PHP Regular expression parsing

Question

I am having an array of strings.

(102) Name3

How can I match for strings starting with (####) and get the Name part of the line easily. I am trying the following which is not working.

if(preg_match("/(d+)/", $myArray[$i], $matches))

Answer 1

You need to escape the parentheses that are in the expected input string:

if(preg_match("/\((\d+)\) (.+)$/", $myArray[$i], $matches))

That's the following regular expression:

• / Start of regex (this can be any character, but is usually a /)
• \( the character (
• ( begin capturing group 1
• \d any digit
• + previous match, 1 or more times
• ) end capturing group 1
• a space
• ( begin capturing group 2
• . almost any character
• + previous match, 1 or more times
• ) end capturing group 2
• $ end of string
• / End of regex (must match first character)

You can find more descriptive definitions on regular-expressions.info.

With the above, matches will contain for example:

array(
    '(123) input string',
    123, // capturing group 1
    'input string' // capturing group 2
)