I have problem when trying to parse with minimum value to map in Android.
There some sample JSON format with more information ex:
[{id:"1", name:"sql"},{id:"2",name:"android"},{id:"3",name:"mvc"}]
This that example most common to use and easy to use just use getString("id")
or getValue("name")
.
But how do I parse to map using this JSON format with just only string and value minimum format to java map collection using looping. And because the string json will always different one with another. ex:
{"1":"sql", "2":"android", "3":"mvc"}
Thank
Use the JavaScript function JSON. parse() to convert text into a JavaScript object: const obj = JSON. parse('{"name":"John", "age":30, "city":"New York"}');
parse() JSON parsing is the process of converting a JSON object in text format to a Javascript object that can be used inside a program. In Javascript, the standard way to do this is by using the method JSON. parse() , as the Javascript standard specifies.
JSON. parse() is used to convert String to Object. JSON. stringify() is used to convert Object to String.
The JSON.stringify() method converts a JavaScript value to a JSON string, optionally replacing values if a replacer function is specified or optionally including only the specified properties if a replacer array is specified.
You need to get a list of all the keys, loop over them and add them to your map as shown in the example below:
String s = "{menu:{\"1\":\"sql\", \"2\":\"android\", \"3\":\"mvc\"}}";
JSONObject jObject = new JSONObject(s);
JSONObject menu = jObject.getJSONObject("menu");
Map<String,String> map = new HashMap<String,String>();
Iterator iter = menu.keys();
while(iter.hasNext()){
String key = (String)iter.next();
String value = menu.getString(key);
map.put(key,value);
}
My pseudocode example will be as follows:
JSONArray jsonArray = "[{id:\"1\", name:\"sql\"},{id:\"2\",name:\"android\"},{id:\"3\",name:\"mvc\"}]";
JSON newJson = new JSON();
for (each json in jsonArray) {
String id = json.get("id");
String name = json.get("name");
newJson.put(id, name);
}
return newJson;
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