Question

我已编写此代码以将json发送到服务器

protected String doInBackground(Integer...params) {
    // Create a new HttpClient and Post Header
    HttpClient httpclient = new DefaultHttpClient();
    HttpPost httppost = new HttpPost("http://juelminapp.visualsparks.net/ws/login");
    try {
        // Add your data
        JSONObject obj = new JSONObject();
        obj.accumulate("method", "login");
        obj.accumulate("email", "adilshafique@hotmail.com");
        obj.accumulate("password", "Adil2014$");
        obj.accumulate("submit", "Login");
        JSONObject o2 = new JSONObject();
        JSONArray array = new JSONArray();
        array.put(obj);
        o2.put("customer", array);
        StringEntity entity = new StringEntity(o2.toString());
        httppost.setEntity(entity);
        // Execute HTTP Post Request
        HttpResponse response = httpclient.execute(httppost);
        InputStream is = response.getEntity().getContent();
        BufferedReader br = new BufferedReader(new InputStreamReader(is));
        data = br.readLine();
        br.close();
    } catch (ClientProtocolException e) {
        // TODO Auto-generated catch block
    } catch (IOException e) {
        // TODO Auto-generated catch block
    } catch (Exception e) {}
    return null;
}

它提供响应，因为方法未定义。服务器的描述如下网络服务网址

http://juelminapp.visualsparks.net/ws/login

在json要求中发送格式。您需要发送方法发布并将所有json数组分配给客户数组密钥，然后Web服务通过post方法从客户处获取。

即在php

array("customer"=>$jsondata)

然后，Web服务接收此客户帖子值中的所有json数据，发布的字段名称如下：

method => 'login',
email => adilshafique@hotmail.com
password => Adil2014$
submit => Login

第一个是发布的字段名称，第二个是数据。请使用相同的帖子字段名称。

将在我们的数据库中找不到将返回Method undefined，Invalid Password，User ID的消息。

您的帐户尚未激活.Entered主电子邮件存在于数据库中，使用另一个用户名。

如果成功登录，会话ID将返回

PHP

$url = 'http://juelminapp.visualsparks.net/ws/login';
$data = array('method' = > 'login', 'email' = > 'adilshafique@hotmail.com', 'password' = > 'Adil2014$', 'submit' = > 'Login');
$ch = curl_init($url);
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false);
$data_string = urlencode(json_encode($data));
curl_setopt($ch, CURLOPT_CUSTOMREQUEST, "POST");
curl_setopt($ch, CURLOPT_POSTFIELDS, array("customer" = > $data_string));
$result = curl_exec($ch);
curl_close($ch);

Answer 1

问题是当服务器期望作为多部分表单数据时，您正在发送一个简单的http实体。试试这个，

protected String doInBackground(Integer...params) {
    // Create a new HttpClient and Post Header
    HttpClient httpclient = new DefaultHttpClient();
    HttpPost httppost = new HttpPost("http://juelminapp.visualsparks.net/ws/login");
    try {
        // Add your data
        JSONObject obj = new JSONObject();
        obj.accumulate("method", "login");
        obj.accumulate("email", "adilshafique@hotmail.com");
        obj.accumulate("password", "Adil2014$");
        obj.accumulate("submit", "Login");
        String urlEncodedData = URLEncoder.encode(obj.toString(), "utf-8");

        // Changed Here to MultiPartEntity

        MultipartEntity reqEntity = new MultipartEntity(HttpMultipartMode.BROWSER_COMPATIBLE);
        reqEntity.addPart("customer", new StringBody(urlEncodedData));
        httppost.setEntity(reqEntity);


        // Execute HTTP Post Request
        HttpResponse response = httpclient.execute(httppost);
        InputStream is = response.getEntity().getContent();
        BufferedReader br = new BufferedReader(new InputStreamReader(is));
        data = br.readLine();
        br.close();
    } catch (ClientProtocolException e) {
        // TODO Auto-generated catch block
    } catch (IOException e) {
        // TODO Auto-generated catch block
    } catch (Exception e) {}
    return null;
}

这对你有用。希望这有帮助！

在android中将json发送到服务器时出错

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