我有一个类似于下面的嵌套地图对象
{12345={{"status":"200","outcome":"Success","message":"Account created"}}
{23121={{"status":"400","outcome":"Exception","message":"Invalid State value"}}
{43563={{"status":"200","outcome":"Success","message":"Account updated"}}
{72493={{"status":"400","outcome":"Exception","message":"Bad Request"}}
我需要将其转换为Map<String, List<Map<String, Map<String, String>>>
,其中外部地图的键是状态( 200 or 400 )
的值,并且该值是状态= 200或400或其他有效值的原始地图的列表从内部地图开始。
因此,新结构看起来像
{{200={[{12345={"status":"200","outcome":"Success","message":"Account created"}},
{43563={"status":"200","outcome":"Success","message":"Account updated"}}
]
},
{400={[{23121={"status":"400","outcome":"Exception","message":"Invalid State value"}},
{72493={"status":"400","outcome":"Exception","message":"Bad Request"}}
]
}
}
最终,我需要根据不同的状态生成报告。
这是我刚开始时遇到的问题,但是很困难。
我想遍历外部地图,获取内部地图,获取状态键的值,然后根据状态码值将地图添加到列表中。
这就是我使用循环的方式
private static Map<String, List<Map<String, Map<String, String>>>> covertToReport(Map<String, Map<String, String>> originalMap) {
Map<String, List<Map<String, Map<String, String>>>> statusBasedListOfMaps = new TreeMap<>();
//loop through the map
//for each key, get the inner map
//get the status value for each inner map
List<Map<String, Map<String, String>>> accountsMapsList;
for (Entry<String, Map<String, String>> entry : originalMap.entrySet()) {
String accNum = entry.getKey();
Map<String, String> childMap = entry.getValue();
String stausVal = childMap.get("status");
accountsMapsList = statusBasedListOfMaps.get(stausVal) == null ? new ArrayList<>() : statusBasedListOfMaps.get(stausVal);
accountsMapsList.add((Map<String, Map<String, String>>) entry);
statusBasedListOfMaps.put(stausVal, accountsMapsList);
}
return statusBasedListOfMaps;
}
当然,下面的代码不会编译,但这就是我想要得到的。
private static void covertToReport(Map<String, Map<String, String>> originalMap) {
Map<String, List<Map<String, Map<String, String>>>> statusBasedListOfMaps;
statusBasedListOfMaps = originalMap.entrySet()
.stream()
.filter(e -> e.getValue()
.values()
.stream()
.map(innerMap -> Collectors.toList())
.collect(Collectors.toMap(Map.Entry::getKey, Collectors.toList(e)));
这可能吗?
答案 0 :(得分:1)
您可以将Collectors.groupingBy()
与Collectors.mapping()
一起使用:
private static Map<String, List<Map<String, Map<String, String>>>> convertToReport(Map<String, Map<String, String>> originalMap) {
return originalMap.entrySet().stream()
.collect(Collectors.groupingBy(e -> e.getValue().get("status"),
Collectors.mapping(Map::ofEntries, Collectors.toList())));
}
您按status
分组,然后使用Map.ofEntries()
将相关条目映射到自己的地图。如果您使用的是Java,则可以使用它代替Map::ofEntries
:
e -> new HashMap<>() {{ put(e.getKey(), e.getValue()); }}
结果将是这样:
200=[
{12345={status=200, message=Account created, outcome=Success}},
{43563={status=200, message=Account created, outcome=Success}}
],
400=[
{72493={status=400, message=Invalid State value, outcome=Exception}},
{23121={status=400, message=Invalid State value, outcome=Exception}}
]
答案 1 :(得分:0)
您的函数返回一个Map<String, List<Map<String, Map<String, String>>>>
,但是您的结构看起来像Map<String, Map<String, Map<String, String>>>
如果您真正喜欢的是Map<String, Map<String, Map<String, String>>>
,则代码如下:
Map<String, Map<String, Map<String, String>>> result= map.entrySet().stream().collect(Collectors.groupingBy(entry -> entry.getValue().get("status"), Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue)));