Regex pattern including all special characters

Question

I want to write a simple regular expression to check if in given string exist any special character. My regex works but I don\'t know why it also includes all numbers, so wh

Answer 1

Use this regular expression pattern ("^[a-zA-Z0-9]*$") .It validates alphanumeric string excluding the special characters

Answer 2

For people (like me) looking for an answer for special characters like Ä etc. just use this pattern:

• Only text (or a space): "[A-Za-zÀ-ȕ ]"

• Text and numbers: "[A-Za-zÀ-ȕ0-9 ]"

• Text, numbers and some special chars: "[A-Za-zÀ-ȕ0-9(),-_., ]"

Regex just starts at the ascii index and checks if a character of the string is in within both indexes [startindex-endindex].

So you can add any range.

Eventually you can play around with a handy tool: https://regexr.com/

Good luck;)

Answer 3

We can achieve this using Pattern and Matcher as follows:

Pattern pattern = Pattern.compile("[^A-Za-z0-9 ]");
Matcher matcher = pattern.matcher(trString);
boolean hasSpecialChars = matcher.find();

Answer 4

Please use this.. it is simplest.

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

https://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html

    StringBuilder builder = new StringBuilder(checkstring);
    String regex = "\\p{Punct}"; //Special character : `~!@#$%^&*()-_+=\|}{]["';:/?.,><
    //change your all special characters to "" 
    Pattern  pattern = Pattern.compile(regex);
    Matcher matcher = pattern.matcher(builder.toString());
    checkstring=matcher.replaceAll("");

Answer 5

To find any number of special characters use the following regex pattern: ([^(A-Za-z0-9 )]{1,})

[^(A-Za-z0-9 )] this means any character except the alphabets, numbers, and space. {1,0} this means one or more characters of the previous block.

Answer 6

(^\W$)

^ - start of the string, \W - match any non-word character [^a-zA-Z0-9_], $ - end of the string