Question

我一直试图让这个正则表达式工作。假设解析URL，如果找到字符串'_skipThis'，则不匹配该字符串。还需要反向引用。例如：

String 1: a_particular_part_of_string/a/b/c/d/e/f
Result: preg_match should return true
Backreference: $1 -> a_particular_part_of_string, $2 -> /a/b/c/d/e/f

String 2: a_particular_part_of_string_skipThis/a/b/c/d/e/f
Result: preg_match should return false.
Backreference: nothing here.

我试过以下正则表达式..

reg1 = ([a-zA-Z0-9_]+)(\/.*)
reg2 = ([a-zA-Z0-9]+(?!_skipThis))(\/.*)
reg3 = ((?!_skipThis).*)(\/.*)
reg4 = ((?!_skipThis)[a-zA-Z0-9_]+)(\/.*)

请帮帮我！在此先感谢!!!

Answer 1

只需匹配_skipThis，如果找到则返回 false 。

 if (strpos($theString, '_skipThis') === false) {
        // perform preg_match
 } else 
        return false;

（当然有一个正则表达式。假设_skipThis仅出现在第一个/之前，

 return preg_match('|^([^/]+)(?<!_skipThis)(/.*)$|', $theString); 
 //                   -------              -----
 //                     $1   --------------  $2
 //                          Make sure it is *not* preceded '_skipThis'

否则，如果需要避免出现在任何地方的_skipThis，

 return preg_match('|^(?!.*_skipThis)([^/]+)(/.*)$|', $theString);
 //                   ---------------
 //                   Make sure '_skipThis' is not found anywhere.

Answer 2

试试这个：

$str1 = 'a_particular_part_of_string/a/b/c/d/e/f'; //1
$str2 = 'a_particular_part_of_string_skipThis/a/b/c/d/e/f'; //0
$keep = 'a_particular_part_of_string';
$skip = '_skipThis';

$m = array();
if(preg_match("/($keep)(?!$skip)(.*)$/", $str1, $m))
    var_dump($m);
else
    echo "Not found\n";

$m = array();
if(preg_match("/($keep)(?!$skip)(.*)$/", $str2, $m))
    var_dump($m);
else
    echo "Not found\n";

输出：

array(3) {
  [0]=>
  string(39) "a_particular_part_of_string/a/b/c/d/e/f"
  [1]=>
  string(27) "a_particular_part_of_string"
  [2]=>
  string(12) "/a/b/c/d/e/f"
}

Not found

PHP Preg_match：尝试匹配字符串中不存在子字符串的字符串

2 个答案: