从字符串 php 正则表达式获取内容

Question

我有下一个字符串：

$string = '["["123456","NAME","1","INFORMATION","15/12/2015","","","0","OTHER ATTRIBUTE","",""]","["["123", "OTHER"]"';

我想提取双引号之间的字母数字信息，我试过这个答案

Regex to get content between single and double quotes php

但输出是这样的，

[0]=> array(12) { [0]=> string(3) ""["" [1]=> string(3) "","" [2]=> string(3) "","" [3]=> string(3) "","" [4]=> string(3) "","" [5]=> string(3) "","" [6]=> string(3) "","" [7]=> string(3) "","" [8]=> string(3) "","" [9]=> string(3) "","" [10]=> string(3) "","" [11]=> string(3) ""]"" } [1]=> array(12) { [0]=> string(1) """ [1]=> string(1) """ [2]=> string(1) """ [3]=> string(1) """ [4]=> string(1) """ [5]=> string(1) """ [6]=> string(1) """ [7]=> string(1) """ [8]=> string(1) """ [9]=> string(1) """ [10]=> string(1) """ [11]=> string(1) """ } [2]=> array(12) { [0]=> string(1) "[" [1]=> string(1) "," [2]=> string(1) "," [3]=> string(1) "," [4]=> string(1) "," [5]=> string(1) "," [6]=> string(1) "," [7]=> string(1) "," [8]=> string(1) "," [9]=> string(1) "," [10]=> string(1) "," [11]=> string(1) "]" }

我怎样才能做到这一点？

Answer 1

您的输入 "["123456"... 与 link you mentioned 不同。而且这里没有单引号。

一个想法是要求\w word character 在模式中至少有一个[A-Za-z0-9_]

if(preg_match_all('/"([^"]*?\w[^"]*)"/', $string, $out) > 0)
  print_r($out[1]);

• [^"]* 匹配任意数量的字符，即not "。 * 之后的 ? 变为 lazy。
• 在$1 中查找第一个capture group（括号括起来的组）捕获的匹配项。

See demo and more explanation at regex101 或 PHP demo at eval.in

Answer 2

用 preg_match_all 试试这个 regey：

\"[^\"\,\[\]]+\"

你可以在这里测试它：http://www.phpliveregex.com/p/eKH