Question

我不知道我在哪里看到它，但是有人能告诉我如何使用php和regex完成这个吗？

'this is a string "that has quoted text" inside.'

我希望能够像这样爆炸它

[0]this
[1]is
[2]a
[3]string
[4]"that has quoted text"
[5]inside

保持报价不变。

Answer 1

需要PHP＆gt; = 5.3.0

$str = 'this is a string "that has quoted text" inside';

$x = str_getcsv($str,' ','"');

var_dump($x);

这会删除引号，但确实会保留引用块的内容。

Answer 2

您可以尝试以下代码：

$str = 'this is a string  "that has quoted text" inside.';
var_dump ( preg_split('#\s*("[^"]*")\s*|\s+#', $str, -1 , PREG_SPLIT_DELIM_CAPTURE | PREG_SPLIT_NO_EMPTY) );

Output: 
array(6) {
  [0]=>
  string(4) "this"
  [1]=>
  string(2) "is"
  [2]=>
  string(1) "a"
  [3]=>
  string(6) "string"
  [4]=>
  string(22) ""that has quoted text""
  [5]=>
  string(7) "inside."
}

以下是above working code on dialpad

的链接

更新：如需转发支持，请尝试：

preg_split('#\s*((?<!\\\\)"[^"]*")\s*|\s+#', $str, -1 , PREG_SPLIT_DELIM_CAPTURE | PREG_SPLIT_NO_EMPTY);

Answer 3

这适用于regexpal.com，带有您的示例字符串：

((".*?")|([\S]*))

Answer 4

如果您不一定需要正则表达式，也可以使用strtok来标记字符串。有关示例，请参阅此tokenizedQuoted function in the comments on the strtok manual page和my enhancement of that tokenizedQuoted function。

Answer 5

这需要回顾并展望未来......

尝试类似：

preg_split('/(?<!(".+)) (?!(.+"))/', $str, -1 ,PREG_SPLIT_NO_EMPTY);

[未经测试]

将字符串分解为标记，保持引用的substr完整

5 个答案:

更新：如需转发支持，请尝试：