Using user input value to display a specific object from an array list

Question

I am trying to create a program where the user inputs animal pertinent values, which then create animal Objects, which are then saved into an array list. The area I am having trouble with is shown below. After the Array list is populated, I cannot figure out how to use the user input (select) to find that values index within the array list. (int index = animalList.indexOf(?))

Would appreciate any help

Scanner in = new Scanner(System.in);
  List <Animal> animalList = new ArrayList <Animal>();
    char ans;

do{  // User input
  Animal animal = new Animal(); // arraylist

   System.out.println("Animal's 'common' name: ");
     animal.setName(in.next());

   System.out.println("Animal's class: ");
     animal.setAnmlClass(in.next());

   System.out.println("Vertabrate or Invertabrate: ");
     animal.setCharVert(in.next());

   System.out.println("Warm or Cold blooded: ");
     animal.setCharBld(in.next());

   System.out.println("Animal's habitat (general): ");
     animal.setCharHab(in.next());

   System.out.println("Would you like to enter in a new animal (y/n)? ");
     String answer = in.next();
        ans = answer.charAt(0);

    animalList.add(animal);

  }while(ans == 'y');

   System.out.println("Enter the animal you wish to view: ");
     String select = in.next();
     System.out.println(select);
     int index =  animalList.indexOf( ? );
     System.out.println(index);

indexOf won't get you what you want. Essentially, there's no way for the user to input a string that will end up satisfying Animal.equal(). — Gus
– Gus, Commented Jun 22, 2015 at 18:11
Does Animal class have overridden equals() based on name? — PM 77-1
– PM 77-1, Commented Jun 22, 2015 at 18:14

FredK · Accepted Answer · 2015-06-22 18:09:11Z

1

You might consider making a HashTable instead of a List, with the animal name as the key.

answered Jun 22, 2015 at 18:09

FredK

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2 Comments

LKat Over a year ago

Thank you, the only problem is one of the specs for this project is using an Array List. If that wasn't a must, I would have gone with the Hash.

Srini Over a year ago

Use a HashMap<K,V> and not a HashTable<K,V>

Roel Strolenberg · Accepted Answer · 2015-06-22 18:12:17Z

0

Unfortunately, you need a custom method for this with a while loop:

public int getIndexWithName(List<Animal> animals, String name){
    for (int index = 0; index < animals.size(); index++){
       if (name.equals(animals.get(index).getName())){
           return index;
       }
    }
    return -1;
}

And use this:

int index =  getIndexWithName(animalList, select);
//Do some validation check
if (index < 0){
    //Give some error message
    System.out.println("That's not a common name I recognize!");
}

answered Jun 22, 2015 at 18:12

Roel Strolenberg

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1 Comment

LKat Over a year ago

Thank you, I used a bit of what you suggested in my own way (had static non-static issues), anyways... I am a novice, so it may not be pretty, but it works... Thanks again

user5033850 · Accepted Answer · 2015-06-22 18:25:51Z

0

 System.out.println("Enter the name of the animal you wish to view: ");
 String select = in.next();
 System.out.println(select); // Check that your toString() its proper
 //There's not a straight way to find each animal's index
// so you 'll have to write an int getIndex(select) method to do that and then you 'll call it
 int index = getIndext(select);
 System.out.println("The animal you chose has index: "+index);

answered Jun 22, 2015 at 18:25

user5033850

Comments

Srini · Accepted Answer · 2015-06-22 18:41:28Z

Create a HashMap<String, Animal> that maps an animal's name to the object.

Scanner in = new Scanner(System.in);
List <Animal> animalList = new ArrayList <Animal>();

Map<String, Animal> map = new HashMap<>();

char ans;

do{  // User input
  Animal animal = new Animal(); // arraylist

 System.out.println("Animal's 'common' name: ");

 String name = in.next();
 animal.setName(name);
 map.put(name, animal);

System.out.println("Animal's class: ");
 animal.setAnmlClass(in.next());

System.out.println("Vertabrate or Invertabrate: ");
 animal.setCharVert(in.next());

System.out.println("Warm or Cold blooded: ");
 animal.setCharBld(in.next());

System.out.println("Animal's habitat (general): ");
 animal.setCharHab(in.next());

System.out.println("Would you like to enter in a new animal (y/n)? ");
 String answer = in.next();
    ans = answer.charAt(0);

animalList.add(animal);

}while(ans == 'y');

System.out.println("Enter the animal you wish to view: ");
 String select = in.next();
 System.out.println(map.get(select));

LKat · Accepted Answer · 2015-06-22 18:52:11Z

With help from Roel & Theodora, I ended up with this. It may not be the cleanest or "prettiest" way to do this, but it works with no issues. System.out.println("Enter the animal you wish to view: "); String select = in.next();

 for (int index = 0; index < animalList.size(); index++){

 if (select.equals(animalList.get(index).getName())){
     System.out.println(index);

     System.out.format("%n"); // aesthetic break
     System.out.print(animalList.get(index).getName()+": ");
     System.out.print(animalList.get(index).getAnmlClass()+", ");
     System.out.print(animalList.get(index).getCharVert()+", ");
     System.out.print(animalList.get(index).getCharBld()+" Blooded, ");
     System.out.print(animalList.get(index).getCharHab()+" Terrain");
     System.out.format("%n"); // aesthetic break
   }
 else{
     System.out.println("The Zoo does not house that animal currently");
 }
 }

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