Question

我有像：

这样的字符串

postGeneratePasswordToken
putUsers_VAL_Calculations_NUMBER _
deleteUsers_VAL_Clients_ID _
getDictionary_VAL _

我期待的结果是数组（上面的例子中的数组），其元素如下所示：

发布，生成，密码，令牌
获取，用户，_VAL_，计算，_NUMBER _
删除，用户，_VAL_，客户端，_ID _
get，Dictionary，_VAL _

下面是我的代码：

$regex = '#(post|get|put|delete)(([A-Z][a-z]+)|(_[A-Z]{2,}_))+#';

$nameArray = preg_match_all($regex, $methodName, $matches);

和示例输出

array(5) {
  [0]=>
  array(1) {
    [0]=>
    string(33) "putUsers_VAL_Calculations_NUMBER_"
  }
  [1]=>
  array(1) {
    [0]=>
    string(3) "put"
  }
  [2]=>
  array(1) {
    [0]=>
    string(8) "_NUMBER_"
  }
  [3]=>
  array(1) {
    [0]=>
    string(12) "Calculations"
  }
  [4]=>
  array(1) {
    [0]=>
    string(8) "_NUMBER_"
  }
}

我知道我的正则表达式工作正常（与不同的测试人员一起检查），但我不知道如何解决这个问题，我已经阅读了preg_match的文档，并且它没有帮助，因为它是olny确实在正则表达式中返回每个组。任何想法我怎么能解决这个问题？

Answer 1

Any Idea how I can solve this problem?

是的，您可以像这样使用preg_split：

$result = preg_split('/(?<=[a-z])(?=[A-Z])|(?=_[A-Z]*_)|(?<=_)(?=[A-Z][a-z])/', $str);

$ result数组包含你想要的字符串。

这个想法是preg_split使用的正则表达式与字符串不匹配;相反，它匹配字符串中的位置。此位置由外观定义，因此它具有零个字符。

此代码生成所需的数组（请参阅online demo底部的结果）：

<?php
$regex = "~(?<=[a-z])(?=[A-Z])|(?=_[A-Z]*_)|(?<=_)(?=[A-Z][a-z])~";
print_r(preg_split($regex,"postGeneratePasswordToken"));
print_r(preg_split($regex,"putUsers_VAL_Calculations_NUMBER_"));
print_r(preg_split($regex,"deleteUsers_VAL_Clients_ID_"));
print_r(preg_split($regex,"getDictionary_VAL_"));

解释正则表达式

(?<=                     # look behind to see if there is:
  [a-z]                  #   any character of: 'a' to 'z'
)                        # end of look-behind
(?=                      # look ahead to see if there is:
  [A-Z]                  #   any character of: 'A' to 'Z'
)                        # end of look-ahead
|                        # OR
(?=                      # look ahead to see if there is:
  _                      #   '_'
  [A-Z]*                 #   any character of: 'A' to 'Z' (0 or more
                         #   times (matching the most amount
                         #   possible))
  _                      #   '_'
)                        # end of look-ahead
|                        # OR
(?<=                     # look behind to see if there is:
  _                      #   '_'
)                        # end of look-behind
(?=                      # look ahead to see if there is:
  [A-Z]                  #   any character of: 'A' to 'Z'
  [a-z]                  #   any character of: 'a' to 'z'
)                        # end of look-ahead

PHP preg_match不会按预期返回所有值

1 个答案: