使用LoopJ AndroidAsyncHttp检索值

罗汉

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

但是,我现在正在尝试使用get请求获得响应,而且我似乎无法理解为什么没有调用onSuccess和onFailure方法。我在这里浏览了所有问题,似乎找不到解决onSuccess方法新实现的问题。有人可以帮忙吗?

在ButtonClick上调用的方法:

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 Client可以正常工作,我将输出发布到EDIT2中):

<?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进行相同的操作,但这也不起作用。真希望能得到答案!

编辑:添加有效的代码,并在同一活动中完美地工作。这是一个发帖请求:

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:响应从php页面返回到POSTMAN客户端:

[{“ id”:“ 7”,“ path”:“ uploads / speed.png”},{“ id”:“ 8”,“ path”:“ uploads / Untitled.png”},{“ id”: “ 9”,“ path”:“ uploads / Untitled.png”},{“ id”:“ 10”,“ path”:“ uploadsspeed_2.png”},{“ id”:“ 11”,“ path”: “ uploads / speed_3.png”}]

马特·奥(Matt O)

看起来您的服务器正在返回JSON数组,因此您应该尝试在JsonHttpResponseHandler中重写onSuccess(JSONArray array)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...
    });

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