CODEWITHSUNDEEP
javavalidationexception
Nadia Rodriguez
Nadia Rodriguez

Reputation: 87

How to validate input through constructor in java

I have read that it is possible to create a constructor that checks whether the input is correct or not and I tried doing so with an IllegalArgumentException. My code works fine as long as I do not create an object, it does not matter if the input is correct or not, the compiler will give an error. This is my code:

public class Knoten{
private String typ;

public Knoten(String typ) throws Exception{
  try{
    if (typ.equals ("BASE")||typ.equals ("WORD")||typ.equals ("ROOT")||typ.equals ("AFFIX")){
      this.typ=typ;
    }}catch(IllegalArgumentException ex){
       System.out.println ("invalid typ");
    }}
  public String getTyp(){
        return typ;
    }

  public String toString(){
        return "typ: "+getTyp();
    }

public static void main (String args[]){
Knoten Knot1= new Knoten("haha");//throws error
Knoten Knot2= new Knoten("BASE");//throws error
}}

I tried also doing this, getting the same result:

public class Knoten{
  private String typ;

  public Knoten(String typ) throws Exception{
    if (typ.equals ("BASE")||typ.equals ("WORD")||typ.equals ("ROOT")||typ.equals ("AFFIX")){
        this.type=type;
    }else{
         throw new IllegalArgumentException("invalid type");
    }

    }
  public String getTyp(){
    return typ;
}

  public String toString(){
    return "typ: "+getTyp();
}

public static void main (String args[]){
Knoten Knot1= new Knoten("haha");//throws error
Knoten Knot2= new Knoten("BASE");//throws error

}

}

So my question is if I have to handle the exception every time I create an object of that class or if there is a nicer way of doing this.

Upvotes: 0

Views: 936

Answers (3)

Joseph Peter
Joseph Peter

Reputation: 133

you have a typo error in your code . Use 

this.typ = typ; // instead of this.type = type;

and also remeber to catch the Exception that is being thrown by the constructor

Upvotes: 0

Omar Silva
Omar Silva

Reputation: 301

You need to catch the exception when you instantiate your new object, in this case your second solution is fine, except on main you would do:

public static void main (String args[]){
    try {
        Knoten Knot1= new Knoten("haha"); // This will throw error
        Knoten Knot2= new Knoten("BASE");
    }
    catch (IllegalArgumentException e) {
        // deal with exception here
    }
}

Also, I would really look into using enums for your solution, and then you would not need to validate your type when creating a new object. For instance, you could have an enum like so:

public enum KnotenType {
     BASE,
     WORD,
     ROOT,
     AFFIX,
}

Then your constructor would look like:

public Knoten(KnotenType type) {
    ...
}

More info on Enums: https://docs.oracle.com/javase/tutorial/java/javaOO/enum.html

Upvotes: 0

Amit Bera
Amit Bera

Reputation: 7315

I believe you are getting compilation error like: Unhandled exception type Exception and you are getting this as you are throwing a Checked Exception (java.lang.Exception) from the constructor. Don't put any Exception in throws or try to throw the unchecked exception like :

public Knoten(String type) throws IllegalArgumentException { // throws unchecked exception is optional
    if (type.equals("BASE") || type.equals("WORD") || type.equals("ROOT") || type.equals("AFFIX")) {
        this.type = type;
    } else {
        throw new IllegalArgumentException("invalid type");
    }

}

Or surround it Constructor try catch/finally. 

try {
    Knoten Knot1 = new Knoten("haha");
} catch (Exception e) {
    e.printStackTrace();
}

Upvotes: 1

Related Questions