Printing a HashMap that contains of String and List of Objects

Question

How can I print a HashMap that containts of String and List of Objects?

HashMap<String, List<Object>> Map = new HashMap<String, List<Object>>();

Each object has a name and a key value so I need to get something like:

StringFirst = [ firstObject 15, second object 31, thirdobject 16]
StringSecond = [ fourthObject 15, fifthObject 31, sixthObject 16]

    public static class Graph {

    public class Edge extends Graph {
        String edgeName;
        int time;

        Edge(String edgeName, int time){
            this.edgeName = edgeName;
            this.time = time;
        }

        public void Out(){
            System.out.println(this.edgeName);
            System.out.println(this.time);
        }
    }

Capitalized Map is an interface, you might be having issues with that as well. — Patrick J Abare II
– Patrick J Abare II, Commented May 14, 2015 at 18:26

shruti1810 · Accepted Answer · 2015-05-26 08:35:31Z

4

Try this out for printing the map:

for(Map.Entry<String, ArrayList<Edge>> e : map.entrySet()){
   for(Edge e1 : e.getValue())
      System.out.println(e.getKey() + " = "+ e1.Out());
}

edited May 26, 2015 at 8:35

answered May 14, 2015 at 18:28

shruti1810

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Comments

Patrick J Abare II · Accepted Answer · 2015-05-14 18:25:51Z

0

The easiest way to iterate through a map would be to use the keyset.

for(String key: map.keyset()) { System.out.println(key + " = " + map.get(key)); }

On the .get(key) you may need to handle it differently, or iterate through the returned list

Alternately this is functionally the same,

for(Map.Entry e : map.entrySet()) { System.out.println(e.getKey() + " = "+ e.getValue()); }

answered May 14, 2015 at 18:25

Patrick J Abare II

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2 Comments

Влад Курочкин Over a year ago

It prints as in comment i answered upper. Problem is that for each object of list i need to call method that return its name

Patrick J Abare II Over a year ago

You need to override the toString() if you want it to display in a different way or iterate through the returned list.

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