String Array Into HashMap Collection Object

Question

I have the following

 String[] temp;

which returns

red
blue
green

I would like to add the string array into a collection obejct like HashMap so that I could retrieve values in any class like

HashMap hash = New HashMap();
hash.get("red");
hash.get("blue");
hash.get("green");

How can I do this?

Thanks

Update 1

     String str = "red,blue,green";
        String[] temp;
        String delimiter = ",";
        temp = str.split(delimiter);
for (int i = 0; i < temp.length; i++) {
System.out.println(temp[i]);
}

With the above code, I would like to retrieve values based on values in array. E.g. I would like to get the values in from another class by calling hash.get("One"), which would return red, hash.get("Two") which would return blue and so forth.

create map using Map hash = New HashMap<String, String>(); Then hash.put("red", "red"), .... will work — nidhin
– nidhin, Commented Aug 2, 2011 at 6:10
If you create a new Hashmap, it will be empty in the beginning. And what do you want to get from the hashmap? The string 'red'? If you have it already - why do you need a HashMap? — user unknown
– user unknown, Commented Aug 2, 2011 at 6:12
What values do you want to associate with each String? Unless you need values associated you can use a List<String> or Set<String> — Peter Lawrey
– Peter Lawrey, Commented Aug 2, 2011 at 6:12
"One", "Two" as strings? Or 1, 2 as numbers? And why would you need strings? — Bozho
– Bozho, Commented Aug 2, 2011 at 17:10

Swagatika · Accepted Answer · 2011-08-02 06:29:07Z

3

Map<String, String> hash = new HashMap<String, String>();    
for(i = 0 ; i < temp.length(); i++)
{
   hash.put(temp[i], temp[i]);
}

Then you can retrieve from map hash.get (temp[i]);

answered Aug 2, 2011 at 6:29

Swagatika

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3 Comments

Jacob Over a year ago

I tried your solution. I have a question, could it be possible to be make more generic i.e. if I calling hash.get from another class I wouldn't be able to call with temp[i].

Swagatika Over a year ago

Yes, you can call from another class, if your hash is public field. But to get the values in the hash you need to know the Key.So what i mean to say whether you call from same class or another class, you need to know the key, probably you do like hash.get("red") , where red you get as an input in another class.

Jacob Over a year ago

I resolved the issue of key value with an integer and iterating integer value inside loop when I am adding values to HashMap. Thanks

ngesh · Accepted Answer · 2011-08-02 06:12:04Z

-1

HashMap<String, String>() map = new HashMap<String, String>();
map.put(temp[i], temp[i]);//here i have considered key as the value itself.. u can use something else //also.

answered Aug 2, 2011 at 6:12

ngesh

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Comments

djna · Accepted Answer · 2011-08-02 06:33:45Z

My doubt how I do map temp[i] with red, blue or green?

Using a hash map won't solve this problem directly. Currently you need to write

 temp[ someNumberHere ];

so

 temp[ 1 ];

yields a String "blue"

If you have a hashMap then instead you might write

 myColourMap.get( someNumberHere );

so

 myColourMap.get( 1 );

Would yield "blue". In either case you are converting a value to a corresponding string, but you do need to know that "someNumber". If you want "blue" you need to know to ask for number 1.

It may be that what you need is to use nicely named constant values:

public Class Colours {

     public static final int RED = 0;
     public static final int BLUE = 1;
     public static final int GREEN = 1; 

     // plus either the array of strings or the hashMap

     public statuc String getColour(int colourNumber ) {
           return myArray[colourNumber]; // or myMap.get(colourNumber)
     }

}

Your clients can now write code such as

 Colours.getColour( Colour.RED );

[It is better to use enums than just raw ints, but let's not divert from arrays and hashMaps right now].

Now when might you prefer a hashMap instead of an array? Consider that you might have more colours, for example 12695295 might be "light pink" and 16443110 might be "lavender".

Now you really don't want an array with 16,443,110 entries when you are only using perhaps 500 of them. Now a HashMap is a really useful thing

 myMap.put( Colour.LAVENDER, 16443110 );

and so on.

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