will be 'x', and the same logic applies for the 'p'.

[bookbuyer2000]

I think I understand.

preg_replace('/(x).(p)/',"......",'example') produces e......le
preg_replace('/(x).(p)/',"$1......$2",'example') produces ex......ple

Thank you!!

[bradgrafelman]

Glad I could help; don't forget to mark this thread resolved (if it is) using the link on the Thread Tools menu above.