Question

我尝试使用POST方法发送数据并获取JSONArray数组，但我没有从服务器获取数据，当我测试具有特定值的php文件时它就没问题了但是当我使用POST方法发送密钥然后不提供任何数据时。

这是我的JSONArray请求：

final JsonArrayRequest jsonArrayRequest = new JsonArrayRequest(Request.Method.POST, jsonURL, (String) null,
            new Response.Listener<JSONArray>() {
                @Override
                public void onResponse(JSONArray response) {
                    int count = 0;
                    while (count < response.length()){
                        try {
                            JSONObject jsonObject = response.getJSONObject(count);
                            DataProvider dataProvider = new DataProvider(jsonObject.getString("Name"),
                                    jsonObject.getString("Url"),

                            arrayList.add(dataProvider);
                            count++;

                        } catch (JSONException e) {

                            Toast.makeText(context,"Error",Toast.LENGTH_SHORT).show();
                            e.printStackTrace();
                        }
                    }
                    CarFragment.adapter.notifyDataSetChanged();
                }
            }, new Response.ErrorListener() {
        @Override
        public void onErrorResponse(VolleyError error) {

        }
    }
    ){
        @Override
        protected Map<String, String> getParams() throws AuthFailureError {

            Map<String,String> param = new HashMap<String,String>();
            param.put("id","01675904620");
            return param;
        }
    };
    Singletone.getSingletone(context).addToRequest(jsonArrayRequest);

这是我的PHP代码：

<?php

$user_id = $_POST["id"];

$db_name = "andr";
$user_name = "user";
$password = "12345";
$server = "localhost";

$con = mysqli_connect($server,$user_name,$password,$db_name);


$result = mysqli_query($con, "SELECT * FROM user WHERE user_id = ".$user_id.";") or die("Error: " . mysqli_error($con));

$response = array();

while($row = mysqli_fetch_array($result))
{
    array_push($response, array("Name"=>$row["name"],"Url"=>$row["image_url"]));
}

echo json_encode($response);

?>

Answer 1

这是使用volley

向服务器发送数据的方法

StringRequest stringRequest = new StringRequest(Request.Method.POST, REGISTER_URL,
            new Response.Listener<String>() {
                @Override
                public void onResponse(String response) {
                    Toast.makeText(MainActivity.this,response,Toast.LENGTH_LONG).show();
                }
            },
            new Response.ErrorListener() {
                @Override
                public void onErrorResponse(VolleyError error) {
                    Toast.makeText(MainActivity.this,error.toString(),Toast.LENGTH_LONG).show();
                }
            }){
        @Override
        protected Map<String,String> getParams(){
            Map<String,String> params = new HashMap<String, String>();
            params.put("ID","1234");
            return params;
        }

    };

    RequestQueue requestQueue = Volley.newRequestQueue(this);
    requestQueue.add(stringRequest);

上面您可以看到，一旦您将ID发布到服务器，服务器就会将其响应发送到您的应用，并在String response处可用。您可以进一步处理此响应并获取您想要的任何内容。

JSONObject jsonObject=null;
    try {
        jsonObject = new JSONObject(json);//json is the response you get from server
        users = jsonObject.getJSONArray(JSON_ARRAY);//JSON_ARRAY is the name of your array

        ids = new String[users.length()];


        for(int i=0;i<users.length();i++){
            JSONObject jo = users.getJSONObject(i);
            ids[i] = jo.getString("ID");

        }
    } catch (JSONException e) {
        e.printStackTrace();
    }

更多参考：

https://www.simplifiedcoding.net/android-volley-tutorial-to-get-json-from-server/ https://www.simplifiedcoding.net/android-volley-post-request-tutorial/

谢谢...... Happie Coding

使用volley获取没有用于使用JSONArray发送POST请求的数据

1 个答案: