package com.alibaba.json.bvt.issue_1400;

import com.alibaba.fastjson.JSON;
import com.alibaba.fastjson.TypeReference;
import junit.framework.TestCase;
import org.junit.Assert;

import java.util.ArrayList;
import java.util.List;

/**
 * Created by kimmking on 11/08/2017.
 */
public class Issue1400 extends TestCase {
    public void test_for_issue() throws Exception {
        TypeReference tr = new TypeReference<Resource<ArrayList<App>>>(){};
        Test test = new Test(tr);
        Resource resource = test.resource;
        Assert.assertEquals(1,resource.ret);
        Assert.assertEquals("ok",resource.message);
        List<App> data =(List<App>) resource.data;
        Assert.assertEquals(2,data.size());
        App app1 = data.get(0);
        Assert.assertEquals("11c53f541dee4f5bbc4f75f99002278c",app1.appId);
    }

    public static class Resource<T> {
        public int ret;
        public String message;
        public T data;
    }

    public static class App {
        public String appId;
    }

    public static class Test {
        String str = "{\"ret\":1,\"message\":\"ok\",\"data\":[{\"appId\":\"11c53f541dee4f5bbc4f75f99002278c\"},{\"appId\":\"c6102275ce5540a59424defa1cccb8ed\"}]}";
        public Resource resource;
        Test(TypeReference tr) {
            resource = (Resource)JSON.parseObject(str, tr);
        }
    }
}

@kimmking hello，你这样写是没问题的，因为在new TypeReference<Resource<ArrayList<App>>>(){}已经指定是ArrayList<App>了，
我这边是用的泛型new TypeReference<Resource<T>>(){}，这样只能解成JSONArray

https://github.com/alibaba/fastjson/wiki/TypeReference
参考其中的单参数和双参数的例子

@wenshao

public static <T> Response<T> parseToMap(String json, Class<T> type) {
    return JSON.parseObject(json,  new TypeReference<Response<T>>(type) {});
}

我在处理数据的过程中怎么拿到第二个参数type呢。

我想到的种方式是
Responseextend一个super class，在super class中拿到泛型的类型

public class BaseResponse<T> {
    private Class<T> genericType;

    public BaseResponse() {
       genericType = (Class<T>) ((ParameterizedType) getClass().getGenericSuperclass()).getActualTypeArguments()[0];
   }

   public Class<T> getGenericType() {
       return genericType;
   }
}

然后在解析时

Response response = new Response();
Response realResp = parseToMap(jsonStr, response.getGenericType());

感觉这样好麻烦，还有别的调用方式吗

@wenshao@kimmking 麻烦看下上面的问题