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

I have defined one pattern to look for any of the ASCII Special Characters ranging between 032 to 126 except the alpha-numeric. You may use something like the one below:

To find any Special Character:

[ -\/:-@\[-\`{-~]

To find minimum of 1 and maximum of any count:

(?=.*[ -\/:-@\[-\`{-~]{1,})

These patterns have Special Characters ranging between 032 to 047, 058 to 064, 091 to 096, and 123 to 126.

Answer 2

Try using this for the same things - StringUtils.isAlphanumeric(value)

Answer 3

That's because your pattern contains a .-^ which is all characters between and including . and ^, which included digits and several other characters as shown below:

If by special characters, you mean punctuation and symbols use:

[\p{P}\p{S}]

which contains all unicode punctuation and symbols.

Answer 4

If you only rely on ASCII characters, you can rely on using the hex ranges on the ASCII table. Here is a regex that will grab all special characters in the range of 33-47, 58-64, 91-96, 123-126

[\x21-\x2F\x3A-\x40\x5B-\x60\x7B-\x7E]

However you can think of special characters as not normal characters. If we take that approach, you can simply do this

^[A-Za-z0-9\s]+

Hower this will not catch _ ^ and probably others.

Answer 5

SInce you don't have white-space and underscore in your character class I think following regex will be better for you:

Pattern regex = Pattern.compile("[^\w\s]");

Which means match everything other than [A-Za-z0-9\s_]

Unicode version:

Pattern regex = Pattern.compile("[^\p{L}\d\s_]");

Answer 6

Here is my regex variant of a special character:

String regExp = "^[^<>{}\"/|;:.,~!?@#$%^=&*\\]\\\\()\\[¿§«»ω⊙¤°℃℉€¥£¢¡®©0-9_+]*$";

(Java code)