正则表达式，用于捕获方括号之间的开始和结束空格(Regular expression to catch start and end whitespaces between square bracket

正则表达式，用于捕获方括号之间的开始和结束空格(Regular expression to catch start and end whitespaces between square brackets)

我需要一个javascript中的正则表达式，在string类似的情况下会失败：

foo[ bar] foo[bar ] foo[ bar ]

foo可以是任何单词（也可以包含特殊字符作为-和. ），同样适用于bar 。

如果string只是：它应该不会失败：

[ bar] [bar ] [ bar ]

如果string是，则不应该失败：

foo[bar]

一点背景信息。 我正在编写一个自定义的linter规则，如果引用的数组在括号中有不需要的空格，它将失败。 任何帮助是极大的赞赏。

这是我到目前为止（ Regex101 ）：

\[([^\]]+)\]

但是这会考虑到整个括号并且不会评估开头（它应该先于）。 此外，它应该只评估空格的存在。

I need a regular expression in javascript which will fail in cases when string is something like:

foo[ bar] foo[bar ] foo[ bar ]

foo can be any word (also can contain special characters as - and .), same goes for bar.

It should not fail if string is just:

[ bar] [bar ] [ bar ]

And should not fail if string is:

foo[bar]

A bit of background info. I am writing a custom linter rule which will fail if the referenced array has unneeded spaces in brackets. Any help is greatly appreciated.

This is what I have so far (Regex101):

\[([^\]]+)\]

But this catches the whole bracket into account and does not evaluate the beginning (it should be preceeded). Also, it should evaluate the presence of whitespaces only.

最满意答案

看来你只想匹配1）之前有一个或多个单词字符的字符串(和2）在[或之前]之前有空格。

你可以用

/\w+\[(\s[^\]]*|[^\]]*\s)]/

请参阅正则表达式演示 。

细节

\w+ - 1个或更多单词字符 \[ - 一个[ (\s[^\]]*|[^\]]*\s) - 两种选择中的任何一种： \s[^\]]* - 一个空格字符后面跟着0 +字符除了] | - 要么 [^\]]*\s - 除了以外的0+字符后跟一个空白字符 ] - a ] char。

JS演示：

var rx = /\w+\[(\s[^\]]*|[^\]]*\s)]/;
var strs = [ "foo[ bar]", "foo[bar ]", "foo[ bar ]", "[ bar]", "[bar ]", "[ bar ]"];
for (var s of strs) {
  console.log(s, "=>", rx.test(s));
} 
  
 
It seems you only want to match strings that 1) have 1 or more word chars before ( and 2) have whitespace right after [ or right before ].
 
You may use
 
/\w+\[(\s[^\]]*|[^\]]*\s)]/
 
See the regex demo.
 
Details
 
 
 \w+ - 1 or more word chars 
 \[ - a [ 
 (\s[^\]]*|[^\]]*\s) - either of the two alternatives: 
   
   \s[^\]]* - a whitespace char followed with 0+ chars other than ] 
   | - or 
   [^\]]*\s - 0+ chars other than ] followed with a whitespace char 
   
 ] - a ] char. 
 
JS demo:
 
 
 
  
  
var rx = /\w+\[(\s[^\]]*|[^\]]*\s)]/;
var strs = [ "foo[ bar]", "foo[bar ]", "foo[ bar ]", "[ bar]", "[bar ]", "[ bar ]"];
for (var s of strs) {
  console.log(s, "=>", rx.test(s));
}