我知道这个问题已经被问到了,但仍然对我不起作用。我无法将Json对象发送到服务器。我没有从服务器得到任何响应。我在这里犯了什么错误。如果我犯了一个愚蠢的错误,请不要将其标记为重复或阻止我的帐户。我绝对是编程新手。我已附上Satck溢出提供的解决方案。seethis
JsonObject response = new JsonObject();
response.add("Description", a1 );/*a1 is Json array of
multiple values*/
response.add("Amount", a2 );/*a2 is Json array of
multiple values*/
System.out.println("Json:"+response);/*Both the list is saved inside this Json object*/
Map<String,String> hm = new HashMap();
hm.put("Values",String.valueOf(response));/*Trying to pass the json object to server respose is the Json object*/
CustomRequest req = new
CustomRequest(Request.Method.POST,ip,hm,
this.createRequestSuccessListener(),
this.createRequestErrorListener());
Mysingleton.getInstance(this).addTorequestque(req);/*This
is a singleton class*/
public Response.Listener<JSONObject> createRequestSuccessListener(){
return new Response.Listener<JSONObject>() {
@Override
public void onResponse(JSONObject response) {
AlertDialog.Builder al = new AlertDialog.Builder(Database.this)
.setTitle("Message")
.setMessage("Sent")
.setPositiveButton("OK",null);
AlertDialog al1 = al.create();
al.show();
}
};
}
public Response.ErrorListener createRequestErrorListener(){
return new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
AlertDialog.Builder al = new AlertDialog.Builder(Database.this)
.setTitle("Error")
.setMessage("Something went wrong")
.setPositiveButton("OK",null);
AlertDialog al1 = al.create();
al.show();
}
};
}
[CustomRequet][1]
My php source code is here
if(!empty($_POST['Values']){
$all_arraylist = json_decode(Values,true);
$ result = print_r($all_arraylist,true);
echo $result;
}
else{
echo "Empty";
}