Question

我有一张表

id int pk auto_inc | created int(11) | amount int | user_id int

我想创建一个按天分组的行列表，总计金额字段。

我试过这个：

SELECT created, sum(amount) as amount, id FROM total_log WHERE user_id = $this->user_id GROUP BY DAY(created)

这不会给出正确的结果。它们被分成一行。

日期从dd / mm / yyyy格式保存到unix时间戳，如1349046000

Answer 1

SELECT 
    DATE(FROM_UNIXTIME(created)) as d, 
    sum(amount) as amount
FROM total_log 
WHERE user_id = $this->user_id 
GROUP BY d

Answer 2

MySQL不喜欢混合day和int列：

mysql> select day(1349046000);
+-----------------+
| day(1349046000) |
+-----------------+
|            NULL |
+-----------------+
1 row in set, 1 warning (0.00 sec)

mysql> show warnings;
+---------+------+----------------------------------------+
| Level   | Code | Message                                |
+---------+------+----------------------------------------+
| Warning | 1292 | Incorrect datetime value: '1349046000' |
+---------+------+----------------------------------------+
1 row in set (0.00 sec)

因此，NULL的所有行都会day(some_int_value)，并且它们都会在同一组中。

我建议改为使用该列的date或datetime类型。

此外，不应在group by语句中引用不在select子句中的列，除非在它们上使用聚合函数。

Answer 3

试

SELECT 
DAY(DATE(FROM_UNIXTIME(created))), 
sum(amount) as amount 
FROM total_log 
WHERE user_id = $this->user_id 
GROUP BY DAY(DATE(FROM_UNIXTIME(created)))

MySQL Group By Day by Day

3 个答案: