php - 如何使用Volley通过JSON从mysql获取多行数据

Question

我正在尝试通过 JSON 将下面提到的 php 文件的输出获取到我的 Android 应用程序中，但我只得到我的 Android 应用程序中的第一个条目。我也尝试执行以下步骤，但我能够仅获取第一个输出，但其余输出在我的应用程序中不可见

<?php 
    define('HOST','XXXX');
    define('USER','XXXX');
    define('PASS','XXXX');
    define('DB','XXXX'); 
    $con = mysqli_connect(HOST,USER,PASS,DB) or die('Unable to Connect'); 
    if($_SERVER['REQUEST_METHOD']=='GET'){ 
        $customerEmail = $_GET['customerEmail']; 
        $sql = "SELECT `amount`, `customerEmail`, `CCAvenueOrder_id` FROM `                 OrderAborted` WHERE customerEmail='".$customerEmail."'"; 
        $r = mysqli_query($con,$sql);
        while($res = mysqli_fetch_array($r))
        { 
            $result = array();
            array_push($result,array(
                            "customerEmail"=>$res['customerEmail'],
                            "CCAvenueOrder_id"=>$res['CCAvenueOrder_id'],
                            "amount"=>$res['amount']
                            )
            );
            echo json_encode(array("result"=>$result));
        }
        mysqli_close($con); 
    }
?>

JSON 代码

     private void getData() {
    String id = editTextId.getText().toString().trim();
    if (id.equals("")) {
        Toast.makeText(this, "Please enter an id",         Toast.LENGTH_LONG).show();
        return;
    }
     loading = ProgressDialog.show(this,"Please wait...","Fetching...",false,false);

    String url = Constants.DATA_URL+editTextId.getText().toString().trim();

    StringRequest stringRequest = new StringRequest(url, new Response.Listener<String>() {
        @Override
        public void onResponse(String response) {
            loading.dismiss();
            showJSON(response);
        }
               },
            new Response.ErrorListener() {
                @Override
                public void onErrorResponse(VolleyError error) {
                     Toast.makeText(OrderHistory.this,error.getMessage().toString(),Toast.LENGTH_LONG).show();
                }
            });

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

private void showJSON(String response){
    String name="";
    String address="";
    String vc = "";
    try {
        JSONObject jsonObject = new JSONObject(response);
        JSONArray result = jsonObject.getJSONArray(Constants.JSON_ARRAY);
        JSONObject collegeData = result.getJSONObject(0);
        name = collegeData.getString(Constants.KEY_NAME);
        address = collegeData.getString(Constants.KEY_ADDRESS);
        vc = collegeData.getString(Constants.KEY_VC);
    } catch (JSONException e) {
        e.printStackTrace();
    }
     textViewResult.setText("Name:\t"+name+"\nAddress:\t" +address+ "\nVice        Chancellor:\t"+ vc);
   }

Answer 1

很简单，在完成循环之前不要发送数据，并将数组初始化移到循环之外

<?php 
    define('HOST','XXXX');
    define('USER','XXXX');
    define('PASS','XXXX');
    define('DB','XXXX'); 
    $con = mysqli_connect(HOST,USER,PASS,DB) or die('Unable to Connect'); 
    if($_SERVER['REQUEST_METHOD']=='GET'){ 
        $customerEmail = $_GET['customerEmail']; 
        $sql = "SELECT `amount`, `customerEmail`, `CCAvenueOrder_id` FROM `OrderAborted` WHERE customerEmail='".$customerEmail."'"; 
        $r = mysqli_query($con,$sql);

        $result = array();
        while($res = mysqli_fetch_array($r))
        { 
            //$result = array();
            $result[] = array(
                            "customerEmail"=>$res['customerEmail'],
                            "CCAvenueOrder_id"=>$res['CCAvenueOrder_id'],
                            "amount"=>$res['amount']
                            );
            //echo json_encode(array("result"=>$result));
        }

        echo json_encode(array("result"=>$result));
        mysqli_close($con); 
    }
?>

如果您使用mysqli_fetch_assoc()而不是mysqli_fetch_array()，您还可以简化循环，并准确地从结果集中获取手动构建的数组。

<?php 
    define('HOST','XXXX');
    define('USER','XXXX');
    define('PASS','XXXX');
    define('DB','XXXX'); 
    $con = mysqli_connect(HOST,USER,PASS,DB) or die('Unable to Connect'); 
    if($_SERVER['REQUEST_METHOD']=='GET'){ 
        $customerEmail = $_GET['customerEmail']; 
        $sql = "SELECT `amount`, `customerEmail`, `CCAvenueOrder_id` FROM `OrderAborted` WHERE customerEmail='".$customerEmail."'"; 
        $r = mysqli_query($con,$sql);

        $result = array();
        while($res = mysqli_fetch_assoc($r))
        { 
            $result[] = $res;
            //echo json_encode(array("result"=>$result));
        }

        echo json_encode(array("result"=>$result));
        mysqli_close($con); 
    }
?>

如果您使用mysqli_fetch_all()，甚至可以再次简化它

<?php 
    define('HOST','XXXX');
    define('USER','XXXX');
    define('PASS','XXXX');
    define('DB','XXXX'); 
    $con = mysqli_connect(HOST,USER,PASS,DB) or die('Unable to Connect'); 
    if($_SERVER['REQUEST_METHOD']=='GET'){ 
        $customerEmail = $_GET['customerEmail']; 
        $sql = "SELECT `amount`, `customerEmail`, `CCAvenueOrder_id` FROM `OrderAborted` WHERE customerEmail='".$customerEmail."'"; 
        $r = mysqli_query($con,$sql);

        $result = mysqli_fetch_all($r, MYSQLI_ASSOC);

        echo json_encode(array("result"=>$result));
    }
?>

And to get rid of the SQL Injection vector

<?php 
    define('HOST','XXXX');
    define('USER','XXXX');
    define('PASS','XXXX');
    define('DB','XXXX'); 
    $con = mysqli_connect(HOST,USER,PASS,DB) or die('Unable to Connect'); 
    if($_SERVER['REQUEST_METHOD']=='GET'){ 
        //$customerEmail = $_GET['customerEmail']; 

        $sql = "SELECT `amount`, `customerEmail`, `CCAvenueOrder_id` 
                FROM `OrderAborted` 
                WHERE customerEmail=?"; 

        $stmt = mysqli_prepare($con,$sql);

        $stmt = mysqli_stmt_bind_param ($stmt, 's', $_GET['customerEmail']);

        $status = mysqli_execute($stmt);

        $r = mysqli_stmt_get_result ($stmt);

        $result = mysqli_fetch_all($r, MYSQLI_ASSOC);

        echo json_encode(array("result"=>$result));
    }
?>