Android http發布+ Web服務PHP

Question

我在PHP中有一個返回字符串的Web服務，他引用了兩個參數id和te。 我已經在mozzila插件海報上測試了它的使用，因此決定將其用於我的android應用程序。

這是我的android代碼：

final String query = null;

AsyncHttpClient client = new AsyncHttpClient();
RequestParams rp = new RequestParams();
rp.put("id", num);
rp.put("te", tab);
Log.i("http","before send\n");
client.post("http://appdomain.hol.es/webService.php",rp, new JsonHttpResponseHandler(){

    public void onSuccess(String jObject)
    {    
        query.replace(query, jObject);
Log.i("http","recived: "+jObject+"\n");

    }   
    public void onFailure(Throwable arg0)
    {
Log.i("http","fail");   
    }
});

我正在調試log.i，我已經看到它沒有顯示既沒有記錄也沒有失敗。 有人可以幫我嗎？

PD：我離開了最相關的webService

$id = $_POST["id"];
$te = $_POST["te"]; 
$query = "SELECT `preg` , `respA` , `respB` , `respC` , `respD` , `respV`FROM `".$te."` WHERE `id` =".$id;

$resultado= mysql_query($query,$link);
$arraySalida = array();
while($registro = mysql_fetch_assoc ($resultado) ):
    $cadena = "{$registro['preg']};{$registro['respA']};{$registro['respB']};{$registro['respC']};{$registro['respD']};{$registro['respV']}";
    $arraySalida[]= $cadena;

endwhile;
echo implode(":",$arraySalida); 

@jaimin的解決方案有效，但是編譯器說：類型不匹配：無法在（！）中從AsyncTask轉換為String

這是代碼：

public String BBDD(int num, String tab)
    {
        HttpAsyncTask httpAsyncTask = new HttpAsyncTask(String.valueOf(num),tab);
        /*(!)*/String resul = httpAsyncTask.execute("http://opofire.hol.es/webServiceOpoFire.php");

        return resul;
}

Answer 1

查看下面的代碼，將數據發布到php

class PlaceOrder extends AsyncTask<Void, Void, Void> {

    @Override

    protected Void doInBackground(Void... params) {

    // TODO Auto-generated method stub

    try {

    HttpClient httpClient = new DefaultHttpClient();

    HttpPost httpPst = new HttpPost(

    "http://appdomain.hol.es/webService.php");

    ArrayList<NameValuePair> parameters = new ArrayList<NameValuePair>(

    2);
    // add ur parameter here
    parameters.add(new BasicNameValuePair("id", value1);
    parameters.add(new BasicNameValuePair("te", value2);
    httpPst.setEntity(new UrlEncodedFormEntity(parameters));

    HttpResponse httpRes = httpClient.execute(httpPst);



    String str=convertStreamToString(httpRes.getEntity().getContent()).toString();

    Log.i("mlog","outfromurl"+str);


    } catch (UnsupportedEncodingException e) {

    // TODO Auto-generated catch block

    e.printStackTrace();

    } catch (ClientProtocolException e) {

    // TODO Auto-generated catch block

    e.printStackTrace();

    } catch (IOException e) {

    // TODO Auto-generated catch block

    e.printStackTrace();

    }

    return null;

    }

    }

    public static String convertStreamToString(InputStream is) {

    BufferedReader reader = new BufferedReader(new InputStreamReader(is));

    StringBuilder sb = new StringBuilder();

    String line = null;

    try {

    while ((line = reader.readLine()) != null) {

    sb.append(line + "\n");

    }

    } catch (Exception e) {

    e.printStackTrace();

    } finally {

    try {

    is.close();

    } catch (IOException e) {

    e.printStackTrace();

    }

    }

    return sb.toString();

    }

Answer 2

對於Http Post，我建議您使用AsyncTask <>，它將在與UI分開的線程中運行，這是我兩個月以來一直在使用的代碼，並且工作正常

 private class HttpAsyncTask extends AsyncTask<String, Void, String> {
 private String id,te;
 public  HttpAsyncTask(String id,String te){
        this.id = id;
        this.te = te;

    }

        @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(), "Data Sent!", Toast.LENGTH_LONG).show();
       }
    }
public static String POST(String url){
    InputStream inputStream = null;
    String result = "";
    try {

        // 1. create HttpClient
        HttpClient httpclient = new DefaultHttpClient();

        // 2. make POST request to the given URL
        HttpPost httpPost = new HttpPost(url);
       // pass parameters in this way

       List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(2);
    nameValuePairs.add(new BasicNameValuePair("id", "value "));
    nameValuePairs.add(new BasicNameValuePair("te", "value"));

    //add data
    httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));

        // 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 {
    // TODO Auto-generated method stub

    BufferedReader bufferedReader = new BufferedReader( new InputStreamReader(inputStream));
    String line = "";
    String result = "";
    while((line = bufferedReader.readLine()) != null)
        result += line;

    inputStream.close();
    return result;

}

為了傳遞值，您可以創建這樣的構造函數。

在您的活動中執行此操作

 HttpAsyncTask httpAsyncTask = new HttpAsyncTask(id,te);//this will pass variables values 
 String  ResultfromServer = httpAsyncTask.execute(urlStr);// String ResultfromServer is your response string

在AsyncTask中我創建了構造函數

這對我很有幫助，希望對您也有幫助