如何以 JSON 格式将 Android 中的数据发布到服务器？

Question

我有这个 JSON 字符串。我想将它发布到服务器（即使用 POST 方法）。如何在 Android 中做到这一点？

JSON 字符串：

{
    "clientId": "ID:1234-1234",
    "device": {
        "userAgent": "myUA",
        "capabilities": {
            "sms": true,
            "data": true,
            "gps": true,
            "keyValue": {
                "Key2": "MyValue2",
                "Key1": "myvalue1"
            }
        },
        "screen": {
            "width": 45,
            "height": 32
        },
        "keyValue": {
            "DevcKey2": "myValue2",
            "DevcKey1": "myValue1"
        }
    },
    "time": 1294617435368
}

如何形成这个 JSON 数组并将其 POST 到服务器？

Answer 1

我自己做的。

JSONObject returnedJObject= new JSONObject();
JSONObject KeyvalspairJObject=new JSONObject ();
JSONObject devcKeyvalspairJObject=new JSONObject ();
JSONObject capabilityJObject=new JSONObject();
JSONObject ScreenDimensionsJObject =new JSONObject();
JSONObject deviceJObject= new JSONObject();
try{
    KeyvalspairJObject.put("key1","val1");
    KeyvalspairJObject.put("key2","val2");
    capabilityJObject.put("sms", false);
    capabilityJObject.put("data", true);
    capabilityJObject.put("gps", true);
    capabilityJObject.put("wifi", true);
    capabilityJObject.put("keyValue", KeyvalspairJObject);
    ScreenDimensionsJObject.put("width", 45);
    ScreenDimensionsJObject.put("height", 45);
    devcKeyvalspairJObject.put("Devckey1","val1");
    devcKeyvalspairJObject.put("DEVCkey2","val2");
    deviceJObject.put("userAgent", "MYUserAgent");
    deviceJObject.put("capabilities", capabilityJObject);
    deviceJObject.put("screen", ScreenDimensionsJObject);
    deviceJObject.put("keyValue", devcKeyvalspairJObject);

    returnedJObject.put("clientId", "ID:1234-1234");
    returnedJObject.put("carrier","TMobile");
    returnedJObject.put("device",deviceJObject);
    returnedJObject.put("time",1294617435);
    returnedJObject.put("msisdn","1234567890");
    returnedJObject.put("timezone","GMT");
}
catch(JSONException e)
{
}

这就是我们向服务器发送 JSON 数据的方式。

public String putDataToServer(String url,JSONObject returnedJObject) throws Throwable
{
    HttpPost request = new HttpPost(url);
    JSONStringer json = new JSONStringer();
    StringBuilder sb=new StringBuilder();


    if (returnedJObject!=null) 
    {
        Iterator<String> itKeys = returnedJObject.keys();
        if(itKeys.hasNext())
            json.object();
        while (itKeys.hasNext()) 
        {
            String k=itKeys.next();
            json.key(k).value(returnedJObject.get(k));
            Log.e("keys "+k,"value "+returnedJObject.get(k).toString());
        }             
    }
    json.endObject();


    StringEntity entity = new StringEntity(json.toString());
                         entity.setContentType("application/json;charset=UTF-8");
    entity.setContentEncoding(new BasicHeader(HTTP.CONTENT_TYPE,"application/json;charset=UTF-8"));
    request.setHeader("Accept", "application/json");
    request.setEntity(entity); 

    HttpResponse response =null;
    DefaultHttpClient httpClient = new DefaultHttpClient();

    HttpConnectionParams.setSoTimeout(httpClient.getParams(), Constants.ANDROID_CONNECTION_TIMEOUT*1000); 
    HttpConnectionParams.setConnectionTimeout(httpClient.getParams(),Constants.ANDROID_CONNECTION_TIMEOUT*1000); 
    try{
        response = httpClient.execute(request); 
    }
    catch(SocketException se)
    {
        Log.e("SocketException", se+"");
        throw se;
    }
    InputStream in = response.getEntity().getContent();
    BufferedReader reader = new BufferedReader(new InputStreamReader(in));
    String line = null;
    while((line = reader.readLine()) != null){
        sb.append(line);
    }
    return sb.toString();
}

Answer 2

如果您已经将 JSON 作为字符串，只需使用常规 HTTP 连接 (URL.openConnection()) 将其发布。无需解析它或类似的东西。

Answer 3

看看这段代码

https://gist.github.com/9457c486af9644cf6b18

查看 retrieveJSONArray(ArrayList jsonArray,String[] key) 和 retrieveJSONString(ArrayList jsonObject)

Answer 4

JSONObject returnedJObject= new JSONObject();

url = "http://49.50.76.75/website/exxx/api.php";

// makeSignupRequest();

try {
    returnedJObject.put("email",eemail);
    returnedJObject.put("password",passwrd);
    returnedJObject.put("type","customer");
    returnedJObject.put("method","register");
    returnedJObject.put("username",usr);

    try {
        putDataToServer(url, returnedJObject);

    }
    catch (Throwable m) {
        m.printStackTrace();
    }
}

Answer 5

如果这不是一个“爱好”应用程序，只需进行一两个服务调用，我会考虑查看异步 HTTP 或 REST 库。

有几个值得考虑的，包括来自 Google 的 Volley：https://developer.android.com/training/volley/index.html

以及从 Square 进行改造：http://square.github.io/retrofit

Volley 是一个很好的通用异步 HTTP 库，而 Retrofit 专门专注于实现 REST 客户端。

这两个库都将处理您在示例中处理的大部分低级细节。

无论您是继续使用 DefaultHttpClient 代码，还是使用 Volley 或 Retrofit，您肯定会希望确保您的网络请求发生在后台线程上，以免阻塞 UI 线程。从您的示例中不清楚您是否正在这样做。

Answer 6

你也可以使用Httpurlconnection 请检查链接，你会得到更多的想法。或Httpsurlconnection