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Level parse() method in Java with Examples
  • Last Updated : 14 Oct, 2019

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)

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