How to check if a number starts with Special Characters using regular expression

Question

Using following regexp I can get if a string starts with a or b.

        Pattern p = Pattern.compile("^(a|b)");
        Matcher m = p.matcher("0I am a string");
        boolean b = m.find();
        System.out.println("....output..."+b);

But I need to check if the string starts with some special characters like * or ^ etc. In that case the following regexp gives Pattern error.

Pattern p = Pattern.compile("^(*|^)");
        Matcher m = p.matcher("0I am a string");
        boolean b = m.find();

        System.out.println("....output..."+b);

How to check if a number starts with * or ^ using regular expression

Answer 1

Characters like |*.[](){}?+^$ have special meaning in regex. Use [\^*] (within a [] only ] and - have special meaning), or escape them like ^(\*|\^).

Answer 2

Thanks to mathematical.coffee for nice response

Finally I found a solution by using ^[*^] as regexp to test whether a string contains * or ^

Source Code

Pattern p = Pattern.compile("^[*^]");
        Matcher m = p.matcher("I am a string");
        boolean b = m.find();

        System.out.println("....output..."+b);

Answer 3

You can try '(\^|\*)[0-9]+' for the purpose.

Since both ^ and * are special characters in regex, you'll need to put \ before them.

If you wish that the whole string must match with it, put the regex between ^ and $.

Answer 4

Use this:

"0I am a string".matches("[\\^*].*");

Answer 5

* and ^ are special characters in regex which need to be escaped, so you will need to use a \ to escape it.

See this link for more information.