6

AndroidでHttpPostメソッドを使用して情報を送信する方法についての情報を探していましたが、常にこれが表示されます:

HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost(posturl);

List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("Name","Var1"));
params.add(new BasicNameValuePair("Name2","Var2"));

httppost.setEntity(new UrlEncodedFormEntity(params));    
HttpResponse resp = httpclient.execute(httppost);
HttpEntity ent = resp.getEntity();

問題は、XML 形式の文字列を受け取るリソースに接続する必要があるため、それができないことです。

を使用せずに文字列のみを送信する方法についてのアイデアList<nameValuePair>

4

3 に答える 3

-1 
// Sending HTTPs Requet to Server

    try {
        Log.v("GG", "Sending sever 1 - try");
        // start - line is for sever connection/communication
        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost(
                "10.0.0.1/abc.php");

        List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(1);
        nameValuePairs.add(new BasicNameValuePair("qrcode", contents));


        httppost.setEntity(new UrlEncodedFormEntity(
                nameValuePairs));
        HttpResponse response = httpclient.execute(httppost);
        HttpEntity entity = response.getEntity();
        // end - line is for sever connection/communication
        InputStream is = entity.getContent();

        Toast.makeText(getApplicationContext(),
                "Send to server and inserted into mysql Successfully", Toast.LENGTH_LONG)
                .show();

        // Execute HTTP Post Request
        response= httpclient.execute(httppost);

        entity = response.getEntity();
        String getResult = EntityUtils.toString(entity);
        Log.e("response =", " " + getResult);



    } catch (Exception e) {
        Log.e("log_tag", "Error in http connection "
                + e.toString());
    }
于 2014-01-22T09:04:22.450 に答える