关于java:Java-8-Stream根据某个相同的key合并两个Map-List其他key相同时优先保留其中一个List的值

// ============== new两条源数据，value值均为一个字，退出list ==================

    Map<String, Object> map1 = new HashMap<>();
    map1.put("id", 1);
    map1.put("name", "甲");
    map1.put("age", 12);
    map1.put("gender", "男");
    Map<String, Object> map2 = new HashMap<>();
    map2.put("id", 2);
    map2.put("name", "乙");
    map2.put("age", 13);
    map2.put("gender", "女");
    List<Map<String, Object>> sourceList = new ArrayList<>();
    sourceList.add(map1);
    sourceList.add(map2);

    // ============== new三条新带并入的数据，value值均为两个字，退出list ==================
    Map<String, Object> newMap1 = new HashMap<>();
    newMap1.put("id", 1);
    newMap1.put("name", "甲");
    newMap1.put("age", 10);
    newMap1.put("birthday", "2011-09-29");
    Map<String, Object> newMap2 = new HashMap<>();
    newMap2.put("id", 2);
    newMap2.put("name", "乙");
    newMap2.put("age", 11);
    newMap2.put("birthday", "2010-09-29");
    Map<String, Object> newMap3 = new HashMap<>();
    newMap3.put("id", 3);
    newMap3.put("name", "丙");
    newMap3.put("age", 19);
    newMap3.put("birthday", "2002-09-29");
    List<Map<String, Object>> newList = new ArrayList<>();
    newList.add(newMap1);
    newList.add(newMap2);
    newList.add(newMap3);
    /**
     * sourceList有两条数据，newList有三条数据
     * 咱们想要的成果是将两个mapList合并，尽量保留各自所有数据，遇到id雷同的map则合并，
     * 当呈现其余key雷同的状况时，sourceList中有值就优先用sourceList中的值，
     * 如果同一个key，sourceList的值为空，此时再用newList中的值
     * (下面例子中name和age是key抵触的，name两个list均雷同，但age不同)
     * 并且最终合并成一个mapList，达到如下成果：
     * [{"birthday":"2011-09-29","gender":"男","name":"甲","id":1,"age":12},
     * {"birthday":"2010-09-29","gender":"女","name":"乙","id":2,"age":13},
     * {"birthday":"2002-09-29","name":"丙","id":3,"age":19}]
     */
    //个别理论开发中后面的两个List都是数据库查问进去而不是手动new进去的，就可能会有List为空的状况，为防止空指针异样不能间接用sourceList来addAllnewList
    // 所以这里要手动New一个合并的list
    List<Map<String, Object>> list = new ArrayList<>();
    list.addAll(sourceList);
    list.addAll(newList);
    //合并
    List<Map<String,Object>> combine = list.stream()
            .collect(Collectors.groupingBy(group -> group.get("id").toString())) // 依据map中id的value值进行分组, 这一步的返回后果Map<String,List<Map<String, Object>>>
            .entrySet() // 失去Set<Map.Entry<String, List<Map<String, Object>>>
            .stream()
            .map(m -> { // 进入映射环境
                // m.getValue()的后果是 List<Map<String, Object>>
                Map<String, Object> collect = m.getValue().stream()
                        // o.entrySet() 的后果是 Set<Map.Entry<String, Object>>
                        .flatMap(o -> o.entrySet().stream()).filter(e -> e.getValue() != null)//过滤下，value须要是不为空的，否则报错
                        // (m1, m2) -> m1 的意思是如果key雷同 m1 == m2 则value应用m1（此处为sourceList中的值）
                        .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (m1, m2) -> m1
                        ));
                return collect;
            }).sorted(Comparator.comparing(m -> m.get("id").toString())).collect(Collectors.toList());
    System.out.println(JSON.toJSONString(combine));
    
    本文相干参考链接：https://segmentfault.com/q/1010000010380854