Question

这并非完全有可能,但希望令人感到惊讶。

我有这样的定期表述:

$pattern =  #post/date/(dddd)-(dd)-(dd)# ;

而且,我甚至有这样的说法:

$string =  post/date/2012-01-01 ;

当然,我不知道确切的格局和前面的描述,但他们将照此看。

在结束发言时,我需要看到这样一个阵列:

$groups = array( 2012 ,  01 ,  01);

阵列中应包含与括号内三个经常表达团体相对应的部分。这是可能的吗?

Answer 1

父母之间的这些事情是父辈,如果你提出申请,你可以取得结果。您可使用“前言”

$string =  post/date/2012-01-01 ;
$pattern =  #post/date/(dddd)-(dd)-(dd)# ;
preg_match_all($pattern, $string, $matches);
$groups = array($matches[1][0], $matches[2][0],$matches[3][0]);
echo  <pre> ;
print_r($groups);
echo  </pre> ;

sure, this is an example that just shows the behavior, you will need to check first if the string matched the pattern and if it did, how many times.. you can see that piece of code working here: http://ideone.com/zVu75

Answer 2

Not a regular expression but will do what you want to achieve

<?php
$string =  post/date/2012-01-01 ;
$date= basename($string);
$array=explode( - ,$date);
print_r($array);
?>

产出阵列将照此办理。

Array
(
    [0] => 2012
    [1] => 01
    [2] => 01
)

Answer 3

$str =  post/date/2012-01-01 ;
preg_match( #post/date/(d+)-(d+)-(d+)# , $str, $m);
array_shift($m);
$group = $m;
print_r($group);

产出

Array
(
    [0] => 2012
    [1] => 01
    [2] => 01
)

Answer 4

请回顾一下:

$matches = sscanf($string,  %*[^0-9]%d-%d-%d );

http://www.un.org。

array(3) {
  [0]=> int(2012)
  [1]=> int(1)
  [2]=> int(1)
}

友情链接