I have a HashMap like so
Map<String , String>
when I get an element from the map by key it returns
["6"] I need only the 6 , not ["6"]
The value is added like so
map.put(k , jsonarray.toString()) ;
Thanks
4 Answers
Then you have 2 options change your map to
Map<String,Integer>
and then you'll need to put an int value inside
or when getting the value use:
Integer.valueOf(map.get("key"));
thats only valid in case your String inside the json is only an int - our you will get an exception!
3 Comments
Yarpole Cosgrave
I feel chaning the map would be best.
Yarpole Cosgrave
is it possible to put a json array there as in Interger ?
J. N
Youll have to parse it first before u put it inside the map
You can use Integer#parseInt(String) to convert a String to an Integer
int value = Integer.parseInt(map.get("key"));
1 Comment
Yarpole Cosgrave
This approach does not work.Unless you parse the string first. on first pass the [ is evaluated and returns an error as [ in not an integer , same for " .. as highlighted by @J. N
You get only what you actually put in as value in the Map. So instead of using toString() while storing, you could put in the just the value which is 6 and for the multiple values you could put in an array of integer values.
Comments
Change your Map to String and JsonArray and store JsonArray directly instead of converting it to String
Map<String , JsonArray>
Using your Map key, iterate your JsonArray object
JsonArray jsonArray;
Iterator<JsonElement> it = jsonArray.iterator();
while(it.hasNext()){
System.out.println(it.next());
}
1 Comment
Yarpole Cosgrave
I think in most cases this would be the one of the best answers, But for my use case it moves json parsing from the parsing layer to the transport layer
jsonarray.toString()sample ?