使用LoopJ AndroidAsyncHttp检索值

Question

我刚刚开始使用LoopJ AndroidAsyncHttp库及其出色的数据上传。

但是，我现在正试图使用get请求获得响应，我似乎不明白为什么我的onSuccess和onFailure方法都没有被调用。我在这里仔细研究了这些问题，似乎找不到一个问题来解决onSuccess方法的新实现。有人能帮忙吗？

方法在ButtonClick上被调用：

public void displayUploaded(View view){

    RequestParams params=new RequestParams();


    try{


        AsyncHttpClient client = new AsyncHttpClient();

        client.get("http://192.168.1.4/clientservertest/returnUploadedImages.php",
                    new JsonHttpResponseHandler() {

                    @Override
                    public void onSuccess(JSONObject jsonObject) {
                    // Display a "Toast" message

                    Toast.makeText(getApplicationContext(), "Success!", Toast.LENGTH_LONG).show();


                    Log.d("android", jsonObject.toString());
                    }

                    @Override
                    public void onFailure(int statusCode, Throwable throwable, JSONObject error) {
                    // Display a "Toast" message

                    Toast.makeText(getApplicationContext(), "Error: " + statusCode + " " + throwable.getMessage(), Toast.LENGTH_LONG).show();

                    // Log error message
                    // to help solve any problems
                     Log.e("android", statusCode + " " + throwable.getMessage());
                     }
                     });




    }
    catch(Exception e){
        Toast toast2=Toast.makeText(getApplicationContext(),"Failed first TRY",Toast.LENGTH_LONG);
        toast2.show();
        e.printStackTrace();
    }


}

这是我的php代码(在chrome上使用Postman客户端很好，我已经在EDIT2中发布了输出)：

<?php 

        #Connect to Database 
        $con = mysqli_connect("localhost","root","", "mytestdatabase"); 

        #Check connection 
        if (mysqli_connect_errno()) { 
            echo 'Database connection error: ' . mysqli_connect_error(); 
            exit(); 
        } 


        #Query the database to get the user details. 
        $userdetails = mysqli_query($con, "SELECT * FROM images"); 

        #If no data was returned, check for any SQL errors 
        if (!$userdetails) { 
            echo 'Could not run query: ' . mysqli_error($con); 
            exit; 
        } 

        #Return the results
        $rows = array();
        while($r = mysqli_fetch_assoc($userdetails)) {
            $rows[] = $r;
        } 


        print(json_encode($rows));  

?>

我也尝试过使用其他ResponseHandlers做同样的事情，但这也不起作用。真希望得到答案！

编辑:添加有效的代码，并在相同的活动中完美地工作。这是员额请求：

File selectedPicture=new File(picturePath);

        RequestParams params=new RequestParams();

        try{

           params.put("UploadedPic",selectedPicture);
           AsyncHttpClient client = new AsyncHttpClient();
           client.post("http://192.168.1.4/clientservertest/imageupload.php", params,new AsyncHttpResponseHandler());

        }

EDIT2:从php页面返回到POSTMAN客户端的响应：

{"id":"7“、”路径“：”上载/快速.path“}、{"id":"8”、“路径”：“上载/取消标题.path”}、{"id":"9“、”路径“：”上载/未标题.path“}、{"id":"10”、“路径”：“uploadsspeed_2.png”}、{"id":"11“、”路径“：”上载/速度_3.png“}}

Answer 1

看起来您的服务器正在返回一个JSON数组，因此您应该尝试在您的onSuccess(JSONArray数组)方法中重写该方法。

client.get("http://192.168.1.4/clientservertest/returnUploadedImages.php",
    new JsonHttpResponseHandler() {
        @Override
        public void onSuccess(JSONArray jsonArray) {
            Toast.makeText(getApplicationContext(), "Success!", Toast.LENGTH_LONG).show();
            Log.d("android", jsonArray.toString());
        }
        //etc...
    });