mongodb聚合过程中有几个code需合并然后再分组的实现

需求：取code中近一个月累计最多的前三名，其中有几个code需合并为一个

实现:

private String[] tempStr = new String[]{"A", "B", "C", "D"};
private Set<String> tempSet = new HashSet<>(Arrays.asList(tempStr));

Aggregation aggregation = Aggregation.newAggregation(
                Aggregation.match(Criteria.where("dateTime").gt(getOneMonthBeforeDate())),
                Aggregation.project("code", "dateTime")
                        .and(DateOperators.DateToString.dateOf("dateTime").toString("%Y-%m-%d").withTimezone(DateOperators.Timezone.valueOf("Asia/Shanghai"))).as("dateTime")
                        .and(ConditionalOperators.when(Criteria.where("code").in(tempSet)).then("XXXX").otherwiseValueOf("$code")).as("code"),
                Aggregation.group("code", "dateTime").count().as("count"),
                Aggregation.group("code").sum("count").as("count"),
                Aggregation.project("count").and("$_id").as("code"),
                Aggregation.sort(Sort.Direction.DESC, "count"),
                Aggregation.limit(3)
        );

/**
     * 获取一个月前的零点日期
     *
     * @return
     */
    private Date getOneMonthBeforeDate() {
        Calendar calendar = Calendar.getInstance();
        calendar.set(Calendar.HOUR_OF_DAY, 0);
        calendar.set(Calendar.MINUTE, 0);
        calendar.set(Calendar.SECOND, 0);
        calendar.set(Calendar.MILLISECOND, 0);
        calendar.add(Calendar.MONTH, -1);
        return calendar.getTime();
    }