连续性问题

#9月用户连续连续查询最大天数
SELECT
        user_id,
        max( date_cnt ) AS max_cnt 
FROM
        (
SELECT
        user_id,
        d - d_ranking AS d_group,--连续日期的组标记 
        count( 1 ) AS date_cnt 
        FROM (
SELECT
        user_id,
        d,
        row_number ( ) over ( PARTITION BY user_id ORDER BY d ) AS d_ranking --与标记日期的日期差的排序 
FROM
        (
SELECT
        user_id,
        datediff( create_date, '2019-01-01' ) AS d --与标记日期的日期差 
FROM
        ( SELECT user_id, date( gmt_update ) AS create_date FROM ods_v_location_query_record where month(gmt_update)=9 and year(gmt_update)=2020 GROUP BY user_id, date( gmt_update )
        ) a --在这一层获取用户的发单日期并去重 
        )b --这一层获取与标记日期的日期差 
        ) c --获取连续日期的排序 
GROUP BY
        user_id,
        d - d_ranking 
        ) d --获取每一个连续日期组的连续天数 
left join  ods_t_user_month_card_product c on d.user_id=e.user_id
where e.user_id  is null
GROUP BY
user_id