Level parse() method in Java with Examples

The parse() method of java.util.logging.Level is used to parse a level name string into a Level. The name of the logging level must be a valid logging name. The argument string may consist of either a level name or an integer value. Level example name: “INFO”, “800”.

Syntax:

public static Level parse(String name)
    throws IllegalArgumentException

Parameters: This method accepts a name which is the string to be parsed.

Return: This method returns parsed value.

Exception: This method throws the following Exception:

• NullPointerException: if the name is null.
• IllegalArgumentException: if the value is not valid. Valid values are integers between Integer.MIN_VALUE and Integer.MAX_VALUE, and all known level names.

Below programs illustrate parse() method:
Program 1:

// Java program to illustrate parse() method

import java.util.logging.Level;

public class GFG {

    public static void main(String[] args)
    {

        // Get level of logger using
        // parse method.
        Level level
            = Level.parse("WARNING");

        // print result
        System.out.println("Level  = "
                           + level.toString());
    }
}

Output:

Level  = WARNING

Program 2:

// Java program to illustrate parse() method

import java.util.logging.Level;

public class GFG {

    public static void main(String[] args)
    {

        // Get level of logger using
        // parse method.
        Level level
            = Level.parse("400");

        // print result
        System.out.println("Level  = "
                           + level.toString());
    }
}

Output:

Level  = FINER

References: https://docs.oracle.com/javase/10/docs/api/java/util/logging/Level.html#parse(java.lang.Object)