Java detect special characters in a string

Question

I use regex

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

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

Then I tried the following:

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

I tested:

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

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

e.g.

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

Anything I miss?

I get matches for "abcAsd10\\" and "abcAsd10^", so here it returns true. Note that I use C# so I think it has to to with Java. It can help if you write a little code sample that demonstrates your problem. — buckley
– buckley, Commented Jun 5, 2012 at 18:47
Sorry, I had r="[^A-za-z0-9]+"; (with little "z"). That causes returning "false" — Eric
– Eric, Commented Jun 5, 2012 at 18:57
So you have your answers now? Not clear if you had a type in your so question or the problem itself. — buckley
– buckley, Commented Jun 5, 2012 at 18:58
I had r="[^A-za-z0-9]+"; in my code (with little "z", i.e. A-z). That causes returning "false". But why it works with other special chars except backsplash \ and caret ^? — Eric
– Eric, Commented Jun 5, 2012 at 19:10

jahroy · Accepted Answer · 2012-06-05 18:51:42Z

2

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

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...

answered Jun 5, 2012 at 18:51

jahroy

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Comments

buckley · Accepted Answer · 2012-06-05 18:37:43Z

1

Shouldn't that be:

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

In your question you write a_z (underscore)

answered Jun 5, 2012 at 18:37

buckley

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Comments

baby boom · Accepted Answer · 2012-06-05 18:53:50Z

1

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

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

true

answered Jun 5, 2012 at 18:53

baby boom

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3 Comments

Paul Vargas Over a year ago

Originally he had [^A-Za_z0-9]+

baby boom Over a year ago

so we need to flag this question... it is nothing but a bug

Eric Over a year ago

The original one ([^A-Za_z0-9]+) should give true, Paul. Aynone try '[^A-za_z0-9]+'?

Paul Vargas · Accepted Answer · 2012-06-05 19:00:32Z

1

You could also change only:

[\w&&^_]+

Where:

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

See more in class Pattern

edited Jun 5, 2012 at 19:00

answered Jun 5, 2012 at 18:43

Paul Vargas

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1 Comment

Paul Vargas Over a year ago

Perhaps it might be interesting to experience another way.

Robert de W · Accepted Answer · 2012-06-05 18:50:07Z

0

Try quoting your input into the matcher.

Pattern.quote(p)

answered Jun 5, 2012 at 18:50

Robert de W

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Comments

elias · Accepted Answer · 2012-06-05 19:09:14Z

0

Simply do:

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

answered Jun 5, 2012 at 19:09

elias

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Eric · Accepted Answer · 2012-06-05 19:11:20Z

0

Sorry, I had

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 ^?

edited Jun 5, 2012 at 19:11

answered Jun 5, 2012 at 19:04

Eric

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