Java 使用正则表达式匹配除 = 之外的任何字符

Using regex to match any character except =

提问by conceptSeeker

I am trying to write a String validation to match any character (regular, digit and special) except =.

我正在尝试编写一个字符串验证来匹配除 = 之外的任何字符（常规、数字和特殊）。

Here is what I have written -

这是我写的——

    String patternString = "[[^=][\w\s\W]]*";
    Pattern p = Pattern.compile(patternString);
    Matcher m = p.matcher(str);

    if(m.matches())
        System.out.println("matches");
    else
        System.out.println("does not");

But, it matches the input string "2009-09/09 12:23:12.5=" with the pattern.

但是，它将输入字符串 "2009-09/09 12:23:12.5=" 与模式匹配。

How can I exclude = (or any other character, for that matter) from the pattern string?

如何从模式字符串中排除 = （或任何其他字符，就此而言）？

回答by tripleee

If the only prohibited character is the equals sign, something like [^=]*should work.

如果唯一禁止的字符是等号，则[^=]*应该可以使用类似的字符。

[^...]is a negated character class; it matches a single character which is any character except one from the list between the square brackets. *repeats the expression zero or more times.

[^...]是一个否定字符类；它匹配单个字符，该字符是方括号之间列表中的任何字符除外。*重复表达式零次或多次。

回答by moodywoody

If you only want to check for occurence of "=" why don't you use the String indexOf() method?

如果您只想检查“=”的出现，为什么不使用 String indexOf() 方法？

if str.indexOf('=')  //...

回答by phihag

First of all, you don't need a regexp. Simply call contains:

首先，您不需要正则表达式。只需致电contains：

if(str.contains("="))
    System.out.println("does not");
else
    System.out.println("matches");

The correct regexp you're looking for is just

您正在寻找的正确正则表达式只是

String patternString = "[^=]*";

回答by Chetter Hummin

If your goal is to not have any = characters in your string, please try the following

如果您的目标是在字符串中不包含任何 = 字符，请尝试以下操作

String patternString = "[^=]*";