Preg_match_all returning array within array?

问题

I am trying to get the information out of this array, but for some reason it is nesting everything into $matches[0].

<?

$file = shell_exec('pdf2txt.py docs/April.pdf');

preg_match_all('/.../',$file,&$matches);
print_r($matches)

?>

Is this working as intended? Is there a way to put this in an array of depth 1?

EDIT:

This is the RegEx:

([A-Z][a-z]+\s){1,5}\s?[^a-zA-Z\d\s:,.\'\"]\s?[A-Za-z+\W]+\s[\d]{1,2}\s[A-Z][a-z]+\s[\d]{4}

回答1:

preg_match_all() always returns an array (if successful, otherwise you get an empty array) where index 0 contains an array with an element for each entire match, and the other indexes become the capturing groups, with an internal array for each match.

This might be easier to understand...

array(2) {
  [0]=>
  array(2) {
    [0]=>
    string(12) "entire match"
    [1]=>
    string(32) "entire match matched second time"
  }
  [1]=>
  array(2) {
    [0]=>
    string(15) "capturing group"
    [1]=>
    string(35) "capturing group matched second time"
  }
}

回答2:

If your objective is to obtain only the captured characters (what was captured by your "([A-Z][a-z]+\s){1,5}") you should look inside $matches[1]. $matches[1][0] contains the first captured character sequence.

Per the preg_match_all docs, if no order flag is specified (as in your example), PREG_PATTERN_ORDER is assumed. Using this pattern, you'll find that $matches[0] is an array, which contains all strings that matched your full pattern, and $matches[1] contains an array of strings captured by your regex.

来源：https://stackoverflow.com/questions/6054442/preg-match-all-returning-array-within-array

标签

php

regex

preg-match-all