MYSQL 具有日期范围 select 的 COUNT 个值与没有日期范围的值不同

Question

我有这个查询来获取日期范围内的计数值（使用独特的日期过滤器到一天以查看选择详细信息）：

SELECT `dates`.`date`, COUNT(*)
FROM (
    SELECT CURDATE() - INTERVAL (units.mul + (10 * tens.mul) + (100 * hundreds.mul) + (200 * thousands.mul)) DAY AS `date`
    FROM       (SELECT 0 AS mul 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 units
    CROSS JOIN (SELECT 0 AS mul 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 tens
    CROSS JOIN (SELECT 0 AS mul 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 hundreds
    CROSS JOIN (SELECT 0 AS mul 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 thousands
) `dates`
LEFT JOIN `prices` ON (`prices`.`date` = `dates`.`date`)
WHERE `dates`.`date` = '2020-07-07'
GROUP BY `dates`.`date`
ORDER BY `dates`.`date` ASC;

+------------+-----------+
| date       | COUNT(*)  |
+------------+-----------+
| 2020-07-07 |    150840 |
+------------+-----------+
1 row in set (0.06 sec)

但同样的查询只针对价格 table 结果是：

SELECT COUNT(*) FROM `prices` WHERE `date` = '2020-07-07';

+----------+
| COUNT(*) |
+----------+
|    37710 |
+----------+
1 row in set (0.01 sec)

为什么第一次查询结果不是：

+------------+----------+
| date       | COUNT(*) |
+------------+----------+
| 2020-07-07 |    37710 |
+------------+----------+
1 row in set (0.06 sec)

谢谢！

Answer 1

子查询 returns 4 次 2020-07-07。然后以价格 table 加入比赛 4 次。尝试不同的：

SELECT `dates`.`date`, COUNT(*)
FROM (
    SELECT distinct CURDATE() - INTERVAL (units.mul + (10 * tens.mul) + (100 * hundreds.mul) + (200 * thousands.mul)) DAY AS `date`
    FROM       (SELECT 0 AS mul 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 units
    CROSS JOIN (SELECT 0 AS mul 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 tens
    CROSS JOIN (SELECT 0 AS mul 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 hundreds
    CROSS JOIN (SELECT 0 AS mul 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 thousands
) `dates`
LEFT JOIN `prices` ON (`prices`.`date` = `dates`.`date`)
WHERE `dates`.`date` = '2020-07-07'
GROUP BY `dates`.`date`
ORDER BY `dates`.`date` ASC;

Answer 2

37710 * 4 = 150840 在没有连接的情况下检查你的查询，你有 4 行 2020-07-07 我怀疑打字错误 200 * thousands.mul 应该是 1000 * thousands.mul

SELECT `dates`.`date`, COUNT(*)
FROM (
    SELECT CURDATE() - INTERVAL (units.mul + (10 * tens.mul) + (100 * hundreds.mul) + (200 * thousands.mul)) DAY AS `date`
    FROM       (SELECT 0 AS mul 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 units
    CROSS JOIN (SELECT 0 AS mul 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 tens
    CROSS JOIN (SELECT 0 AS mul 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 hundreds
    CROSS JOIN (SELECT 0 AS mul 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 thousands
) `dates`
#LEFT JOIN `prices` ON (`prices`.`date` = `dates`.`date`)
WHERE `dates`.`date` = '2020-07-07'
GROUP BY `dates`.`date`
ORDER BY `dates`.`date` ASC;

+------------+----------+
| date       | COUNT(*) |
+------------+----------+
| 2020-07-07 |        4 |
+------------+----------+
1 row in set (0.011 sec)

MYSQL 具有日期范围 select 的 COUNT 个值与没有日期范围的值不同

MYSQL COUNT values with dates range select is different than without dates range

mysql

date

count

range