RegEx提取括号中的单词不起作用

I am using RegEx and exec to extract words but it is returning the full first match, not all the matches separately.

我使用RegEx和exec来提取单词,但它返回完整的第一个匹配,而不是单独的所有匹配。

The string is 'one two [three four] five [six] seven [nine ten]'. The result should be 'three four', 'six', 'nine ten', instead of '[three four] five [six] seven [nine ten]'.

字符串是'一二[三四]五[六]七[九十]。结果应该是“三四”,“六”,“九十”,而不是“[三四]五[六]七[九十]”。

var text = "one two [three four] five [six] seven [nine ten]" 
var brackets = /\[([^]+)\]/g; 
var match; 
while (match = brackets.exec(text)) {
   console.log(match);
}

What am I missing?

我错过了什么?

1 个解决方案

#1

The problem is with the capturing group ([^]+).

问题在于捕获组([^] +)。

[^]+ matches any character, including newline as there is nothing specified in the negated character class.

[^] +匹配任何字符,包括换行符,因为否定字符类中没有指定任何内容。

Use the below regex

使用下面的正则表达式

/\[([^[\]]+)\]/g

[^[\]]+: will match one or more characters except square brackets [ and ].

[^ [\]] +:将匹配除方括号[和]之外的一个或多个字符。

var text = "one two [three four] five [six] seven [nine ten]"
var brackets = /\[([^[\]]+)\]/g;
var match;
while (match = brackets.exec(text)) {
    console.log(match[1]);
}

You can also use /\[(.*?)\]/g where .*? will match anything except ].

你也可以使用/\ [(。*?)\] / g where。*?将匹配除了]之外

var text = "one two [three four] five [six] seven [nine ten]"
var brackets = /\[(.*?)\]/g;
var match;
while (match = brackets.exec(text)) {
    console.log(match[1]);
}

#1