我需要将一些参数传递给我需要传递的服务器,如下所示 格式
{
"k2": {
"mk1": "mv1",
"mk2": [
"lv1",
"lv2"
]
}
}
那么如何在android中生成这种格式。
我使用As shown in example 5.3尝试了此操作,但在obj.writeJSONString(out);
此行显示错误。任何人都可以帮忙解决这个问题。
提前致谢
答案 0 :(得分:38)
尽管如此,你想要的输出是JSONObject中的JSONArray和另一个JSONObject中的JSONObject。所以,你可以单独创建它们然后可以放在一起。如下。
try {
JSONObject parent = new JSONObject();
JSONObject jsonObject = new JSONObject();
JSONArray jsonArray = new JSONArray();
jsonArray.put("lv1");
jsonArray.put("lv2");
jsonObject.put("mk1", "mv1");
jsonObject.put("mk2", jsonArray);
parent.put("k2", jsonObject);
Log.d("output", parent.toString(2));
} catch (JSONException e) {
e.printStackTrace();
}
<强>输出 - 强>
{
"k2": {
"mk1": "mv1",
"mk2": [
"lv1",
"lv2"
]
}
}
答案 1 :(得分:6)
您可以使用JSONObject
并使用它构建数据。
jsonObject.toString() // Produces json formatted object
答案 2 :(得分:2)
首先,您必须在代码
之后创建单独的类HttpUtil.java.Seepublic class HttpUtil {
// lat=50.2911 lon=8.9842
private final static String TAG = "DealApplication:HttpUtil";
public static String get(String url) throws ClientProtocolException,
IOException {
Log.d(TAG, "HTTP POST " + url);
HttpGet post = new HttpGet(url);
HttpResponse response = executeMethod(post);
return getResponseAsString(response);
}
public static String post(String url, HashMap<String, String> httpParameters)
throws ClientProtocolException, IOException {
Log.d(TAG, "HTTP POST " + url);
List<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>(
httpParameters.size());
Set<String> httpParameterKeys = httpParameters.keySet();
for (String httpParameterKey : httpParameterKeys) {
nameValuePairs.add(new BasicNameValuePair(httpParameterKey,
httpParameters.get(httpParameterKey)));
}
HttpPost method = new HttpPost(url);
UrlEncodedFormEntity urlEncodedFormEntity = new UrlEncodedFormEntity(nameValuePairs);
System.out.println("**************Request=>"+urlEncodedFormEntity.toString());
method.setEntity(urlEncodedFormEntity);
HttpResponse response = executeMethod(method);
return getResponseAsString(response);
}
private static HttpResponse executeMethod(HttpRequestBase method)
throws ClientProtocolException, IOException {
HttpResponse response = null;
HttpClient client = new DefaultHttpClient();
response = client.execute(method);
Log.d(TAG, "executeMethod=" + response.getStatusLine());
return response;
}
private static String getResponseAsString(HttpResponse response)
throws IllegalStateException, IOException {
String content = null;
InputStream stream = null;
try {
if (response != null) {
stream = response.getEntity().getContent();
InputStreamReader reader = new InputStreamReader(stream);
BufferedReader buffer = new BufferedReader(reader);
StringBuilder sb = new StringBuilder();
String cur;
while ((cur = buffer.readLine()) != null) {
sb.append(cur + "\n");
}
content = sb.toString();
System.out.println("**************Response =>"+content);
}
} finally {
if (stream != null) {
stream.close();
}
}
return content;
}
}
答案 3 :(得分:0)
这个例子可以帮助你调用这个函数它将JSON作为字符串值返回..试试吧
public String getResult() {
JSONObject userResults = null;
try {
userResults = new JSONObject();
userResults.put("valueOne",str_one);
userResults.put("valueTwo", str_two);
userResults.put("valueThree" ,str_three);
userResults.put("valueFour", str_four);
} catch (Exception e) {
e.printStackTrace();
}
return userResults.toString();
}