我正在尝试将我上传的视频的视频ID和其他信息存储在android中的不同字符串中。现在我已经创建了一个fql查询来获取视频详细信息。我正在使用json解析来提取像这样的值 -
String fqlQuery = "SELECT vid, owner, title, description,updated_time, created_time FROM video WHERE owner=me()";
Bundle params = new Bundle();
params.putString("q", fqlQuery);
Session session = Session.getActiveSession();
Request request = new Request(session,"/fql",params,HttpMethod.GET,new Request.Callback()
{
public void onCompleted(Response response)
{
try
{
JSONObject json = Util.parseJson(response.toString());
JSONArray data = json.getJSONArray( "data" );
for ( int i = 0, size = data.length(); i < size; i++ )
{
JSONObject getVideo = data.getJSONObject( i );
userNameView.setText(getVideo.getString("vid"));
}
}
catch(Exception e){userNameView.setText(e.toString());}
}
});
Request.executeBatchAsync(request);
}
});
但它让我异常 -
org.json.JSONException:第25个字符的未终止对象 {响应:responseCode:200,graphObject:GraphObject {graphObjectClass = GraphObject,状态= { “数据”:[{ “所有者”:...}]}}}
这是我第一次使用android,facebook sdk以及json解析,所以我会非常感谢所提供的任何帮助。谢谢。
答案 0 :(得分:3)
queryButton.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View v) {
String fqlQuery = "SELECT uid, name, pic_square, status FROM user WHERE uid IN " +
"(SELECT uid2 FROM friend WHERE uid1 = me() LIMIT 25)";
Bundle params = new Bundle();
params.putString("q", fqlQuery);
Session session = Session.getActiveSession();
Request request = new Request(session,
"/fql",
params,
HttpMethod.GET,
new Request.Callback(){
@SuppressWarnings("deprecation")
public void onCompleted(Response response) {
GraphObject graphObject = response.getGraphObject();
if (graphObject != null)
{
if (graphObject.getProperty("data") != null)
{
try {
String arry = graphObject.getProperty("data").toString();
JSONArray jsonNArray = new JSONArray(arry);
for (int i = 0; i < jsonNArray.length(); i++) {
JSONObject jsonObject = jsonNArray.getJSONObject(i);
String name = jsonObject.getString("name");
String uid = jsonObject.getString("uid");
String pic_square = jsonObject.getString("pic_square");
String status = jsonObject.getString("status");
Log.i("Entry", "uid: " + uid + ", name: " + name + ", pic_square: " + pic_square + ", status: " + status);
}
} catch (JSONException e) {
// TODO Auto-generated catch block
e.printStackTrace();
}
}
}
}
});
Request.executeBatchAsync(request);
}
});
答案 1 :(得分:1)
最后我能够在这里得到问题,response.toString()没有给出一个纯JSONObject字符串,所以我不得不从response.toString()的输出中获取子字符串,使其看起来像一个JSONObject字符串
int indexex=nthOccurrence(response.getGraphObject().toString(),'{',1);
int index=response.getGraphObject().toString().length()-1;
String edit=response.getGraphObject().toString().substring(indexex, index);
JSONObject json = Util.parseJson(edit);
JSONArray data = json.getJSONArray( "data" );
for ( int i = 0, size = data.length(); i < size; i++ )
{
JSONObject getVideo = data.getJSONObject( i );
userNameView.setText(getVideo.getString("vid"));
}
我想这可能不是正确的方法,但这是我能想到的唯一方法。如果有人知道更好的方式,请回复。谢谢