我想阅读此JSON
行,但因为它以JSONArray
开头,我有点困惑
"abridged_cast": [
{
"name": "Jeff Bridges",
"id": "162655890",
"characters": [
"Jack Prescott"
]
},
{
"name": "Charles Grodin",
"id": "162662571",
"characters": [
"Fred Wilson"
]
},
{
"name": "Jessica Lange",
"id": "162653068",
"characters": [
"Dwan"
]
},
{
"name": "John Randolph",
"id": "162691889",
"characters": [
"Capt. Ross"
]
},
{
"name": "Rene Auberjonois",
"id": "162718328",
"characters": [
"Bagley"
]
}
],
我只需要使用“名称”并将所有内容保存为一个字符串。 (字符串值将是:Jeff Bridges,Charles Grodin,Jessica Lange,John Randolph,Rene Auberjonois)。
这是我的代码:
try {
//JSON is the JSON code above
JSONObject jsonResponse = new JSONObject(JSON);
JSONArray movies = jsonResponse.getJSONArray("characters");
String hey = movies.toString();
} catch (JSONException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
答案 0 :(得分:58)
如果您在“名称”之后,为什么您的代码段看起来像是试图获取“字符”?
无论如何,这与任何其他列表或类似数组的操作没什么不同:你只需要遍历数据集并获取你感兴趣的信息。检索所有名称应该看起来像这样:
List<String> allNames = new ArrayList<String>();
JSONArray cast = jsonResponse.getJSONArray("abridged_cast");
for (int i=0; i<cast.length(); i++) {
JSONObject actor = cast.getJSONObject(i);
String name = actor.getString("name");
allNames.add(name);
}
(直接输入浏览器,因此未经测试)。
答案 1 :(得分:7)
getJSONArray(attrname)将从该给定属性名称的对象中获取一个数组 在你的情况下,正在发生的是
{"abridged_cast":["name": blah...]}
^ its trying to search for a value "characters"
但你需要进入数组,然后搜索“characters”
试试这个
String json="{'abridged_cast':[{'name':'JeffBridges','id':'162655890','characters':['JackPrescott']},{'name':'CharlesGrodin','id':'162662571','characters':['FredWilson']},{'name':'JessicaLange','id':'162653068','characters':['Dwan']},{'name':'JohnRandolph','id':'162691889','characters':['Capt.Ross']},{'name':'ReneAuberjonois','id':'162718328','characters':['Bagley']}]}";
JSONObject jsonResponse;
try {
ArrayList<String> temp = new ArrayList<String>();
jsonResponse = new JSONObject(json);
JSONArray movies = jsonResponse.getJSONArray("abridged_cast");
for(int i=0;i<movies.length();i++){
JSONObject movie = movies.getJSONObject(i);
JSONArray characters = movie.getJSONArray("characters");
for(int j=0;j<characters.length();j++){
temp.add(characters.getString(j));
}
}
Toast.makeText(this, "Json: "+temp, Toast.LENGTH_LONG).show();
} catch (JSONException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
检查它:)
答案 2 :(得分:1)
这是一种更好的方法。希望这有帮助
protected void onPostExecute(String result) {
Log.v(TAG + " result);
if (!result.equals("")) {
// Set up variables for API Call
ArrayList<String> list = new ArrayList<String>();
try {
JSONArray jsonArray = new JSONArray(result);
for (int i = 0; i < jsonArray.length(); i++) {
list.add(jsonArray.get(i).toString());
}//end for
} catch (JSONException e) {
Log.e(TAG, "onPostExecute > Try > JSONException => " + e);
e.printStackTrace();
}
adapter = new ArrayAdapter<String>(ListViewData.this, android.R.layout.simple_list_item_1, android.R.id.text1, list);
listView.setAdapter(adapter);
listView.setOnItemClickListener(new OnItemClickListener() {
@Override
public void onItemClick(AdapterView<?> parent, View view, int position, long id) {
// ListView Clicked item index
int itemPosition = position;
// ListView Clicked item value
String itemValue = (String) listView.getItemAtPosition(position);
// Show Alert
Toast.makeText( ListViewData.this, "Position :" + itemPosition + " ListItem : " + itemValue, Toast.LENGTH_LONG).show();
}
});
adapter.notifyDataSetChanged();
...