Regex pattern including all special characters

To create a regular expression that includes all special characters, you can use the following pattern:

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

This regular expression will match any single special character in the list.

For example, you can use this regular expression to test if a string contains any special characters:

import java.util.regex.Matcher;
import java.util.regex.Pattern;

String input = "Hello World!";
Pattern pattern = Pattern.compile("[!@#%^&*()_+-=[]{}|;':\",./<>?~`]");
Matcher matcher = pattern.matcher(input);
if (matcher.find()) {
    System.out.println("Input contains special characters.");
} else {
    System.out.println("Input does not contain special characters.");
}

This will output "Input does not contain special characters."

Note that the regular expression includes several characters that must be escaped with a backslash (\) because they have special meaning in a regular expression. These characters are [, ], ^, -, \, and ].

Also, be aware that this regular expression will not match all possible special characters. It only includes the special characters that are commonly used in the ASCII character set. If you need to match all possible special characters, you can use the \p{Punct} Unicode character property, which will match any Unicode punctuation character.