Question

我正在尝试使用一些参数向服务器发送json请求。请求正在进行，异步任务正常工作，但它在服务器上抛出异常，并说无效的URL

这就是我正在做的事情

 protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity);   

    Button btnChart = (Button) findViewById(R.id.btn_chart);

    // Defining click event listener for the button btn_chart
    OnClickListener clickListener = new OnClickListener() {

        @Override
        public void onClick(View v) {
            new HttpAsyncTask().execute("https://tt.student.com/back.json");
              }
    };


    // Setting event click listener for the button btn_chart of the MainActivity layout
    btnChart.setOnClickListener(clickListener);
 }

public static String POST(String url){
    InputStream inputStream = null;
    String result = "";
    try {

        // 1. create HttpClient
        HttpClient httpclient = getNewHttpClient();

        // 2. make POST request to the given URL
        HttpPost httpPost = new HttpPost(url);

        String json = "";

        // 3. build jsonObject
        JSONObject jsonObject = new JSONObject();
        jsonObject.accumulate("user", 1);
        jsonObject.accumulate("student_id", 1);
        jsonObject.accumulate("user_email", "test@test.com");
        jsonObject.accumulate("from", "Fri Oct 10 12:38:00 2014 GMT+0200");
        jsonObject.accumulate("to", "Sat Oct 11 12:38:00 2014 GMT+0200");

        // 4. convert JSONObject to JSON to String
        json = jsonObject.toString();

        // ** Alternative way to convert Person object to JSON string usin Jackson Lib
        // ObjectMapper mapper = new ObjectMapper();
        // json = mapper.writeValueAsString(person);

        // 5. set json to StringEntity
        StringEntity se = new StringEntity(json);

        // 6. set httpPost Entity
        httpPost.setEntity(se);

        // 7. Set some headers to inform server about the type of the content
        httpPost.setHeader("Accept", "application/json");
        httpPost.setHeader("Content-type", "application/json");

        // 8. Execute POST request to the given URL
        HttpResponse httpResponse = httpclient.execute(httpPost);

        // 9. receive response as inputStream
        inputStream = httpResponse.getEntity().getContent();

        // 10. convert inputstream to string
        if(inputStream != null)
            result = convertInputStreamToString(inputStream);
        else
            result = "Did not work!";

    } catch (Exception e) {
        Log.d("InputStream", e.getLocalizedMessage());
    }

    // 11. return result
    return result;
}
private static String convertInputStreamToString(InputStream inputStream) throws IOException{
    BufferedReader bufferedReader = new BufferedReader( new InputStreamReader(inputStream));
    String line = "";
    String result = "";
    while((line = bufferedReader.readLine()) != null)
        result += line;

    inputStream.close();
    return result;

}
private class HttpAsyncTask extends AsyncTask<String, Void, String> {
    @Override
    protected String doInBackground(String... urls) {

        return POST(urls[0]);
    }
    // onPostExecute displays the results of the AsyncTask.
    @Override
    protected void onPostExecute(String result) {
        Toast.makeText(getBaseContext(), "Received!", Toast.LENGTH_LONG).show();
        Log.d(TAG,result);
    }
}

在网址中，我尝试了两种方式，一种是在上面的代码中，另一种是在url本身传递参数https://tt.student.com/back.json?user=1&student_id=1&user=testh@test.com&from=Fri Oct 10 12:38:00 2014 GMT+0200&to=Sat Oct 11 12:38:00 2014 GMT+0200 为此，它说网址中的非法字符......

Answer 1

我认为，问题可能出在服务器或Wifi Supplicate状态。您的设备仅连接到wifi，但尚未通过互联网的真实性或技术上更新，以便从服务器交换数据包。

我建议你使用方法来检查设备的连接性，我认为就是这种情况，因为我有类似的状态，我花了将近1-2个小时来使它工作。这里是互联网连接检查链接

Internet COnnection

我希望它会有所帮助。

Answer 2

链接https://tt.student.com/back.json需要一个应该被android信任的证书。可能您可以通过代码接受证书，但您需要该证书。我试图在浏览器中打开，它显示我不受信任的证书....

使用异步任务android发送带参数的json请求

2 个答案: