Java see if arraylist contains string

Question

I have a class called Paragens like this:

public class Paragens {
    static int contadorParagens = 1;

    String nomeParagem;
    int id;

    public Paragens(String nomeParagem) {
        this.nomeParagem = nomeParagem;
        this.id = contadorParagens++;
    }

    // getters and setters for nomeParagem
}

Every Paragens object has a name and an Id.

This class has a main method where I create several Paragens objects and store them inside an ArrayList like this:

public static void main(String[] args) {
    ArrayList<Paragens> paragens = new ArrayList<Paragens>();
    paragens.add(new Paragens("name1");
    // ... add more paragens
}

This is working ok. If I insert a bunch of paragens and print them I can see tat is all ok.

What I am trying to do is to ask the user to input a paragem name and then I want to see if that paragem is already in the ArrayList.

String name;
System.out.println("Insert paragem name: ");
pickName = sc.nextLine();
System.out.println(paragens.contains(pickName));

What am I doing wrong?

T.J. Crowder · Accepted Answer · 2011-11-27 10:47:44Z

10

contains checks to see if the list contains the actual thing you handed it. In this case, you're passing in a String name, but comparing it to Paragem instances. contains can't magically guess that it's supposed to look at a given property on the Paragem instances to compare the string.

You can easily loop the list to find out for yourself:

boolean found = false;
for (Paragem p : paragems) {
    if (p.nomeParagem.equals(pickName)) { // Or use an accessor function for `nomeParagem` if appropriate
        found = true;
        break;
    }
}

...or as a function:

boolean containsParagemWithName(ArrayList<Paragem> paragems, String pickName) {
    for (Paragem p : paragems) {
        if (p.nomeParagem.equals(pickName)) {
            return true;
        }
    }
    return false;
}

edited Nov 27, 2011 at 10:47

answered Nov 27, 2011 at 10:34

T.J. Crowder

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5 Comments

Thilo Over a year ago

And if you make that into a method of its own, you can do away with the found flag and just return true in the loop or return false after it.

Thilo Over a year ago

@Anonymouse: Don't override equals to make a Paragem equal a String. That would be very unconventional (for example equality would then no longer be symmetric). Equals, hashCode and these methods are very general-purpose, expect to follow certain contracts, and should not do unexpected things to accommodate some special purposes.

Has QUIT--Anony-Mousse Over a year ago

Yeah, for equals he should create a new Paragem, and then check if that one is equal, obviously.

Thilo Over a year ago

Ah, okay, that would be fine (if name equality does indeed mean Paragem equality).

Favolas Over a year ago

Thanks for the explanation. That makes all the sense. Unfortunately I did'n remember doing the for loop. Will implement that.

Igor Popov · Accepted Answer · 2011-11-27 10:38:31Z

1

Well, you need to implement the contains method yourself. Do a for loop over the entire array and check if the name of one of the elements is equal with what you're trying to add. If not, add a new Paragens(pickName).

answered Nov 27, 2011 at 10:38

Igor Popov

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1 Comment

Has QUIT--Anony-Mousse Over a year ago

There is nothing wrong with the contains method. He just needs to either implement equals or manually inspect the instances in a loop.

T.J. Crowder · Accepted Answer · 2011-11-27 10:54:41Z

1

Objects by default are compared by their memory location. So if you have two Paragem with the same name, they are still not equal.

So either, you check the name of each one:

boolean checkDuplicate(String pickName) {
    for (Paragem p : paragems) {
      if (p.nomeParagem.equals(pickName)) return true;
    }
    return false;
}

or implement (override) the equals method to compare names (you should be calling contains on a new Paragem object then instead of a String, though).

edited Nov 27, 2011 at 10:54

T.J. Crowder

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answered Nov 27, 2011 at 10:41

Has QUIT--Anony-Mousse

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3 Comments

T.J. Crowder Over a year ago

"Objects by default are compared by their memory location. So if you have two Paragem with the same name, they are still not equal." Yes, but it's not comparing two Paragem instances. He's comparing String to Paragem.

Has QUIT--Anony-Mousse Over a year ago

@T.J. Crowder: Correct. Which is why it even works less. ;-) Unfortunately, contains in Java Collections is defined for Objects, which introduces this kind of bugs.

Favolas Over a year ago

Understood the for loop but the last thing (implement (override) the equals method to compare names.) didn't understand because I'm new at Java. Nevertheless will implement the for loop

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