Question

I want to create an array from a string that contains brackets like {! !}. However the whitespaces at the beginning and at the end of the encapsulated string should not be displayed.

$string = "{! This should be in the output !} this should not be in the output {!show_in_output!} don't show {!   show   !}";
preg_match_all("/{!(.*)!}/Us", $string , $output);

The resulting array looks like this:

Array
(
    [0] => Array
        (
            [0] => {! This should be in the output !}
            [1] => {!show_in_output!}
            [2] => {!   show   !}
        )

    [1] => Array
        (
            [0] =>  This should be in the output 
            [1] => show_in_output
            [2] =>    show   
        )

)

But it should look like this:

Array
(
    [0] => Array
        (
...
        )

    [1] => Array
        (
            [0] => This should be in the output 
            [1] => show_in_output
            [2] => show   
        )

)

Is there a way to achieve this with a modified regex?
Thank you!

Answer 1

The (.*) in the middle of /{!(.*)!}/ matches any characters between your {! and !}. If you want to NOT capture spaces before and after that, you have to match whitespace and not include the whitespace in your group, so in your case: /{!\s*(.*?)\s*!}/. The ? says to make a minimal match of the .* so that it doesn’t include the whitespace that you want matched by the second \s*.