如何将JSON解析为int? [英] How do I parse JSON into an int?
问题描述
我已经解析了一些JSON数据,只要我将它存储在String变量中就可以正常工作。
I have Parsed some JSON data and its working fine as long as I store it in String variables.
我的问题是我需要int int varibable中的ID而不是在String中。
i试图进行投射 int id =(int)jsonObj.get();
My problem is that I need the ID in an int varibable and not in String.
i have tried to make a cast int id = (int) jsonObj.get("");
但是它给出了一条错误消息,我无法将对象转换为int。
所以我尝试使用以下方式进行转换:
But it gives an error message that I cannot convert an object to an int. So I tried to convert by using:
String id = (String) jsonObj.get("id");
int value = Integer.parseInt(id);
但这也无效。怎么了。 JSON如何使用int?
我的字符串工作得很好,只有当我尝试将它们作为int我才会遇到问题。
But also that is not working. What is wrong. How is JSON working with int? My strings are working just fine its only when I try to make them as an int I get problems.
这是我的代码:
public void parseJsonData() throws ParseException {
JSONParser parser = new JSONParser();
Object obj = parser.parse(jsonData);
JSONObject topObject = (JSONObject) obj;
JSONObject locationList = (JSONObject) topObject.get("LocationList");
JSONArray array = (JSONArray) locationList.get("StopLocation");
Iterator<JSONObject> iterator = array.iterator();
while (iterator.hasNext()) {
JSONObject jsonObj = (JSONObject) iterator.next();
String name =(String) jsonObj.get("name");
String id = (String) jsonObj.get("id");
Planner.getPlanner().setLocationName(name);
Planner.getPlanner().setArrayID(id);
}
}
推荐答案
您可以使用 parseInt
:
int id = Integer.parseInt(jsonObj.get("id"));
或更好,更直接 getInt 方法:
int id = jsonObj.getInt("id");
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