在 Java 中的 Json 对象中获取 Json 对象
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Getting Json object inside a Json object in Java
提问by Exception
So I have some code that is able to send this out:
所以我有一些代码可以将其发送出去：
{"id":1,
"method":"addWaypoint",
"jsonrpc":"2.0",
"params":[
{
"lon":2,
"name":"name",
"lat":1,
"ele":3
}
]
}
The server receives this JSON object as a string named "clientstring":
服务器接收此 JSON 对象作为名为“clientstring”的字符串：
JSONObject obj = new JSONObject(clientstring); //Make string a JSONObject
String method = obj.getString("method"); //Pulls out the corresponding method
Now, I want to be able to get the "params" value of {"lon":2,"name":"name","lat":1,"ele":3} just like how I got the "method". however both of these have given me exceptions:
现在，我希望能够获得 {"lon":2,"name":"name","lat":1,"ele":3} 的“params”值，就像我获得“方法”一样”。然而，这两个都给了我例外：
String params = obj.getString("params");
and
和
JSONObject params = obj.getJSONObject("params");
I'm really at a loss how I can store and use {"lon":2,"name":"name","lat":1,"ele":3} without getting an exception, it's legal JSON yet it can't be stored as an JSONObject? I dont understand.
我真的不知道如何存储和使用 {"lon":2,"name":"name","lat":1,"ele":3} 没有例外，它是合法的 JSON 但它不能存储为 JSONObject？我不明白。
Any help is VERY appreciated, thanks!
非常感谢任何帮助，谢谢！
采纳答案by Lokesh
paramsin your case is nota JSONObject, but it is a JSONArray.
params你的情况是不是一个JSONObject的，但它是一个JSONArray。
So all you need to do is first fetch the JSONArrayand then fetch the first element of that array as the JSONObject.
因此，您需要做的就是首先获取JSONArray，然后获取该数组的第一个元素作为JSONObject。
JSONObject obj = new JSONObject(clientstring);
JSONArray params = obj.getJsonArray("params");
JSONObject param1 = params.getJsonObject(0);
回答by Wundwin Born
How try like that
怎么试试
JSONObject obj = new JSONObject(clientstring);
JSONArray paramsArr = obj.getJSONArray("params");
JSONObject param1 = paramsArr.getJSONObject(0);
//now get required values by key
System.out.println(param1.getInt("lon"));
System.out.println(param1.getString("name"));
System.out.println(param1.getInt("lat"));
System.out.println(param1.getInt("ele"));
回答by incr3dible noob
Here "params" is not an object but an array. So you have to parse using:
这里的“params”不是一个对象而是一个数组。所以你必须解析使用：
JSONArray jsondata = obj.getJSONArray("params");
for (int j = 0; j < jsondata.length(); j++) {
JSONObject obj1 = jsondata.getJSONObject(j);
String longitude = obj1.getString("lon");
String name = obj1.getString("name");
String latitude = obj1.getString("lat");
String element = obj1.getString("ele");
}
