How to create json format java

Question

I would like to create JSON format and I have some problem about this

I have no idea to create Jsonarray and JsonObject in JsonObject

{
   "users": [7, 16, 35],
   "group_id": "sdkfjsdkljflds"
}

I try

JSONObject jsonParams = new JSONObject();

try {
    jsonParams.put("group_id", "dlfsdds");
    jsonParams.put("users", list);

} catch (JSONException e) {
  e.printStackTrace();
}

and my log is show

{
   "users":"[7, 16, 35]",
   "group_id":"dlfsdds"
}

ps. list is from

for (int k=0;k<allMember.size();k++){
    list.add(allMember.get(k));
}

what is my mistake? and How to fix it?

thank for your help:D

You probably need to create a JSONArray from the list, and then add that. Otherwise the list will be converted to a string when it is added to the JSONObject. — Will Richardson
– Will Richardson, Commented Aug 12, 2017 at 6:26

BackSlash · Accepted Answer · 2017-08-12 06:29:47Z

2

You need to convert the list to JSONArray first:

try {
    jsonParams.put("group_id", "dlfsdds");

    JSONArray listJson = new JSONArray();

    for(int i=0; i<list.size(); i++) {
        listJson.put(list.get(i));
    }


    jsonParams.put("users", listJson);

} catch (JSONException e) {
  e.printStackTrace();
}

Most libraries also allow you to avoid the for-loop and simply do:

JSONArray listJson = new JSONArray(list);

answered Aug 12, 2017 at 6:29

BackSlash

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2 Comments

ARR.s Over a year ago

umm , sorry but it's not correct ,I want to set "users": [7, 16, 35] not "users":["7", "16", "35"]

BackSlash Over a year ago

@ARR.s It means that list is a List<String>. Change it to List<Integer>.

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