我正在尝试将数据发送到我的服务器。我创建一个JsonObject,并在创建JsonObjectRequest时将其作为参数传递。它没有给出任何错误,但它没有返回任何东西。和邮递员试过了,效果很好。
这是我的密码:
JSONObject jsonBody = new JSONObject();
try {
jsonBody.put("firstname", "asd");
jsonBody.put("lastname", "asd");
jsonBody.put("id", "1");
} catch (JSONException e) {
e.printStackTrace();
}
//creating a JsonObjectRequest
JsonObjectRequest jsonObjectRequest = new JsonObjectRequest(Request.Method.POST, showPlayersUrl,
jsonBody, new Response.Listener<JSONObject>() {
@Override
public void onResponse(JSONObject response) {
JSONArray players;
try{
players = response.getJSONArray("Players");
} catch (JSONException e) {
e.printStackTrace();
}
}
},
new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
}
});
requestQueue.add(jsonObjectRequest);
}发布于 2018-12-22 15:46:31
好吧,发现问题了。服务器端我并没有接受json格式的数据。只要加上这个就行了:
$_POST = json_decode(file_get_contents('php://input'), true);发布于 2018-12-22 10:16:09
试一试:
RequestQueue queue = Volley.newRequestQueue(this);
private void makeJsonObjReq() {
showProgressDialog();
Map<String, String> postParam= new HashMap<String, String>();
postParam.put("un", "xyz@gmail.com");
postParam.put("p", "somepasswordhere");
JsonObjectRequest jsonObjReq = new JsonObjectRequest(Method.POST,
Const.URL_LOGIN, new JSONObject(postParam),
new Response.Listener<JSONObject>() {
@Override
public void onResponse(JSONObject response) {
Log.d(TAG, response.toString());
msgResponse.setText(response.toString());
hideProgressDialog();
}
}, new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
VolleyLog.d(TAG, "Error: " + error.getMessage());
hideProgressDialog();
}
}) {
@Override
public Map<String, String> getHeaders() throws AuthFailureError {
HashMap<String, String> headers = new HashMap<String, String>();
headers.put("Content-Type", "application/json; charset=utf-8");
return headers;
}
};
jsonObjReq.setTag(TAG);
queue.add(jsonObjReq);
}https://stackoverflow.com/questions/53894696
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