如何在 Java 中合并两个对象的属性？（用户定义类的属性和映射的键值）

Question

我正在开发 spring boot 应用程序并且有两个对象，一个是自定义类的列表，另一个是字符串键和字符串值的映射。前任。DTO object with fixed fields fieldA,fieldB,fieldc. [{"fieldA": "lookupval1","fieldB": "val2","fieldc":"val3"},{"fieldA": "lookupval1","fieldB": "val5","fieldc":"val6"}{"fieldA": "lookupval2","fieldB": "val8","fieldc":"val9"}]另外请注意，此字段 A 的前两个对象具有重复值然后我有Map<String,Map<String,String>如下图的地图{    lookupval1= {fieldA:"lookupval1",fieldD:"xxx",.............},    lookupval2= {fieldA:"lookupval2",fieldD:"yyy",.............}.    .    .    .}我想要的是，我想将列表中的每一项中的字段与Map中的字段组合起来，我可以通过获取fieldA的查找值在map中找到相应的键。字段A：lookupval1然后我想将该响应发送到 REST 客户端，执行此操作的有效方法和数据结构是什么？预期结果：[{"fieldA": "lookupval1","fieldB": "val2","fieldc":"val3",fieldD:"xxx"},{"fieldA": "lookupval1","fieldB": "val2","fieldc":"val3",fieldD:"xxx"}{"fieldA": "lookupval2","fieldB": "val8","fieldc":"val9",fieldD:"yyy"}]]更新：下面是 DTO 对象的结构和 Map 的确切类型。class DTO {    int id;    String name;    String desc;}List<DTO> dtoList = getDTOList(); // returns list of DTO objectMap<String,Map<String,String> mapCollection    = getMapCollection(); // return String, Map<String,String> pairs此处 mapCollection 中的键始终等于 DTO 的 id，并且 DTO 列表将具有重复值。我已经按照我的方式发布了答案，但正在寻找更好的替代方法来减少计算量。

Answer 1

List<DTO> dtoList = getDTOList(); // returns list of DTO object

Map<String,Map<String,String> mapCollection

    = getMapCollection(); // return String, Map<String,String> pairs

List<Map<String,String>> result =  new HashMap<>();

// iterate over dto list

for(DTO singleDTO : dtoList) {

    Map<String,String> singleResult = new HashMap<>();

    // Add all fields in map

    singleResult.put("id",singleDTO.getId());

    singleResult.put("name",singleDTO.getName());

    singleResult.put("desc",singleDTO.getDesc());

    // fetch single map by id lookup

    Map<String,Strig> singleMap = mapCollection.get(singleDTO.getId());

    // fetch all entries from map

    Set<Map.Entry<String,String>> entrySet = singleMap.entrySet();

   // add them in hashmap

    for(Map.Entry<String,String> singleEntry : entrySet) {

        singleResult.put(singleEntry.getKey(),singleEntry.getValue());

    Add map in the list

    result.add(singleEntry);

}}

//result

 [

    {

        "id":"1", "name":"abc", "desc":"some desc abc", "ID":"1", "mapField1":"field1 value",  "mapField2":"field1 value"

    },

    {

        "id":"1", "name":"xyz", "desc":"some desc xyz", "ID":"1", "mapField1":"field1 value",  "mapField2":"field1 value"

    },

    {

        "id":"2", "name":"pqr", "desc":"some desc pqr", "ID":"2", "mapField1":"field1 value 2",  "mapField2":"field1 value2"

    }     

 ]

Answer 2

假设

假设您的初始对象的类型Item如下所述。

class Item {

    String fieldA,fieldB,fieldC;

    // Getters / Setter

}

假设您的初始列表Item被称为items

假设您的初始地图被调用map并且是类型Map<String, Map<String,String>>

解决方案

向 Item 类添加一个方法以返回Map<String, String>对象的表示

public Map<String, String> toMap() {

    return new HashMap<String, String>() {{

        put("fieldA", fieldA);

        put("fieldB", fieldB);

        put("fieldC", fieldC);

    }};

}

创建一个方法来获取fieldD给定fieldA值的值

private String getFieldDValueForFieldAValue(String fieldAValue) {

    return map.get(fieldAValue).get("fieldD");

}

创建一种方法以将正确的fieldD值添加到项映射

private Map<String, String> addFieldD(Map<String, String> itemMap) {

    itemMap.put("fieldD", getFieldDValueForFieldAValue(itemMap.get("fieldA")));

    return itemMap;

}

fieldD然后您可以为每个具有正确值的对象生成一个映射列表

List<Map<String, String>> list = items.stream()

    .map(Item::toMap) // Converts all Items to Maps

    .map(m -> addFieldD(m)) // Adds the correct fieldD value to each Map

    .collect(Collectors.toList()); // make it a List

最后用它制作一个 JSON 字符串：

Type listType = new TypeToken<List<Map<String, String>>>() {}.getType();

Gson gson = new Gson();

String json = gson.toJson(list, listType);

System.out.println(json);

这将输出以下 JSON：

[

  {"fieldA":"lookupval1","fieldC":"val3","fieldB":"val2","fieldD":"xxx"},

  {"fieldA":"lookupval1","fieldC":"val6","fieldB":"val5","fieldD":"xxx"},

  {"fieldA":"lookupval2","fieldC":"val9","fieldB":"val8","fieldD":"yyy"}

]