Java中具有特殊字符的模式匹配器问题

Question

i want to validate string pattern .its work with below code if not any special char in string .

for example :

    Pattern p = Pattern.compile("Dear User, .* is your One Time Password(OTP) for registration.",Pattern.CASE_INSENSITIVE);

    Matcher m = p.matcher("Dear User, 999 is your One Time Password(OTP) for registration.");

    if (m.matches()){
        System.out.println("truee");                
    }else{
        System.out.println("false");
    }  // output false 

and below is working fine if i remove ( and ) .

Pattern p = Pattern.compile("Dear User, .* is your One Time Password OTP for registration.",Pattern.CASE_INSENSITIVE);

    Matcher m = p.matcher("Dear User, 999 is your One Time Password OTP for registration.");

    if (m.matches()){
        System.out.println("truee");                
    }else{
        System.out.println("false");
    }  // output true

Answer 1

Try this:

Pattern p = Pattern.compile("Dear User, .* is your One Time Password\\(OTP\\) for registration\\.",Pattern.CASE_INSENSITIVE);

Matcher m = p.matcher("Dear User, 999 is your One Time Password(OTP) for registration.");

if (m.matches()){
   System.out.println("truee");                
}else{
  System.out.println("false");
} 

( ,) and . are used in regex expressions. You need to escape them if you want normal behaviour.

Answer 2

In regex, you must always beware of special characters when you need to match them literally .

In this case, you have 3 characters: ( (used to open a group), ) (used to close a group) and . (matches any character).

To match them literally, you need to escape them, or place into a character class [...] .

See fixed demo

Answer 3

package com.ramesh.test;

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

public class PatternMatcher {

public boolean containsPattern(String input, String pattern) {
    if (input == null || input.trim().isEmpty()) {
        System.out.println("Incorrect format of string");
        return false;
    }
    // “ * ”is replaced by “ .* ” in replaceAll()
    //“ .* ” Use this pattern to match any number, any string (including the empty string) of any characters.
    String inputpattern = pattern.replaceAll("\\*", ".*");
    System.out.println("first input" + inputpattern);
    Pattern p = Pattern.compile(inputpattern);
    Matcher m = p.matcher(input);
    boolean b = m.matches();
    return b;
}

public boolean containsPatternNot(String input1, String pattern1) {
    return (containsPattern(input1, pattern1) == true ? false : true);
}

public static void main(String[] args) {

    PatternMatcher m1 = new PatternMatcher ();
    boolean a = m1.containsPattern("ma5&*%u&^()k5.r5gh^", "m*u*r*");
    System.out.println(a);// returns True

    boolean d = m1.containsPattern("mur", "m*u*r*");
    System.out.println(d);// returns True

    boolean c = m1.containsPatternNot("ma5&^%u&^()k56r5gh^", "m*u*r*");
    System.out.println(c);// returns false

    boolean e = m1.containsPatternNot("mur", "m*u*r*");
    System.out.println(e);// returns false

}

}

Output: true true false false