Question

我有一个非常简单的表格：

mysql> desc stats;
+-------------+-------------+------+-----+---------+----------------+
| Field       | Type        | Null | Key | Default | Extra          |
+-------------+-------------+------+-----+---------+----------------+
| entry_id    | int(11)     | NO   | PRI | NULL    | auto_increment | 
| entry_date  | date        | NO   |     | NULL    |                | 
| show_name   | varchar(40) | NO   |     | NULL    |                | 
| month_total | int(11)     | NO   |     | NULL    |                | 
+-------------+-------------+------+-----+---------+----------------+

我想要一个select语句，它会在一个语句中提供当前日期和前一天数据的结果，例如，结果将类似于

show_name month_total(Current day) monthly_total(Previous Day)

在单个sql语句中有一种简单的方法吗？

Answer 1

更新：将entry_date添加到输出

是有意义的

SELECT c.show_name, 
       c.month_total current_month_total, 
       p.month_total prev_month_total
  FROM stats c INNER JOIN
       stats p ON p.entry_date = c.entry_date - INTERVAL 1 DAY

假设您有样本数据，如

+----------+------------+-----------+-------------+
| entry_id | entry_date | show_name | month_total |
+----------+------------+-----------+-------------+
|        1 | 2013-03-07 | Name1     |          10 |
|        2 | 2013-03-08 | Name2     |          20 |
|        3 | 2013-03-09 | Name3     |          30 |
|        4 | 2013-03-10 | Name4     |          40 |
+----------+------------+-----------+-------------+

查询的输出是

+-----------+------------+---------------------+------------------+
| show_name | entry_date | current_month_total | prev_month_total |
+-----------+------------+---------------------+------------------+
| Name2     | 2013-03-08 |                  20 |               10 |
| Name3     | 2013-03-09 |                  30 |               20 |
| Name4     | 2013-03-10 |                  40 |               30 |
+-----------+------------+---------------------+------------------+

这是sqlfiddle example

如果您需要特定日期或日期间隔的输出，只需添加WHERE条款今天

...
WHERE c.entry_date = CURDATE();

03/09至03/10

...
WHERE c.entry_date BETWEEN '2013-03-09' AND '2013-03-10'

Answer 2

如果今天是指今天，请在select语句中使用SUBDATE和CURDATE函数以及子查询：

SELECT 
    show_name, 
    month_total AS current_day_total, 
    (SELECT month_total FROM stats WHERE entry_date = SUBDATE(CURDATE(), 1) LIMIT 1) AS previous_day_total
FROM stats
WHERE entry_date = CURDATE();

请参阅fiddle。

如果您希望每天都这样，而不仅仅是当天，请加入。

Answer 3

假设这个测试数据：

+----------+------------+-----------+-------------+
| entry_id | entry_date | show_name | month_total |
| 1        | 2013-03-07 | test1     |           1 |
| 2        | 2013-03-07 | test2     |          11 |
| 3        | 2013-03-08 | test1     |           2 |
| 4        | 2013-03-08 | test2     |          22 |
| 5        | 2013-03-08 | test3     |         222 |
| 6        | 2013-03-09 | test1     |           3 |
| 7        | 2013-03-09 | test2     |          33 |
| 8        | 2013-03-07 | test1     |           5 |
+----------+------------+-----------+-------------+

如果每天有多个不同名称的条目，那么这应该效果很好：

SELECT c.show_name, c.entry_date, 
       c.month_total current_day_month_total, 
       IFNULL(p.month_total,0) previous_day_month_total
  FROM test.stats c LEFT JOIN test.stats p
       ON p.entry_date = c.entry_date - INTERVAL 1 DAY
          AND c.show_name = p.show_name
  GROUP BY c.entry_date, c.show_name;

+-----------+------------+-------------------------+--------------------------+
| show_name | entry_date | current_day_month_total | previous_day_month_total |
| test1     | 2013-03-07 |                       1 |                        0 |
| test2     | 2013-03-07 |                      11 |                        0 |
| test1     | 2013-03-08 |                       2 |                        1 |
| test2     | 2013-03-08 |                      22 |                       11 |
| test3     | 2013-03-08 |                     222 |                        0 |
| test1     | 2013-03-09 |                       3 |                        2 |
| test2     | 2013-03-09 |                      33 |                       22 |
+-----------+------------+-------------------------+--------------------------+

如果每天甚至有多个条目和名称，那么它们将组合它们的值：（参见test1的结果表第一行 - id1：1 + id8：5 = 6）

SELECT c.show_name, c.entry_date,
       c.month_total current_day_month_total, 
       IFNULL(p.month_total,0) previous_day_month_total
  FROM (SELECT show_name, entry_date, SUM(month_total) month_total
        FROM test.stats
        GROUP BY entry_date, show_name) c
     LEFT JOIN
       (SELECT show_name, entry_date, SUM(month_total) month_total
        FROM test.stats
        GROUP BY entry_date, show_name) p
     ON p.entry_date = c.entry_date - INTERVAL 1 DAY
     AND c.show_name = p.show_name;

+-----------+------------+-------------------------+--------------------------+
| show_name | entry_date | current_day_month_total | previous_day_month_total |
| test1     | 2013-03-07 |                       6 |                        0 |
| test2     | 2013-03-07 |                      11 |                        0 |
| test1     | 2013-03-08 |                       2 |                        6 |
| test2     | 2013-03-08 |                      22 |                       11 |
| test3     | 2013-03-08 |                     222 |                        0 |
| test1     | 2013-03-09 |                       3 |                        2 |
| test2     | 2013-03-09 |                      33 |                       22 |
+-----------+------------+-------------------------+--------------------------+

如何从单个表中选择当前日期和上一个日期？

3 个答案: