Wrong data type java lang illegalargumentexception

java.lang.IllegalArgumentException – Reasons and How to Solve?

Here you will learn possible causes of Exception in thread “main” java.lang.IllegalArgumentException and ways to solve it.

I hope you know the difference between error and exception. Error means programming mistake that can be recoverable only by fixing the application code. But exceptions will arise only when exceptional situations occurred like invalid inputs, null values, etc. They can be handled using try catch blocks.

java.lang.IllegalArgumentException will raise when invalid inputs passed to the method. This is most frequent exception in java.

Reasons for java.lang.IllegalArgumentException

• When Arguments out of range. For example percentage should lie between 1 to 100. If user entered 200 then illegalarugmentexcpetion will be thrown.
• When arguments format is invalid. For example if our method requires date format like YYYY/MM/DD but if user is passing YYYY-MM-DD. Then our method can’t understand. Then illegalargument exception will raise.
• When closed files has given as argument for a method to read that file. That means argument is invalid.
• When a method needs non empty string as a parameter but null string is passed.

Like this when invalid arguments given then it will raise illegal argument exception.

How to Handle java.lang.IllegalArgumentException?

IllegalArgumentException extends RuntimeException and it is unchecked exception. So we don’t need to catch it. We have to fix the code to avoid this exception.

• java.lang.Object
  • java.lang.Throwable
    • java.lang.Exception
      • java.lang.RuntimeException
        • java.lang.illegalArgumentException

        Example program which will raise illegalArgumentException.

        Источник

        IllegalArgumentException: argument type mismatch in Hibernate

        Translation: Hibernate provides an argument of wrong type when trying to invoke a setter method.

        My first step would be to find out which setter that is (for instance by debugging the application in eclipse, setting an exception break point, and inspecting the stack variables once the breakpoint is reached).

        Edit: What is the signature of the setter for the mapped property qs ? It should take a Set .

        Solution 2

        The solution is the use of «addScalar» in your query execution.

        Suppose your Entity is a user :

        @Entity @Table(name="user") @DynamicUpdate(value=true) public class UserEntity implements Serializable

        List users = sessionFactory.getCurrentSession() .createQuery(query) .list(); 

        List users = sessionFactory.getCurrentSession() .createQuery(query) .addScalar("id",LongType.INSTANCE) .addScalar("name",StringType.INSTANCE) .list(); 

        Note : prior to LongType.INSTANCE, there was Hibernate.LONG and Hibernate.STRING (they are deprecated now).

        Solution 3

        So, you modified an hibernate mapping file without modifying the Entity? I guess that the qs property was already there then. But is it a java.util.Set (as you used a to map your collection)?

        Solution 4

        I’ve heard of this happening due to underlying database field changes (such as date to timestamp). It might be worth reverting the database changes if you’re able and testing it, or checking the .hbm or annotations as Sands suggested.

        Related videos on Youtube

        

        

        

        

        

        

        

        

        

        Oleksandr

        Quintessential learner and programming newbie..

        Comments

        

        Out of the blue I started getting “IllegalArgumentException: argument type mismatch” in hibernate. The hibernate entity was working for quite some time and svn logs confirm the code to be intact. What might be the case?
        Here’s part of the exception

        Jan 16, 2010 10:47:09 PM org.apache.catalina.core.StandardWrapperValve invoke SEVERE: Servlet.service() for servlet Faces Servlet threw exception java.lang.IllegalArgumentException: argument type mismatch at sun.reflect.NativeMethodAccessorImpl.invoke0(Native Method) at sun.reflect.NativeMethodAccessorImpl.invoke(NativeMethodAccessorImpl.java:39) at sun.reflect.DelegatingMethodAccessorImpl.invoke(DelegatingMethodAccessorImpl.java:25) at java.lang.reflect.Method.invoke(Method.java:597) at org.hibernate.property.BasicPropertyAccessor$BasicSetter.set(BasicPropertyAccessor.java:42) at org.hibernate.tuple.entity.AbstractEntityTuplizer.setPropertyValues(AbstractEntityTuplizer.java:337) at org.hibernate.tuple.entity.PojoEntityTuplizer.setPropertyValues(PojoEntityTuplizer.java:200) at org.hibernate.persister.entity.AbstractEntityPersister.setPropertyValues(AbstractEntityPersister.java:3566) at org.hibernate.engine.TwoPhaseLoad.initializeEntity(TwoPhaseLoad.java:129) at org.hibernate.loader.Loader.initializeEntitiesAndCollections(Loader.java:854) at org.hibernate.loader.Loader.doQuery(Loader.java:729) at org.hibernate.loader.Loader.doQueryAndInitializeNonLazyCollections(Loader.java:236) at org.hibernate.loader.Loader.doList(Loader.java:2220) at org.hibernate.loader.Loader.listIgnoreQueryCache(Loader.java:2104) at org.hibernate.loader.Loader.list(Loader.java:2099) at org.hibernate.loader.hql.QueryLoader.list(QueryLoader.java:378) at org.hibernate.hql.ast.QueryTranslatorImpl.list(QueryTranslatorImpl.java:338) at org.hibernate.engine.query.HQLQueryPlan.performList(HQLQueryPlan.java:172) at org.hibernate.impl.SessionImpl.list(SessionImpl.java:1121) at org.hibernate.impl.QueryImpl.list(QueryImpl.java:79) at org.springframework.orm.hibernate3.HibernateTemplate$30.doInHibernate(HibernateTemplate.java:930) at org.springframework.orm.hibernate3.HibernateTemplate.doExecute(HibernateTemplate.java:419) at org.springframework.orm.hibernate3.HibernateTemplate.executeWithNativeSession(HibernateTemplate.java:374) at org.springframework.orm.hibernate3.HibernateTemplate.find(HibernateTemplate.java:921) 

        Источник

        java.lang.IllegalArgumentException – How to solve Illegal Argument Exception

        In this tutorial, we will discuss how to solve the java.lang.illegalargumentexception – IllegalArgumentException in Java. This exception is thrown in order to indicate that a method has been passed an illegal or inappropriate argument. For example, if a method requires a non-empty string as a parameter and the input string equals null, the IllegalArgumentException is thrown to indicate that the input parameter cannot be null. You can also check this tutorial in the following video:

        

        This exception extends the RuntimeException class and thus belongs to those exceptions that can be thrown during the operation of the Java Virtual Machine (JVM). It is an unchecked exception and thus, it does not need to be declared in a method’s or a constructor’s throws clause. Finally, the IllegalArgumentException exists since the first version of Java (1.0).

        1. The IllegalArgumentException in Java

        The IllegalArgumentException is a good way of handling possible errors in your application’s code. This exception indicates that a method is called with incorrect input arguments. Then, the only thing you must do is correct the values of the input parameters. In order to achieve that, follow the call stack found in the stack trace and check which method produced the invalid argument.

        The following example indicates a sample usage of the java.lang.IllegalArgumentException – IllegalArgumentException . IllegalArgumentExceptionExample.java

        Источник

        How to solve an IllegalArgumentException in Java?
        

        An IllegalArgumentException is thrown in order to indicate that a method has been passed an illegal argument. This exception extends the RuntimeException class and thus, belongs to those exceptions that can be thrown during the operation of the Java Virtual Machine (JVM). It is an unchecked exception and thus, it does not need to be declared in a method’s or a constructor’s throws clause.

        Reasons for java.lang.IllegalArgumentException

        • When Arguments out of range. For example, the percentage should lie between 1 to 100. If the user entered 101 then an IllegalArugmentExcpetion will be thrown.
        • When argument format is invalid. For example, if our method requires date format like YYYY/MM/DD but if the user is passing YYYY-MM-DD. Then our method can’t understand then IllegalArugmentExcpetion will be thrown.
        • When a method needs non-empty string as a parameter but the null string is passed.

        Example1

        public class Student < int m; public void setMarks(int marks) < if(marks < 0 || marks >100) throw new IllegalArgumentException(Integer.toString(marks)); else m = marks; > public static void main(String[] args) < Student s1 = new Student(); s1.setMarks(45); System.out.println(s1.m); Student s2 = new Student(); s2.setMarks(101); System.out.println(s2.m); >>

        Output

        45 Exception in thread "main" java.lang.IllegalArgumentException: 101 at Student.setMarks(Student.java:5) at Student.main(Student.java:15)

        Steps to solve IllegalArgumentException

        • When an IllegalArgumentException is thrown, we must check the call stack in Java’s stack trace and locate the method that produced the wrong argument.
        • The IllegalArgumentException is very useful and can be used to avoid situations where the application’s code would have to deal with unchecked input data.
        • The main use of this IllegalArgumentException is for validating the inputs coming from other users.
        • If we want to catch the IllegalArgumentException then we can use try-catch blocks. By doing like this we can handle some situations. Suppose in catch block if we put code to give another chance to the user to input again instead of stopping the execution especially in case of looping.

        Example2

        import java.util.Scanner; public class Student < public static void main(String[] args) < String cont = "y"; run(cont); >static void run(String cont) < Scanner scan = new Scanner(System.in); while( cont.equalsIgnoreCase("y")) < try < System.out.println("Enter an integer: "); int marks = scan.nextInt(); if (marks < 0 || marks >100) throw new IllegalArgumentException("value must be non-negative and below 100"); System.out.println( marks); > catch(IllegalArgumentException i) < System.out.println("out of range encouneterd. Want to continue"); cont = scan.next(); if(cont.equalsIgnoreCase("Y")) run(cont); >> > >

        Output

        Enter an integer: 1 1 Enter an integer: 100 100 Enter an integer: 150 out of range encouneterd. Want to continue y Enter an integer:

        Источник

        Читайте также:  Sql database and java