Question

我编写了一个程序，将JSON从我的Android设备发送到我的服务器（XAMPP）。我使用表单测试了我的PHP代码，并且它正确地接收了数据。

另一方面，我的应用程序不向我的服务器发送任何数据。执行var_dump($_POST)会在服务器端返回array(0)。

这是我的安卓代码：

public BackGround(Context context){
    this.context=context;
}

@Override
protected String doInBackground(String... params){

    String location_url ="http://192.168.1.90/server_connection.php";
    try{

        URL url1=new URL(location_url);
        HttpURLConnection httpURLConnection =(HttpURLConnection)url1.openConnection();
        httpURLConnection.setRequestMethod("POST");
       // httpURLConnection.setRequestProperty("Content-Type", "application/json");
        httpURLConnection.setDoOutput(true);
        httpURLConnection.setDoInput(true);
        httpURLConnection.setRequestProperty("Content-Type", "application/json;charset=UTF-8");
        httpURLConnection.setRequestProperty("Accept", "application/json");

        OutputStream stream_upload=httpURLConnection.getOutputStream();
        BufferedWriter buffer_writer=new BufferedWriter(new OutputStreamWriter(stream_upload,"UTF-8"));
        String PostData= URLEncoder.encode(String.valueOf(params));

        buffer_writer.write(PostData);

        buffer_writer.flush();
        buffer_writer.close();
        stream_upload.close();

        int HttpResult = httpURLConnection.getResponseCode();
        if (HttpResult == HttpURLConnection.HTTP_OK) {
            InputStream stream_dawnload = httpURLConnection.getInputStream();

            BufferedReader bufferreader = new BufferedReader(new InputStreamReader(stream_dawnload, "iso-8859-1"));
            String result = "";
            String line;
            while ((line = bufferreader.readLine()) != null) {
                result += line;
            }
            bufferreader.close();
            stream_upload.flush();
            stream_dawnload.close();
            httpURLConnection.disconnect();

            return result;
        }

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

    return null;
}

我使用此代码发送数据，其中send_json是我的JSON对象：

backGround.execute(String.valueOf(send_json));

我认为问题是由JSON对象未正确插入POST请求正文引起的。

我知道有类似问题的问题，但没有一个解决方案帮助过我。

我感谢任何帮助。

Answer 1

这里有几点你应该注意。

首先，正如您所提到的，问题似乎主要是因为您的JSON数据未正确格式化为您的请求。您正在将数据传递给具有AsyncTask参数的varargs doInBackground(String... params).：String这会将您在调用execute(send_json)时传递的String[]对象转换为一个String.valueOf(params)个对象。调用AsyncTask不会返回与您最初传递给String的数据类似的内容，而是返回包含您的数据的HTTP数组的文本表示。

其次，我强烈建议您使用第三方库来为您处理网络代码。为了正确管理HTTP连接，您必须编写的代码既不简单也不简洁。那里有更好的选择。 OkHttp是一个很好的例子，我强烈建议您使用它。

顺便说一下，我并不劝你知道如何在java / android中创建和管理function firstNonRepeatingLetter(s) { //input string //return first character that doesn't repeat anywhere else. //parent for loop points to the char we are analyzing //child for loop iterates over the remainder of the string //if child for loop doesnt find a repeat, return the char, else break out of the child for loop and cont if (s.length == 1) { return s;} parent_loop: for (var i = 0; i < s.length - 1; i++){ //parent loop var elem = s[i]; child_loop: for (var j = i + 1; j < s.length; j++){ if (elem == s[j]){ break child_loop; } } return s[i]; } return ""; } console.log(firstNonRepeatingLetter('stress')); // should output t, getting s.连接。您绝对应该尝试了解发送和接收数据时发生的情况。我只是说你不需要自己处理所有这些复杂性，因为有些库可以做得非常好。

Answer 2

阅读this question's solution后，我编辑了我的代码：首先，我从我的服务中删除了backGround.execute(String.valueOf(send_json));，并将vriables作为参数发送到asyncTask，如下所示：

backGround.execute(String.valueOf(mylocation.getLatitude()),String.valueOf(mylocation.getLongitude()));

然后在asyncTask类中创建我的json对象。这是我的doInbackground的代码：

 @Override
protected String doInBackground(String... params){


    String location_url ="http://192.168.1.90/connection.php";


try{
        URL url1=new URL(location_url);

        HttpURLConnection httpURLConnection =(HttpURLConnection)url1.openConnection();

        httpURLConnection.setRequestMethod("POST");
        httpURLConnection.setDoOutput(true);
        httpURLConnection.setDoInput(true);
        httpURLConnection.setRequestProperty("Content-Type", "application/json");
        httpURLConnection.setRequestProperty("Accept", "application/json");
        httpURLConnection.setRequestProperty("Content-Type", "application/json;charset=UTF-8");



        OutputStream stream_upload=httpURLConnection.getOutputStream();
        BufferedWriter buffer_writer=new BufferedWriter(new OutputStreamWriter(stream_upload,"UTF-8"));

    JSONObject obj = new JSONObject();

    try {
        obj.put("Latitude", params[0]);
        obj.put("Longitude", params[1]);



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

    httpURLConnection.connect();

    buffer_writer.write(obj.toString());
    buffer_writer.flush();
    buffer_writer.close();
    stream_upload.close();





        int status = httpURLConnection.getResponseCode();
        if(status ==httpURLConnection.HTTP_OK) {
            InputStream stream_dawnload = httpURLConnection.getInputStream();
            BufferedReader bufferreader = new BufferedReader(new InputStreamReader(stream_dawnload, "iso-8859-1"));
            String result = "";
            String line;
            while ((line = bufferreader.readLine()) != null) {
                result += line;
            }
            bufferreader.close();
            stream_upload.flush();
            stream_dawnload.close();
            httpURLConnection.disconnect();

            return result;
        }


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


    return null;

}

并且最重要的一点是：如果你想将json对象发送到服务器，你应该添加这个代码：

$json = file_get_contents('php://input');

echo (file_get_contents('php://input'));//for being aware of what is in the object.
 $request = json_decode($json, true);

并使用json的值：

$location_lat=$request['Latitude'];
$location_long=$request['Longitude'];

还有一件事，我尝试了$location_lat=$_POST['Latitude']，并且我再次提出了这个错误：未定义的索引。因此我使用了json对象，这种方法不正确。希望有用。

未定义的索引，android不会将数据发送到php代码

2 个答案: