java Java正则表达式验证特殊字符
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Java regex validating special chars
提问by brakebg
This seems like a well known title, but I am really facing a problem in this.
这似乎是一个众所周知的标题，但我确实面临着这个问题。
Here is what I have and what I've done so far.
这是我所拥有的以及到目前为止我所做的。
I have validate input string, these chars are not allowed :
我有验证输入字符串，这些字符是不允许的：
&%$##@!~
&%$##@!~
So I coded it like this:
所以我这样编码：
String REGEX = "^[&%$##@!~]";
String username= "jhgjhgjh.#";
Pattern pattern = Pattern.compile(REGEX);
Matcher matcher = pattern.matcher(username);
if (matcher.matches()) {
System.out.println("matched");
}
回答by Narendra Yadala
Change your first line of code like this
像这样改变你的第一行代码
String REGEX = "[^&%$#@!~]*";
And it should work fine. ^outside the character class denotes start of line. ^inside a character class []means a negation of the characters inside the character class. And, if you don't want to match empty usernames, then use this regex
它应该可以正常工作。^在字符类之外表示行的开始。^字符类内部[]意味着字符类内部字符的否定。而且，如果您不想匹配空用户名，请使用此正则表达式
String REGEX = "[^&%$#@!~]+";
回答by Kent
i think you want this:
我想你想要这个：
[^&%$##@!~]*
回答by smp7d
To match a valid input:
匹配一个有效的输入：
String REGEX = "[^&%$#@!~]*";
To match an invalid input:
匹配无效输入：
String REGEX = ".*[&%$#@!~]+.*";
