获取两个日期之间的星期一mysql答案

【问题标题】：get Mondays between two dates mysql获取两个日期之间的星期一mysql
【发布时间】：2017-03-08 16:52:51
【问题描述】：

我想用 select 语句获取两个日期之间的星期一，但我只在 sql server 中搜索过。

我试过了，但它不起作用：

SELECT DATE_ADD('2017-01-01', INTERVAL ROW DAY) AS Date 
FROM (
            SELECT @row := @row + 1 AS row 
            FROM (
                        SELECT 0 UNION ALL 
                        SELECT 1 UNION ALL 
                        SELECT 3 UNION ALL 
                        SELECT 4 UNION ALL 
                        SELECT 5 UNION ALL 
                        SELECT 6
            ) t1,
            (
                SELECT 0 UNION ALL 
                SELECT 1 UNION ALL 
                SELECT 3 UNION ALL 
                SELECT 4 UNION ALL 
                SELECT 5 UNION ALL 
                SELECT 6
            ) t2, 
            (
                SELECT @row:=-1
            ) t3 LIMIT 31
        ) b
WHERE DATE_ADD('2017-01-01', INTERVAL ROW DAY) BETWEEN '2017-01-01' AND '2017-12-31'
AND DAYOFWEEK(DATE_ADD('2017-01-01', INTERVAL ROW DAY)) = 2

这个查询只给了我一个月的星期一日期。

希望你能帮到我

问候！

【问题讨论】：

标签： mysql sql date select

【解决方案1】：

如果你真的只想解决这个 SQL，你必须更深入地研究并生成一个天序列，这是迄今为止解决这个问题的最困难的辅助任务。除此之外，正如您已经猜到的那样，您可以使用DAYOFWEEK 或WEEKDAY 来获得您想要的日子。

SELECT  * 
   FROM (
        SELECT  DATE_ADD('2013-01-01', 
            INTERVAL n4.num*1000+n3.num*100+n2.num*10+n1.num DAY ) AS DATE 
          FROM  (
              SELECT 0 AS num
              UNION ALL SELECT 1
              UNION ALL SELECT 2
              UNION ALL SELECT 3
              UNION ALL SELECT 4
              UNION ALL SELECT 5
              UNION ALL SELECT 6
              UNION ALL SELECT 7
              UNION ALL SELECT 8
              UNION ALL SELECT 9
         ) AS n1,
         (
              SELECT 0 AS num
              UNION ALL SELECT 1
              UNION ALL SELECT 2
              UNION ALL SELECT 3
              UNION ALL SELECT 4
              UNION ALL SELECT 5
              UNION ALL SELECT 6
              UNION ALL SELECT 7
              UNION ALL SELECT 8
              UNION ALL SELECT 9
        ) AS n2,
        (
              SELECT 0 AS num
              UNION ALL SELECT 1
              UNION ALL SELECT 2
              UNION ALL SELECT 3
              UNION ALL SELECT 4
              UNION ALL SELECT 5
              UNION ALL SELECT 6
              UNION ALL SELECT 7
              UNION ALL SELECT 8
              UNION ALL SELECT 9
        ) AS n3,
        (
              SELECT 0 AS num
              UNION ALL SELECT 1
              UNION ALL SELECT 2
              UNION ALL SELECT 3
              UNION ALL SELECT 4
              UNION ALL SELECT 5
              UNION ALL SELECT 6
              UNION ALL SELECT 7
              UNION ALL SELECT 8
              UNION ALL SELECT 9
        ) AS n4
    ) AS a
WHERE DATE >= '2017-01-01' AND DATE < NOW()
  AND WEEKDAY(DATE) = 0
ORDER BY DATE

其他一些 RDBMS 有内置函数。

如果您有一个已经存在所有日期的日历表，那会更容易。

【讨论】：

【解决方案2】：

我认为这个解决方案的性能会比以前更好

SET @date_start = date('2019-07-06');
SET @date_end = date('2019-09-18');

SELECT CASE WHEN MOD(DAYOFWEEK(@date_start), 7) <= 2 AND
      MOD(DAYOFWEEK(@date_start), 7) + DATEDIFF(v.selected_date, @date_start) - 7 * FLOOR(DATEDIFF(v.selected_date, @date_start) / 7) >= 2 OR
      MOD(DAYOFWEEK(@date_start), 7) > 2 AND
      MOD(DAYOFWEEK(@date_start), 7) + DATEDIFF(v.selected_date, @date_start) - 7 * FLOOR(DATEDIFF(v.selected_date, @date_start) / 7) >= 9 THEN 1 ELSE 0 END
  + FLOOR(DATEDIFF(v.selected_date, @date_start) / 7) num_of_mo
 ,  @date_start start_date,v.selected_date end_date
FROM (select adddate(@date_start,t4*10000 + t3*1000 + t2*100 + t1*10 + t0) selected_date from
 (select 0 t0 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t0,
 (select 0 t1 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t1,
 (select 0 t2 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t2,
 (select 0 t3 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t3,
 (select 0 t4 union select 1 union select 2 union select 3 union select 4 union select 5 union select 6 union select 7 union select 8 union select 9) t4) v
WHERE v.selected_date <= @date_end;

【讨论】：