我正在尝试解析如下所示的单个JSON对象:
{
"message":"Request Successful",
"data":{
"id":"g8nEDt",
"name":"Twins Bazil Twins",
"nameDisplay":"Twins Bazil Twins",
"abv":"6.75",
"isOrganic":"N",
"description":"Beers",
}
},
"status":"success"
}
这是我正在使用的代码。
public class randomBeer extends Activity {
TextView name1;
TextView description1;
TextView abv1;
TextView ibu1;
Button Btngetdata;
//URL to get JSON Array
private static String urlRandom = "http://api.brewerydb.com/v2/beer/random?key=mykey";
//JSON Node Names
private static final String TAG_DATA = "data";
private static final String TAG_NAME = "name";
private static final String TAG_DESCRIPTION = "description";
private static final String TAG_ABV = "abv";
private static final String TAG_IBU = "ibu";
JSONObject data = null;
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.randombeer);
Btngetdata = (Button)findViewById(R.id.getdata);
Btngetdata.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View view) {
new JSONParse().execute();
}
});
}
private class JSONParse extends AsyncTask<String, String, JSONObject> {
private ProgressDialog pDialog;
@Override
protected void onPreExecute() {
super.onPreExecute();
name1 = (TextView)findViewById(R.id.name);
description1 = (TextView)findViewById(R.id.description);
abv1 = (TextView)findViewById(R.id.abv);
ibu1 = (TextView)findViewById(R.id.ibu);
pDialog = new ProgressDialog(randomBeer.this);
pDialog.setMessage("Getting Data ...");
pDialog.setIndeterminate(false);
pDialog.setCancelable(true);
pDialog.show();
}
@Override
protected JSONObject doInBackground(String... args) {
JSONParser jParser = new JSONParser();
// Getting JSON from URL
JSONObject json = jParser.getJSONFromUrl(urlRandom);
return json;
}
@Override
protected void onPostExecute(JSONObject json) {
pDialog.dismiss();
try {
// Getting JSON Array
data= json.getJSONArray(TAG_DATA);
JSONObject c = data.getJSONObject(0);
// Storing JSON item in a Variable
String name = c.getString(TAG_NAME);
String ibu;
if(c.has("ibu")) {
ibu = c.getString(TAG_IBU);
} else {
ibu = "No ibu value";
}
String abv;
if(c.has("abv")) {
abv = c.getString(TAG_ABV);
} else {
abv = "No abv value";
}
String description;
if(c.has("description")) {
description = c.getString(TAG_DESCRIPTION);
} else {
description = "No description available";
}
//Set JSON Data in TextView
name1.setText(name);
description1.setText(description);
abv1.setText(abv);
ibu1.setText(ibu);
} catch (JSONException e) {
e.printStackTrace();
}
}
}
}
问题是它试图获取一个数组,但我只需要和对象。
如何转换此代码,使其获取对象而不是尝试获取数组?
我试图改变
data = json.getJSONArray(TAG_DATA);
至
data = json.getJSONObject(TAG_DATA);
但是然后就行了
JSONObject c = data.getJSONObject(0);
我得到一个错误:
(87,51)错误:不兼容的类型:int无法转换为
串。
最佳答案
它应该是
data= json.getJSONObject(TAG_DATA);
代替
data= json.getJSONArray(TAG_DATA);
在onPostExecute中。现在有
"data":{
"id":"g8nEDt",
"name":"Twins Bazil Twins",
"nameDisplay":"Twins Bazil Twins",
"abv":"6.75",
"isOrganic":"N",
"description":"Beers",
}
在里面。要获取即“名称”,请使用
String name = data.getString("name");
此外:
使用此链接时,请确保执行HttpGet而不是HttpPost
http://api.brewerydb.com/v2/beer/random?key=mykey