Question

我是Android程序员的新手，我希望以下列格式通过JSON向服务器发送数据，并以这种格式实现Json ......还想从服务器获取数据..

网址：http://fort.example.com
注册

{

    "signup": [

    {
    "username": "test1264",
    "password": "1234",
    "email": "example@gmail.com",
    "phoneno": "223344556",
    "altphoneno": "12345678",
    "firstname": "abc",
    "lastname": "xyz"
    }    ]
}

响应：

成功：{"status":1}
失败时：{"status":0}

用户登录的JSON：

{"login":[
      {"username":"test1234",
       "password":"1234"}
       ]}
Response:
on success:  {
"user": [
{
"firstname": "abc",
"lastname": "xyz",
"email": "example@gmail.com",
"phone": "99887766",
"username": "test1234"
}
]
}

失败时：{"error":["Auth error"]}

Answer 1

您可以使用此类代码

// Create a new HttpClient and Post Header
        HttpClient httpclient = new DefaultHttpClient();
        HttpPost httppost = new HttpPost(
                "http://fort.example.com");
        JSONObject json = new JSONObject();

        try {
            // Add your data
            json.put("key", "value");
            StringEntity se = new StringEntity( json.toString());
            httppost.setEntity(se);

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

            BufferedReader reader = new BufferedReader(
                    new InputStreamReader(
                            response.getEntity().getContent(), "UTF-8"));
            String jsonString = reader.readLine();
            JSONTokener tokener = new JSONTokener(jsonString);

            JSONObject finalResult = new JSONObject(tokener);

不要忘记将此添加到nonUI线程。

Answer 2

您可以使用droidQuery轻松完成此操作：

$.ajax(new AjaxOptions().url("http://fort.example.com")
                        .type("POST")
                        .data("\"signup\": [{" +
                              "\"username\": \"test1264\"," +
                              "\"password\": \"1234\"," +
                              "\"email\": \"example@gmail.com\"," +
                              "\"phoneno\": \"223344556\"," +
                              "\"altphoneno\": \"12345678\"," +
                              "\"firstname\": \"abc\"," +
                              "\"lastname\": \"xyz\"" +
                        "\"}]")
                        .dataType("json")
                        .success(new Function() {
                            @Override
                            public void invoke($ d, Object... args) {
                                JSONObject json = (JSONObject) args[0];
                                boolean success = json.getBoolean("status");
                                if (success) {
                                    //handle success
                                }
                                else {
                                    //handle error
                                }
                            }
                        })
                        .error(new Function() {
                            @Override
                            public void invoke($ d, Object... args) {
                                AjaxError error = (AjaxError) args[0];
                                Log.i("Ajax", "Error " + error.status + ": " + error.reason);
                            }
                        }));

您可以针对其他情况重复此逻辑，例如您的登录Web服务。

Answer 3

PostMethod method = new PostMethod();
org.apache.commons.httpclient.URI newUri = new org.apache.commons.httpclient.URI(URIstring, true);
method.setURI(newUri);
method.setRequestBody(JsonObj().toString());
method.setRequestHeader("Content-Type", "application/json");
org.apache.commons.httpclient.HttpClient newhttpClient = new org.apache.commons.httpclient.HttpClient();
int statusCode = newhttpClient.executeMethod(method);
return method.getResponseBodyAsString();

通过JSON将数据发送到服务器

3 个答案: