Question

我已经开发了HTTP Post和Get方法。首先，Http post方法清楚地工作并且身份验证成功但第二步我想获取数据信息，我想要选择Jason对象返回null。

遵循主要代码：

    @Override
    protected Void doInBackground(Void... unused) {
        List<NameValuePair> params=new ArrayList<NameValuePair>();
    params.add(new BasicNameValuePair("login","userno"));
    params.add(new BasicNameValuePair("password", "pass"));
    params.add(new BasicNameValuePair("Accept","application/json"));
    params.add(new BasicNameValuePair("Content-type","x-www-form-urlencoded; charset=UTF-8"));


        veri_string=post.httpPost(url, "GET", params, 2000);
        try {
            veri_json=new JSONObject(veri_string);





        } catch (Exception e) {
            e.printStackTrace();
        }
        Log.d("HTTP POST CEVAP",""+veri_json);


        return null;
    }

我的邮政课程代码：

 try {

        if (method == "POST") {

            HttpParams httpParameters = new BasicHttpParams();
            int timeout1 = time;
            int timeout2 = time;
            HttpConnectionParams.setConnectionTimeout(httpParameters, timeout1);
            HttpConnectionParams.setSoTimeout(httpParameters, timeout2);
            DefaultHttpClient httpClient = new DefaultHttpClient(httpParameters);
            HttpPost httpPost = new HttpPost(url); 

            httpPost.setEntity(new UrlEncodedFormEntity(params,"UTF-8"));
            HttpResponse httpResponse = httpClient.execute(httpPost);
            HttpEntity httpEntity = httpResponse.getEntity();
            veri =  httpEntity.getContent();

        } else if (method == "GET") {

            DefaultHttpClient httpClient = new DefaultHttpClient();
            String paramString = URLEncodedUtils.format(params, HTTP.UTF_8);
            url += "?" + paramString;
            HttpGet httpGet = new HttpGet(url);
            HttpResponse httpResponse = httpClient.execute(httpGet);
            HttpEntity httpEntity = httpResponse.getEntity();
            veri =  httpEntity.getContent();            
        }

    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    } catch (ClientProtocolException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }
    try {
        BufferedReader reader = new BufferedReader(new InputStreamReader(
                veri, "iso-8859-1"), 8);
        StringBuilder sb = new StringBuilder();
        String line = null;
        while ((line = reader.readLine()) != null) {
            sb.append(line + "\n");
        }
        veri.close();
        veri_string = sb.toString();
    } catch (Exception e) {
        Log.e("Buffer Error", "Hata " + e.toString());
             }

    return veri_string; // Aldığımız cevabın string halini geri dönüyoruz

MY web api url：http://ogrenci.izmir.edu.tr/api/ogrenciders

{ “ID”：2602， “OgrenciNo”： “”， “OgrenciAdiSoyadi”：NULL， “OgretimYili”：2014， “OgretimDonemi”：1， “AldigiDersIdNo”：0.0 “AldigiSube”：NULL， “OgretimYiliAdi” ：“2014-2015”，“OgretimDonemiAdi”：“GüzDönemi”，“DersKodu”：“CEN 431”，“DersAdi”：“VERİİLETİŞİMİ”，“Sinif”：4，“SinifAdi”：“4.sınıf”， “Vize1”：0 “Vize2”：0 “Vize3”：0 “Vize4”：0 “Vize5”：0 “Vize6”：0 “Vize7”：0 “Vize8”：0，“Vize9 “：0，” Vize10 “：0，” 最终 “：0”，Butunleme “：0”，EkSinav “：0”，Vize1Adi “：” 维泽 70" ， “Vize2Adi”：“测验 64" ， “Vize3Adi”：“分配

最后我就是这样的错误 enter image description here

Answer 1

在你的Asynctask doInBackground方法中，你必须返回你想要获得的东西。在您的情况下，您需要json字符串

try {
    veri_json=new JSONObject(veri_string);
}

目前你正在返回null，所以这个变量永远不会传回给调用者类。

此外，您还希望在＆＃34; onPostExecute＆＃34;中返回此变量。方法。所以doInBackground返回一些东西（一个String），然后onPostExecute接收字符串（它自动，因为它们是Asynctask的Override方法），然后相同的onPostExecute方法将结果（接收到的String）返回给调用者类，通常是接口

Android Http GET方法请求

1 个答案: