java 将 Json 解析为 String android studio
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Parse Json to String android studio
提问by cofudayffy
I have this JSON object:
我有这个 JSON 对象：
{
"1":{
"id_module":"f83d6101cc",
"adresse_mac":"00:6A:8E:16:C6:26",
"mot_de_passe":"mp0001","name":"a"
},
"2":{
"id_module":"64eae5403b",
"adresse_mac":"00:6A:8E:16:C6:26",
"mot_de_passe":"mp0002",
"name":"a"
}
}
And I would like to parse and get to string id_module, adresse_mac, mot_de_passeand namefor each thing the 1 and the 2.
我想解析并获取 string id_module, adresse_mac,mot_de_passe以及name1 和 2 的每一件事。
So I made this but it's not working :
所以我做了这个，但它不起作用：
TextView txt1=(TextView) findViewById(R.id.textView);
String ajout1 = "http://";
JSONObject json = null;
String str = "1";
HttpResponse response;
HttpClient myClient = new DefaultHttpClient();
HttpPost myConnection = new HttpPost(ajout1);
try {
response = myClient.execute(myConnection);
str = EntityUtils.toString(response.getEntity(), "UTF-8");
} catch (ClientProtocolException e) {
e.printStackTrace();
} catch (IOException e) {
e.printStackTrace();
}
try {
JSONObject jsonObject = new JSONObject(str);
String MAC = jsonObject.getString("id_module");
txt1.setText(MAC);
} catch (JSONException e) {
e.printStackTrace();
}
回答by Shaishav Jogani
You should try this:
你应该试试这个：
String str = "your json string";
JSONObject json = new JSONObject(str);
String module = json.getJSONObject("1").getString("id_module");
String address = json.getJSONObject("1").getString("adresse_mac");
String module2 = json.getJSONObject("2").getString("id_module"); //from 2nd object
回答by damson
You may also check Hymansonfor a performance bonus.
您也可以查看Hymanson的绩效奖金。
Tutorial: HymanSon in 5 minutes
教程：Hyman逊 5 分钟
or Gsonby Google for a kinda more elegant api.
或Google 的Gson以获得更优雅的 api。
回答by tim687
回答by Daniel Krzyczkowski
Try this:
试试这个：
String json = {"1":{"id_module":"f83d6101cc","adresse_mac":"00:6A:8E:16:C6:26","mot_de_passe":"mp0001","name":"a"},"2":{"id_module":"64eae5403b","adresse_mac":"00:6A:8E:16:C6:26","mot_de_passe":"mp0002","name":"a"}}
Then:
然后：
JSONObject obj = new JSONObject(json);
Hope it will help.
希望它会有所帮助。
