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:
import java.util.logging.Level;
public class GFG {
public static void main(String[] args)
{
Level level
= Level.parse( "WARNING" );
System.out.println( "Level = "
+ level.toString());
}
}
|
Program 2:
import java.util.logging.Level;
public class GFG {
public static void main(String[] args)
{
Level level
= Level.parse( "400" );
System.out.println( "Level = "
+ level.toString());
}
}
|
References: https://docs.oracle.com/javase/10/docs/api/java/util/logging/Level.html#parse(java.lang.Object)
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