将JPA查询结果封装成Map返回以及Map排序

一、查询表中个别字段&＃xff0c;并将这些字段封装为键值对map的形式返回给前端

sql语句&＃xff1a;select b.userid,b.name from table_name b where b.userid in (1,2,3);

如果对返回的结果不做任何处理&＃xff0c;返回的结果是这样的&＃xff1a;

[[1,"张三"],[2,"李四"],[3,"王二"]
]


理想中的结果应该是这样的&＃xff1a;

[{"userid":1,"truename": "张三"},{"userid":2,"truename": "李四"},{"userid":3,"truename": "王二"},…
]


思路&＃xff1a;List—>Map—>List

List<Object> list &＃61; userService.findById(xxx);//查询结果原始列表
List<Object> arrayList &＃61; new arrayList();
for (Object obj : list) {Map<String, Object> map &＃61; new HashMap<>(); //这个Map一定要放在循环里面&＃xff0c;如果放在循环外面&＃xff0c;你add进去的map指向的是同一个内存地址&＃xff0c;不管你add进去多少次&＃xff0c;数据都是一样的。Object[] arr &＃61; (Object[]) obj;map.put("userid",arr[0]);map.put("name",arr[1]);arrayList.add(map);
}


此时&＃xff0c;可以正常返回目标map列表。

二、若期望返回&＃xff1a;
“content”: {
“广东”: 55261&＃xff0c;
“浙江”&＃xff1a;2000&＃xff0c;
“安徽”&＃xff1a;2100&＃xff0c;
…
}
代码如下&＃xff1a;

List<Object> organizationProvinceCountList &＃61; sampleService.findGroupByOrganizationAndProvince(organizationName);Map<String, Integer> map &＃61; new HashMap<>();for (Object obj : organizationProvinceCountList) {Object[] arr &＃61; (Object[]) obj;if(arr[0] &＃61;&＃61; null || "".equals(arr[0]))map.put("未知", Integer.valueOf(arr[1].toString()));elsemap.put((String) arr[0], Integer.valueOf(arr[1].toString()));}


如果想对上述结果按照键值对的value倒序排列&＃xff0c;则&＃xff1a;

LinkedHashMap<String, Integer> result &＃61; map.entrySet().stream().sorted(Map.Entry.comparingByValue(Comparator.reverseOrder())).collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (oldValue, newValue) -> oldValue,LinkedHashMap::new));//按value倒序排列return JsonViewObject.success(result.entrySet().iterator().next());//返回倒序排列后的第一个元素&＃xff0c;即按value排序后最大的省份及其数量


参考链接&＃xff1a;
http://t.zoukankan.com/BeenTogether-p-14566444.html jpa中将查询的字段返回为Map键值对类型

https://segmentfault.com/a/1190000020276225 Java8 对 Map 排序