我有这个示例字符串
[can be anything here %+^-_][can be anything here %+^-_][can be anything here %+^-_][can be anything here %+^-_][can be anything here %+^-_][can be anything here %+^-_][can be anything here %+^-_][can be anything here %+^-_]
我的模式是(\[[^\]]+\])
我得到了这个结果
(
[0] => Array
(
[0] => [can be anything here %+^-_]
[1] => [can be anything here %+^-_]
[2] => [can be anything here %+^-_]
[3] => [can be anything here %+^-_]
[4] => [can be anything here %+^-_]
[5] => [can be anything here %+^-_]
[6] => [can be anything here %+^-_]
[7] => [can be anything here %+^-_]
)
[1] => Array
(
[0] => [can be anything here %+^-_]
[1] => [can be anything here %+^-_]
[2] => [can be anything here %+^-_]
[3] => [can be anything here %+^-_]
[4] => [can be anything here %+^-_]
[5] => [can be anything here %+^-_]
[6] => [can be anything here %+^-_]
[7] => [can be anything here %+^-_]
)
)
为什么结果有两个数组？无论如何，这不是什么大问题，但我想知道。
如何仅使用正则表达式删除每个数组值的开头和末尾的括号。像这样。
[0] => Array
(
[0] => can be anything here %+^-_
[1] => can be anything here %+^-_
[2] => can be anything here %+^-_
[3] => can be anything here %+^-_
[4] => can be anything here %+^-_
[5] => can be anything here %+^-_
[6] => can be anything here %+^-_
[7] => can be anything here %+^-_
)
答案 0 :(得分：1)
为什么结果有两个数组？
因为在括号中加上一些东西（()）会为它分配一个组号，这就是组1的含义。第0组是你的整场比赛。有关详细信息，请参阅this。
如何在每个数组值的开头和结尾删除括号？
将正则表达式更改为：
\[([^\]]+)\]
对于上述情况，匹配的[]位于()之外。这将使第1组成为你想要的。
为了使第0组成为您想要的，没有第1组，您必须使用look-around：
(?<=\[)[^\]]+(?=\])
但这在很大程度上是不必要的。
(?<=\[)是肯定的后卫，检查前一个字符是[，但不会在匹配中包含它。
(?=\])是正向前瞻，检查下一个字符是]，但不会在匹配中包含它。
答案 1 :(得分：0)
只需将大括号从捕获组中删除：
\[([^\]]+)\]
您有两个数组，因为一个是正则表达式的完全匹配，另一个是()捕获的数组。