Remove everything after 1st opening bracket incl. the opening bracket - in a string - PHP

Question

I am pulling location off an API.

The variable $location sometimes contains:

Paris
New York (Manhattan)

Paris is OK.

I only want the city name. So, New York (Manhattan) should be New York.

How do I remove everything after the 1st bracket (and including the 1st bracket) in variable $location?

And will the code that does that affect the content of $location if it only contains 1 word (e.g. Paris)?

Thank you.

score 2 · Answer 1 · edited Aug 28 '15 at 03:22

2

$pos = strpos($location, "(");

if ($pos) $location = substr($location, 0, $pos);

What about this?

edited Aug 28 '15 at 03:22

Jonny 5

12,171
2
25
42

answered Aug 27 '15 at 23:15

Fappie.

482
6
18

3

And why not `$location = trim(explode('(', $location)[0]);`? In this way you don't need to test anything. – Casimir et Hippolyte Aug 27 '15 at 23:24

score 1 · Accepted Answer · answered Aug 27 '15 at 23:13

1

You can use a regex like below and use an empty replacement string:

\(.*

Working demo

$re = "/\\(.*/"; 
$str = "Paris New York (Manhattan)\n\nParis is OK. I only want the city name. So, New York (Manhattan) should be New York.\n"; 

$result = preg_replace($re, "", $str);

Check the substitution panel below:

answered Aug 27 '15 at 23:13

Federico Piazza

30,085
15
87
123

This worked for me. I then used rtrim to remove the extra space it left (from the space between the word and the bracket). Awesome. Thx. – Cody Raspien Aug 27 '15 at 23:28
@CodyRaspien glad to help. Btw, you can also use `$re = "/\\s*\\(.*/";` to remove the blanks too. – Federico Piazza Aug 27 '15 at 23:29

Remove everything after 1st opening bracket incl. the opening bracket - in a string - PHP

2 Answers2