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