解析JSON似乎是一个非常常见的讨论话题.我环顾四周,仍然没有找到我要找的东西.这是我的HttpClient的代码
public class CreateJsonRequest {
public static String SendJsonRequest(String URL, Map<String,Object> params){
try{
DefaultHttpClient httpClient = new DefaultHttpClient();
HttpPost httpPost = new HttpPost(URL);
JSONObject holder = new JSONObject();
for (Map.Entry<String, Object> m : params.entrySet()){
try {
holder.put(m.getKey(), m.getValue());
}
catch (JSONException e) {
Log.e("Hmmmm", "JSONException : "+e);
}
}
StringEntity se;
se = new StringEntity(holder.toString());
httpPost.setEntity(se);
httpPost.setHeader("Accept", "text/json");
httpPost.setHeader("Content-type", "text/json");
HttpResponse response = httpClient.execute(httpPost);
HttpEntity entity = response.getEntity();
if(entity != null){
final JSONObject respObject = new JSONObject(EntityUtils.toString(entity));
String result = respObject.toString();
parseJSON(result);
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我正在使用HttpClient向服务器发送JSON请求.然后,服务器以JSON方式返回响应.这非常有效.现在这里是我遇到麻烦的地方.我从服务器收到HttpEntity.然后我把它变成一个看起来像这样的字符串.{"Make":"Ford","Year": 1975, "Model":"Mustang"} 我希望能够将此字符串发送到我的parseJSON(String jString)方法,并返回键值映射.我认为这与其他帖子的不同之处在于我希望parse方法能够为我发送的任何JSON字符串创建键值映射.所以如果我发送它{"Engine":"v8","Cylinders": 8, "Transmission":"Manual","Gears": 4}仍然可以工作.这可行吗?如果是这样,你能给我一些正确方向的推动吗?
Cri*_*ian 20
在这种情况下,您可以使用该类的keys方法JSONObject.它基本上会返回一个Iterator键,然后您可以迭代以获取并将值放入映射中:
try {
JSONObject jsonObject = new JSONObject(theJsonString);
Iterator keys = jsonObject.keys();
Map<String, String> map = new HashMap<String, String>();
while (keys.hasNext()) {
String key = (String) keys.next();
map.put(key, jsonObject.getString(key));
}
System.out.println(map);// this map will contain your json stuff
} catch (JSONException e) {
e.printStackTrace();
}
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