【发布时间】:2015-05-03 08:29:30
【问题描述】:
JSONObject 是这样的:
{"tag":"login","success":1,"error":0,"name":"pb","dir":"DH","br":"LL","gr":"IW","email":"empty"}
logcat 说:
Value Access of type java.lang.String cannot be converted to JSONObject
在旁边运行一个 AsyncTask,它可以工作。为什么 volley-code 说它是一个字符串?来自服务器的答案肯定是 JSONObject.... 为什么说它是 String?
private void makeJsonObjectRequest() {
JsonObjectRequest jsonObjReq = new JsonObjectRequest(Method.POST,
url, null, new Response.Listener<JSONObject>() {
@Override
public void onResponse(JSONObject response) {
Log.d(TAG, response.toString());
try {
String dir = response.getString("dir");
Log.d("dir", dir);
} catch (JSONException e) {
e.printStackTrace();
Log.d("JSONException", e.getMessage().toString());
}
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
Log.d("onErrorResponse", error.getMessage().toString());
}
})
{
@Override
protected Map<String, String> getParams() {
Map<String, String> params = new HashMap<String, String>();
params.put("tag", "login");
params.put("name", name);
params.put("password", password);
return params;
}
@Override
public Map<String, String> getHeaders() throws AuthFailureError {
Map<String,String> params = new HashMap<String, String>();
params.put("Content-Type","application/x-www-form-urlencoded");
return params;
}
};
AppController.getInstance().addToRequestQueue(jsonObjReq);
}
【问题讨论】: