PHP/Regex如何獲取此字符串的值？

Question

你好，我敢肯定，我可以找到誰可以解決這個..

我希望得到「2013年4月21日」數值超出此字符串：提前

$string = "Issue Date: Sunday April 21, 2013/week 10/2003week 11/2003week 12/2003week 13.." 

感謝。 ..

Answer 1

使用preg_match()有以下的正則表達式/^Issue Date: [A-z]+ (.*) \//：

// prepare string and pattern 
$string = 'Issue Date: Sunday April 21, 2013/week 10/2003week 11/2003week 12/2003week 13..'; 
$pattern = '/^Issue Date: [A-z]+ (.*) \//'; 

// exec regex and check for results 
if(!preg_match($pattern, $string, $matches)) { 
    die('the pattern was not found'); 
} 

// output the results: 
$date = $matches[1]; 
echo $date; // output: 'April 21, 2013' 

Answer 2

如果你（在 「簽發日期」）找什麼首先出現，那麼這將工作：

Issue Date: (.*) /

Answer 3

無需正則表達式：|：（。*？）

$string = 'Issue Date: Sunday April 21, 2013/week 10/2003week 11/2003week 12/2003week 13..'; 
$array = explode(' ',$string); 
$date = $array[3] . ' ' . $array[4] . ' ' . $array[5]; 
echo $date;