How do I iterate through a Java Linked Hash map?

Question

I searched this question and found answers that used the Map.Entry like here, however the getValue() method returned an Object object instead of the type of object in the map. Like in the example below, I need it to return a User object so I can use a method from that class. When I tried using the while loop below however, it never leaves the loop. I was wondering the correct way to do this.

Map<String, User> users = new LinkedHashMap<String, User>();
users.put(name, user);
while(users.values().iterator().hasNext()){
 currentUser = users.values().iterator().next();
 currentUser.someMethod();
}

• Matt Ball: The Map.Entry way is the correct way when you need both keys and values. If getValue()'s compile-time return type is an Object then you are likely missing type parameters on the Map.Entry declaration.
• loveToCode: did you try to debug it and see where it gets stuck?
• user1875021: @MattBall Alright I'll try again.
• user1875021: @loveToCode Yeah it just keeps going though the while loop and the .next() method returns the same object everytime

Answer 1

I was wondering the correct way to do this.

You should use the Map.Entry type; you just need to provide type parameters to use with generics:

for (Map.Entry<String,User> entry : users.entrySet()) {
 // entry.getValue() is of type User now
}

• user1875021: Then would I be able to do entry.getValue().someUserMethod();
• user207421: @user1875021 As he's already told you that entry.getValue() is of type User, the answer to that should be obvious. You could always try it.
• Sergey Kalinichenko: @user1875021 Absolutely! entry.getValue() is of type User, you can do anything you would like with it.
• findusl: What is supposed to be the advantage of using entrySet()? He doesn't use the key, so iterating over values() totally get's the job done or?
• Sergey Kalinichenko: @findusl If OP isn't planning on using the key, of course he should use values(); perhaps it would be slightly more economical in terms of CPU cycles. I was under impression that the code inside the loop wasn't a complete fragment, and wanted to leave OP some flexibility to access the key along with the value.

Answer 2

You're misusing the Iterator, and you're omitting the Generics specifications.

Iterator<User> it = users.values().iterator();
while (it.hasNext())
{
 User currentUser = it.next();
 currentUser.someMethod();
}

• user1875021: Ah I see. I tried that, but wasn't familiar about Generics so I kept getting an Object. Thanks

Answer 3

The easiest way to iterare is using for. In a LinkedHashMap it would be like this:

 private static LinkedHashMap<Integer, String> dataBase = new LinkedHashMap<Integer, String>();
 for(Integer key : dataBase.keySet()) {
 String ret = dataBase.get(key);
 }

Source

Answer 4

Each time you call .iterator(), it gives you a brand new iterator at the first element.

You need to call .iterator() once and store it in a local variable.

• user1875021: So like Iterator temp = users.values.iterator();?

Answer 5

change your code for this:

Iterator<User> it = users.values().iterator();
 while (it.hasNext())
 {
 currentUser = it.next();
 currentUser.someMethod(); 
 }

Answer 6

You have to remove "next" from the collection after it has been process. with your code, it is stock at "next"; For example: if your "users.values()" returns a collection i.e

{{john, 1}, {jen, 2}, {peter, 3}}

Your while loops keeps return true and because it keeps returning "{john,1}" which is at the fist index. and so it is an infinite "loop". This is how Iterator works. hasNext() will return true if it has an element at the first index and next() will return that element that hasNext() confirmed it has.

For your loop to progress i think you have to add a line of code or 2. see below:

 Map<String, String> users = new HashMap<String, String>();
 users.put("1", "John");
 users.put("2", "peter");
 users.put("3", "uche");
 Iterator<String> coll = users.values().iterator();
 while(coll.hasNext()){
 String currentUser = coll.next();
 System.out.print(currentUser);
 //coll.remove(currentUser);
 }

hope this helps..

My former code might not be the "right" way to do it but did the job and did not SKIP any second element @ EJP. thanks any way for putting me straight.. :)

• user207421: You don't have to remove anything. You just have to iterate correctly. This code will remove and skip every second element. -1