Question

在我的应用程序中，有一个愿望清单，用户可以通过手机添加一些带有姓名，备注和图像的心愿单项目。我想将这些数据发送到DB的服务器端。格式低于..

 {
  "post": {
    "product name": "somename",
    "note": "description of the product",
    "image": "http://localhost/someimage.jpg",
   },
 }

所以我的问题是如何构建这个JSON数组并将其POST到服务器数据库... 以及如何编写php以从客户端接收POST请求？提前谢谢....

Answer 1

上课：

public class HttpClass
{
public static String postData(String url,List<NameValuePair> params) {
    // Create a new HttpClient and Post Header

    String responseString = "";
    String responsemsg = "";

    try {
        HttpClient httpclient = new DefaultHttpClient();
        //String tempUrl = HungryPagesConfig.registrationAPI;
        HttpPost httppost = new HttpPost(url);
        httppost.setEntity(new UrlEncodedFormEntity(params));
        HttpResponse response = httpclient.execute(httppost);
        responseString = EntityUtils.toString(response.getEntity());
        // Log.e("Rsponse", EntityUtils.toString(response.getEntity()));

        Log.e("Rsponse", responseString);
    } 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 responseString;
}
}

在()：

中发帖mainActivity

public void Fun()

{
JSONObject Json,Mainjson;
String Data;
try {
            Json.put("product name",  "somename");
            Json.put("note", "description of the product");
            Json.put("image","http://localhost/someimage.jpg");
            Mainjson.put("post",Json);
        } catch (JSONException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

        Data = Mainjson.toString();
        Log.e("Rsponse", Data);

        PostData.nameValuePairs = new ArrayList<NameValuePair>();
         PostData.add(new BasicNameValuePair("data", Data));
}

现在调用Fun()函数，无论您希望成功完成任何帖子。

制作另一个课程PostData：

public class PostData 
{

String url;
JSONObject add;
public HttpClass jParser = new HttpClass();
public void post()
{
tempUrl = HungryPagesConfig.AddMenuItemAdmin;
        try {
            add = new JSONObject(jParser.postData(url,
                    nameValuePairs));
            Log.e("Rsponse", add.toString());
        } catch (JSONException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }

}
}

Answer 2

试试这个

HttpClient httpclient = new DefaultHttpClient();
HttpPost httppost = new HttpPost("http://yourwebsite.com.au/index.jsp");
String trackid= getTrackid();
List<NameValuePair> namevaluepairs = new ArrayList<NameValuePair>(2);
namevaluepairs.add(new BasicNameValuePair("columval1","123"));
namevaluepairs.add(new BasicNameValuePair("columval2","145"));
namevaluepairs.add(new BasicNameValuePair("columval3","445"));
httppost.setEntity(new UrlEncodedFormEntity(namevaluepairs));
HttpResponse response = httpclient.execute(httppost);
HttpEntity rp = response.getEntity();
String origresponseText = EntityUtils.toString(rp);
String htmlTextStr = Html.fromHtml(origresponseText).toString();
pass=htmlTextStr.trim();

Android - 如何将数据发布到服务器数据库

2 个答案: