问题
I am hitting a service and sometimes getting back something like this:
{ "param1": "value1", "param2": "value2" }
and sometimes getting return like this:
[{ "param1": "value1", "param2": "value2" },{ "param1": "value1", "param2": "value2" }]
How do I tell which I'm getting? Both of them evaluate to a String when I do getClass() but if I try to do this:
json = (JSONObject) new JSONParser().parse(result);
on the second case I get an exception
org.json.simple.JSONArray cannot be cast to org.json.simple.JSONObject
How to avoid this? I would just like to know how to check which I'm getting back. (The first case will sometimes have []
in it so I can't do index of and I'd like a cleaner way than just checking the first character.
There has got to be some sort of method that checks this?
回答1:
Simple Java:
Object obj = new JSONParser().parse(result);
if (obj instanceof JSONObject) {
JSONObject jo = (JSONObject) obj;
} else {
JSONArray ja = (JSONArray) obj;
}
You could also test if the (purported) JSON starts with a [
or a {
if you wanted to avoid the overhead of parsing the wrong kind of JSON. But be careful with leading whitespace.
回答2:
Though it's similar to above one, but their is not default constructor of JSONParser. Error coming was: The constructor JSONParser() is undefined
Use this instead
JsonElement jsonElement = new JsonParser().parse(jsonString);
if (jsonElement.isJsonArray()) {
//Your Code
} else {
//Your Code
}
来源:https://stackoverflow.com/questions/16410421/how-to-tell-if-return-is-jsonobject-or-jsonarray-with-json-simple-java