Java 正则表达式中有 not (!) 运算符吗？
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Is there a not (!) operator in regexp?
提问by Roman
I need to remove all characters from the given string except for several which should left. How to do that with regexp?
我需要从给定的字符串中删除所有字符，除了几个应该留下的字符。如何用正则表达式做到这一点？
Simple test: characters[1, a, *] shouldn't be removed, all other should from string "asdf123**".
简单测试：不应删除字符[1, a, *]，所有其他字符都应来自字符串“asdf123**”。
采纳答案by Jon Skeet
There is: ^ in a set.
集合中有：^。
You should be able to do something like:
您应该能够执行以下操作：
text = text.replaceAll("[^1a*]", "");
Full sample:
完整样本：
public class Test
{
public static void main(String[] args)
{
String input = "asdf123**";
String output = input.replaceAll("[^1a*]", "");
System.out.println(output); // Prints a1**
}
}
回答by jjnguy
When used inside [and ]the ^(caret) is the notoperator.
时所使用的内部[和]所述^（脱字符号）是not操作员。
It is used like this:
它是这样使用的：
"[^abc]"
That will match any character except for abor c.
这将匹配除abor之外的任何字符c。
回答by eldarerathis
There's a negated character class, which might work for this instance. You define one by putting ^at the beginning of the class, such as:
有一个否定的字符类，它可能适用于这个实例。您可以通过放置^在类的开头来定义一个，例如：
[^1a\*]
[^1a\*]
for your specific case.
对于您的具体情况。
回答by Cfreak
In a character class the ^ is not. So
在字符类中， ^ 不是。所以
[^1a\*]would match all characters except those.
[^1a\*]将匹配除那些字符之外的所有字符。
回答by user318904
You want to match against all characters except: [asdf123*], use ^
您想匹配除 [asdf123*] 之外的所有字符，请使用 ^
回答by Steve Emmerson
There's no "not" operator in Java regular expressions like there is in Perl.
Java 正则表达式中没有像 Perl 中那样的“非”运算符。
