PHP Preg_match：尝试匹配字符串中不存在子字符串的字符串

我一直在努力让这个正则表达式工作。假设要解析URL，如果找到字符串“\u skipThis”，则与字符串不匹配。此外，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.

我试过下面的正则表达式

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

---

请帮帮我！提前感谢

只需匹配

\u跳过此

，如果找到则返回false

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

（当然，这里有一个正则表达式。假设

\u skipt这个

只出现在第一个

/

之前

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

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

---

非常感谢您的帮助。字符串的“特定部分”必须始终保持不同。谢谢！

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