如何在 RegExp javascript 中放置 []（方括号）？
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How can I put [] (square brackets) in RegExp javascript?
提问by ilyo
I'm trying this:
我正在尝试这个：
str = "bla [bla]";
str = str.replace(/\[\]/g,"");
console.log(str);
And the replace doesn't work, what am I doing wrong?
而且替换不起作用，我做错了什么？
UPDATE: I'm trying to remove any square brackets in the string, what's weird is that if I do
更新：我正在尝试删除字符串中的任何方括号，奇怪的是，如果我这样做了
replace(/\[/g, '')
replace(/\]/g, '')
it works, butreplace(/\[\]/g, '');
doesn't.
它有效，但replace(/\[\]/g, '');
无效。
回答by Rudu
It should be:
它应该是：
str = str.replace(/\[.*?\]/g,"");
You don't need double backslashes (\) because it's not a stringbut a regex statement, if you build the regex from a string you doneed the double backslashes ;).
您不需要双反斜杠 (\)，因为它不是字符串而是正则表达式语句，如果您从字符串构建正则表达式，则确实需要双反斜杠 ;)。
It was also literally interpreting the 1 (which wasn't matching). Using .*says any value between the square brackets.
它也从字面上解释了 1（不匹配）。Using.*表示方括号之间的任何值。
The new RegExp string build version would be:
新的 RegExp 字符串构建版本将是：
str=str.replace(new RegExp("\[.*?\]","g"),"");
UPDATE:To remove square brackets only:
更新：仅删除方括号：
str = str.replace(/\[(.*?)\]/g,"");
Your above code isn't working, because it's trying to match "[]" (sequentially without anything allowed between). We can get around this by non-greedy group-matching ((.*?)) what's between the square brackets, and using a backreference ($1) for the replacement.
您上面的代码不起作用，因为它试图匹配“[]”（顺序不允许任何东西）。我们可以通过非贪婪的组匹配 ( (.*?)) 方括号之间的内容来解决这个问题，并使用反向引用 ( $1) 进行替换。
UPDATE 2:To remove multiple square brackets
更新 2：删除多个方括号
str = str.replace(/\[+(.*?)\]+/g,"");
// bla [bla] [[blaa]] -> bla bla blaa
// bla [bla] [[[blaa] -> bla bla blaa
Note this doesn't match open/close quantities, simply removes all sequential opens and closes. Also if the sequential brackets have separators (spaces etc) it won't match.
请注意，这与打开/关闭数量不匹配，只是删除所有顺序打开和关闭。此外，如果连续括号有分隔符（空格等），它将不匹配。
回答by Hristo
You have to escape the bracket, like \[and \]. Check out http://regexpal.com/. It's pretty useful :)
您必须转义括号，例如\[和\]。查看http://regexpal.com/。它非常有用:)
To replace all brackets in a string, this should do the job:
要替换字符串中的所有括号，这应该可以完成以下工作：
str.replace(/\[|\]/g,'');
I hope this helps.
Hristo
我希望这有帮助。
赫里斯托
回答by sidyll
What exactly are you trying to match?
你到底想匹配什么？
If you don't escape the brackets, they are considered character classes. This:
如果您不转义括号，它们将被视为字符类。这个：
/[1\]/
Matches either a 1 or a backslash. You may want to escape them with one backslash only:
匹配 1 或反斜杠。您可能只想用一个反斜杠来转义它们：
/\[1\]/
But this won't match either, as you don't have a [1]in your string.
但这也不匹配，因为您[1]的字符串中没有 a 。
回答by WEBjuju
I stumbled on this question while dealing with square bracket escaping within a character classthat was designed for use with password validation requiring the presence of special characters.
我在处理一个字符类中的方括号转义时偶然发现了这个问题，该字符类旨在用于需要存在特殊字符的密码验证。
Note the double escaping:
注意双重转义：
var regex = new RegExp('[\\]]');
var regex = new RegExp('[\\]]');
As @rudu mentions, this expression iswithin a string so it must be double escaped. Note that the quoting type (single/double) is not relevant here.
作为@rudu提到，这种表达是一个字符串中，因此必须进行双重转义。请注意，引用类型（单/双）在这里不相关。
Here is an example of using square brackets in a character class that tests for all the special characters found on mykeyboard:
这里是一个字符类，对所有特殊字符的测试中发现用方括号的例子我的键盘：
var regex = new RegExp('[-,_,\',",;,:,!,@,#,$,%,^,&,*,(,),[,\],\?,{,},|,+,=,<,>,~,`,\\,\,,\/,.]', 'g')
回答by Marc B
Two backslashes produces a single backslash, so you're searching for "a backslash, followed by a character class consisting of a 1or a right bracket, and then you're missing an closing bracket.
两个反斜杠产生一个反斜杠，因此您正在搜索“一个反斜杠，后跟由 a1或 a组成的字符类right bracket，然后您缺少一个右括号。
Try
尝试
str.replace(/\[1\]/g, '');
