Java检测字符串中的特殊字符

I use regex

我用正则表达式

r="[^A-Za-z0-9]+";

to detect if a string has one or more chars other than letters and digits;

检测字符串是否有一个或多个字母和数字以外的字符;

Then I tried the following:

然后我尝试了以下内容:

Pattern.compile(r).matcher(p).find();

I tested:

! @ # $ % ^ & * ( ) +  =- [ ] \ ' ; , . / { } | " : < > ? ~ _ `

Most of the time, it works except backsplash \ and caret ^.

大多数情况下,它的工作原理除了backsplash \和caret ^。

e.g.

String p = "abcAsd10^"    (return false)
String p = "abcAsd10\\"   (return false)

Anything I miss?

我想念的一切?

7 个解决方案

#1

The following code prints "Found: true" when I compile and run it:

当我编译并运行它时,以下代码打印“Found:true”:

class T
{
    public static void main (String [] args)
    {
        String thePattern = "[^A-Za-z0-9]+"; 
        String theInput = "abskKel35^";
        boolean isFound = Pattern.compile(thePattern).matcher(theInput).find();
        System.out.println("Found: " + isFound);
    }
}

Not sure why you would see a different result...

不确定为什么你会看到不同的结果......

#2

Shouldn't that be:

不应该是:

r="[^A-Za-z0-9]+";

In your question you write a_z (underscore)

在你的问题中你写a_z(下划线)

#3

When I run this code on my machine it prints true..

当我在我的机器上运行此代码时,它打印为真..

        String r="[^A-Za-z0-9]+"; 
        String p = "abcAsd10\\" ;
        Matcher m = Pattern.compile(r).matcher(p);
        System.out.println(m.find());

true

#4

You could also change only:

你也可以只改变:

[\w&&^_]+

Where:

\w A word character with underscore: [a-zA-Z_0-9]

\ w带下划线的单词字符:[a-zA-Z_0-9]

\W A non-word character: [^\w]

\ W一个非单词字符:[^ \ w]

See more in class Pattern

在课程模式中查看更多内容

#5

Try quoting your input into the matcher.

尝试将您的输入引用到匹配器中。

Pattern.quote(p)

#6

Simply do:

p.matches(".*\\W.*"); //return true if contains one or more special character

#7

Sorry, I had

对不起,我有

r="[^A-za-z0-9]+"; (with little "z", i.e. A-z).

That causes returning "false". But why it works with other special chars except backsplash \ and caret ^?

这导致返回“假”。但是为什么它与其他特殊字符一起工作,除了backsplash \和caret ^?

#1