Javascript 正则表达式 - 匹配除 + 之外的任何字符，也应匹配空字符串
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Regular Expression - Match any character except +, empty string should also be matched
提问by zaq
I am having a bit of trouble with one part of a regular expression that will be used in JavaScript. I need a way to match any character other than the +character, an empty string should also match.
我在 JavaScript 中使用的正则表达式的一部分遇到了一些麻烦。我需要一种方法来匹配字符以外的任何+字符，空字符串也应该匹配。
[^+]is almost what I want except it does not match an empty string. I have tried [^+]*thinking: "any character other than +, zero or more times", but this matches everything including +.
[^+]几乎是我想要的，除了它不匹配空字符串。我试过[^+]*想：“除+0 次或多次以外的任何字符”，但这与包括+.
回答by chown
Add a {0,1} to it so that it will only match zero or one times, no more no less:
添加一个 {0,1} 到它，这样它只会匹配零次或一次，不多不少：
[^+]{0,1}
Or, as FailedDev pointed out, ?works too:
或者，正如 FailedDev 指出的那样，?也可以：
[^+]?
As expected, testing with Chrome's JavaScript console shows no match for "+"but does match other characters:
正如预期的那样，使用 Chrome 的 JavaScript 控制台测试显示不匹配"+"但匹配其他字符：
x = "+"
y = "A"
x.match(/[^+]{0,1}/)
[""]
y.match(/[^+]{0,1}/)
["A"]
x.match(/[^+]?/)
[""]
y.match(/[^+]?/)
["A"]
回答by Code Jockey
[^+]means "match any single character that is not a+"[^+]*means "match any number of characters that are not a+" - which almost seems like what I think you want, except that it will match zero characters if the first character (or even all of the characters) are+.
[^+]意思是“匹配任何不是+”的单个字符[^+]*意思是“匹配任意数量的不是 a 的字符+” - 这几乎看起来像我认为你想要的，除了如果第一个字符（甚至所有字符）是+.
use anchors to make sure that the expression validates the ENTIRE STRING:
使用锚点来确保表达式验证整个字符串：
^[^+]*$
means:
方法：
^ # assert at the beginning of the string
[^+]* # any character that is not '+', zero or more times
$ # assert at the end of the string
回答by Scott Rippey
If you're just testing the string to see if it doesn't contain a +, then you should use:
如果您只是测试字符串以查看它是否不包含 a +，那么您应该使用：
^[^+]*$
This will match only if the ENTIRE string has no +.
仅当 ENTIRE 字符串没有+.
