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  1. Prepare
  2. Java
  3. Strings
  4. Pattern Syntax Checker
  5. Discussions

Pattern Syntax Checker

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387 Discussions

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  • SandraShari
     + 0 comments

    It ensures accuracy in pattern recognition, preventing errors in applications like regex validation. Ekbet 16

  • geethikanavya48
     + 0 comments

    import java.util.Scanner; import java.util.regex.Pattern; import java.util.regex.PatternSyntaxException;

    public class Solution {

    public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int testCases = sc.nextInt(); // Number of test cases
    sc.nextLine(); // Consume the newline character

    for (int i = 0; i < testCases; i++) {
        String pattern = sc.nextLine(); // Input regex pattern
        try {
            Pattern.compile(pattern); // Attempt to compile the pattern
            System.out.println("Valid");
        } catch (PatternSyntaxException e) {
            System.out.println("Invalid");
        }
    }
}

    }

  • Prince_R
     + 0 comments

    import java.util.Scanner; import java.util.regex.; / * @Author : Rehan */ public class SolutionDump { public static void main(String[] args){ Scanner in = new Scanner(System.in); int testCases = Integer.parseInt(in.nextLine());

        while(testCases>0){
        String pattern = in.nextLine();
        try{
            Pattern.compile(pattern);
            System.out.println("Valid");
        }catch(PatternSyntaxException e){
            System.out.println("Invalid");
        }
        testCases--; 
    }
    in.close();
}

    }

  • samimteyaz176
     + 0 comments

    import java.util.*; import java.util.regex.Pattern; import java.util.regex.PatternSyntaxException;

    public class Solution {

    public static void main(String[] args) {

    Scanner sc = new Scanner(System.in);
     int N = Integer.parseInt(sc.nextLine());

     for(int i=0;i<N;i++){
        String pattern = sc.nextLine();
        try {
            Pattern.compile(pattern); // Attempt to compile the pattern
            System.out.println("Valid");
        } catch (PatternSyntaxException e) {
            System.out.println("Invalid");
        }
    }
  }  
}

  • apisit_baworn1
     + 0 comments
    public class Solution
{
	public static void main(String[] args){
		Scanner in = new Scanner(System.in);
		while(in.hasNextLine()){
			String pattern = in.nextLine();
          	try{
              Pattern.compile(pattern);
              System.out.println("Valid");
              }
              catch(Exception e ){
                  System.out.println("Invalid");

              }
    }
    in.close();
}

    }