Question

我试图在一个查询中获取一列的总和以及两个日期之间相同列的总和。这有可能吗？

我的表格如下：

uid|amount|date

我试图制作以下两个查询：

SELECT sum(amount) as `keys` FROM tbl_keys WHERE uid = 1

SELECT sum(amount) as `keys` FROM tbl_keys WHERE uid = 1 AND YEAR(`date`) = YEAR(CURRENT_DATE)
AND MONTH(`date`) = MONTH(CURRENT_DATE)

Answer 1

您可以使用UNION查询：

SELECT 'All' AS cnt, sum(amount) as `keys` FROM tbl_keys WHERE uid = 1
UNION ALL
SELECT 'Current_month' AS cnt, sum(amount) as `keys`
FROM tbl_keys
WHERE
  uid = 1
  AND `date`<= last_day(current_date)
  `date`>= current_date - interval (day(current_date)-1) day

（我更喜欢在日期列中使用>=和<=，因为它可以使用索引（如果存在），而MONTH()或YEAR()等函数则不能，我还假设date是一个日期列，并且它不包含时间信息。）

如果您希望将结果放在一行中，则可以使用内联查询：

SELECT
  (SELECT sum(amount) as `keys` FROM tbl_keys WHERE uid = 1) AS total,
  (SELECT sum(amount) as `keys`
   FROM tbl_keys
   WHERE
     uid = 1
     AND `date`<= last_day(current_date)
    `date`>= current_date - interval (day(current_date)-1) day
   ) AS current_month

Answer 2

这样的事情：

SELECT sum(amount) as `keys`,
(
    SELECT sum(t.amount)
    FROM tbl_keys as t 
    WHERE t.uid = tbl_keys.uid AND YEAR(t.`date`) = YEAR(CURRENT_DATE)
    AND MONTH(t.`date`) = MONTH(CURRENT_DATE)
) as `keys2`
FROM tbl_keys 
WHERE uid = 1

Answer 3

SELECT sum(amount) AS `keys`
FROM (
SELECT amount FROM tbl_keys
UNION ALL
SELECT amount FROM tbl_keys
WHERE uid = 1 
AND YEAR(`date`) = YEAR(CURRENT_DATE)
AND MONTH(`date`) = MONTH(CURRENT_DATE)
) AS new_table;

使用UNION子句，您将获得所需的输出。

Answer 4

使用CASE仅计算指定日期的金额：

SELECT SUM(amount) AS `keys`,
    SUM(CASE WHEN YEAR(`date`) = YEAR(CURRENT_DATE)
        AND MONTH(`date`) = MONTH(CURRENT_DATE) THEN amount ELSE 0 END) AS 'keys2'
    FROM tbl_keys
    WHERE uid = 1
;

我的猜测是，这比使用UNION SELECT的解决方案更有效。

MYSQL获得1个查询

4 个答案: