How to use multiple REGEX on one variable?

Question

I have a function which is containing multiple REGEX. Here is my function:

function generate($str) {
  $str = preg_replace(/\*\*(.*?)\*\*/, "<b>$1</b>", $str);
  $str = preg_replace(/__(.*?)__/, "<u>$1</u>", $str);
  $str = preg_replace(/\*(.*?)\*/, "<i>$1</i>", $str);
  $str = preg_replace(/--(.*?)--/, "<del>$1</del>", $str);
  $str = preg_replace(/`(.*?)`/, "<code>$1</code>", $str);
  return $str;
}

Now I want to know is this ^ the standard way? Overwriting several times on one variable? Is there any better approach?

RTM: php.net/manual/en/function.preg-replace.php You can do everything in one call — Rizier123
– Rizier123, Commented Dec 22, 2015 at 14:23
@Rizier123 Do you mean something like what the followed answer said? — stack
– stack, Commented Dec 22, 2015 at 14:36

Ben Cook · Accepted Answer · 2015-12-22 14:40:37Z

2

The documentation for preg_replace says that the first two parameters can be arrays or strings. If they are arrays, each match in the first array will be replaced by the corresponding string in the second array. E.g.:

function generate($str) {
  return preg_replace(
    array(
      '/\*\*(.*?)\*\*/',
      '/__(.*?)__/',
      '/\*(.*?)\*/',
      '/--(.*?)--/',
      '/`(.*?)`/'
    ),
    array(
      "<b>$1</b>",
      "<u>$1</u>",
      "<i>$1</i>",
      "<del>$1</del>",
      "<code>$1</code>"
    ),
    $str
  );
}

edited Dec 22, 2015 at 14:40

answered Dec 22, 2015 at 14:29

Ben Cook

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2 Comments

F. Müller Over a year ago

Good solution. return preg_replace(...) should be enough here.

Ben Cook Over a year ago

Yes, that's a good simplification. I edited the answer to reflect this.

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