Php 如何在每年每个月的结束日期前获取数据?
这是我表格中的示例数据Php 如何在每年每个月的结束日期前获取数据?,php,mysql,sql,eloquent,Php,Mysql,Sql,Eloquent,这是我表格中的示例数据 id_item | qty | t_in | t_out | created_at 1 5 1 0 2018-07-05 10:41:00 1 5 1 0 2018-08-03 10:41:00 1 5 0 1 2018-08-05 10:41:00 1 5 1 0 2018-09-05 10:41
id_item | qty | t_in | t_out | created_at
1 5 1 0 2018-07-05 10:41:00
1 5 1 0 2018-08-03 10:41:00
1 5 0 1 2018-08-05 10:41:00
1 5 1 0 2018-09-05 10:41:00
1 5 1 0 2018-09-20 10:41:00
1 5 0 1 2018-10-31 10:41:00
我的预期结果将是
id_item | qty | year | month
1 5 2018 07
1 5 2018 08
1 15 2018 09
1 10 2018 10
我尝试过的方法是有效的,但当我想按月份分组时,效果并不理想
$date = '2018-10-31'
$test = Model::whereDate('created_at','<=',$date)->select(DB::raw('(SUM(CASE T_IN WHEN 1 THEN qty ELSE qty * - 1 END)) as total'))->groupBy('id_item')->get();
从上面的查询中,我只能得到一行记录。我还尝试按月份和年份分组,但由于日期条件导致结果不正确。如何包含多个这是一个滚动求和问题。在较新版本的Mariadb/MySQL中,可以使用带框架的窗口函数来解决此问题。然而 我们可以使用用户定义的变量来解决这个问题。在派生表中,我们首先确定一个月的总数量变化。然后,我们使用此结果集计算月底的最终数量,方法是将上月行的数量与当月行的数量变化相加
我还扩展了查询,以考虑存在多个IDI项值时的情况。
尝试以下原始查询:SELECT
@roll_qty := CASE WHEN @id_itm = dt.id_item
THEN @roll_qty + dt.qty_change
ELSE dt.qty_change
END AS qty,
@id_itm := dt.id_item AS id_item,
dt.year,
dt.month
FROM
(
SELECT
t.id_item,
SUM(t.qty * t.t_in - t.qty * t.t_out) AS qty_change,
YEAR(t.created_at) AS `year`,
LPAD(MONTH(t.created_at), 2, '0') AS `month`
FROM your_table AS t
GROUP BY t.id_item, `year`, `month`
ORDER BY t.id_item, `year`, `month`
) AS dt
CROSS JOIN (SELECT @roll_qty := 0,
@id_itm := 0
) AS user_init_vars;
| id_item | year | month | qty |
| ------- | ---- | ----- | --- |
| 1 | 2018 | 07 | 5 |
| 1 | 2018 | 08 | 5 |
| 1 | 2018 | 09 | 15 |
| 1 | 2018 | 10 | 10 |
如果要使用变量,则需要正确地使用。MySQL不保证SELECT中表达式的求值顺序。因此,变量不应该在一个表达式中赋值,然后在另一个表达式中使用 这会产生复杂的表达式,但也有可能:
select yyyy, mm, total,
(@t := if(@ym = concat_ws('-', yyyy, mm), @t + total,
@ym := concat_ws('-', yyyy, mm), total
)
) as running_total
from (select year(created_at) as yyyy, month(created_at) as mm,
id_item,
sum(case T_IN when 1 then qty else - qty end) as total
from transactions
where created_at < '2018-11-01'
group by id_item
order by id_item, min(created_at)
) i cross join
(select @ym := '', @n := 0);
add->groupBy'YEARcreated_at'->groupBy'MONTHcreated_at'检查这个问题您的主键是什么?您可以将总和简化为SUMqty*t\u in-qty*t_out@MadhurBhaiya我只想显示今年和今年12个月的数量。如果使用组,那么它将显示所有不同的月份和年份。我在使用我自己的查询,选择今年每个月的12个结束日,然后将join保留到我的事务表中,结果看起来不错。选择“2018-01-31”作为月日,选择“2018-02-28”作为月日,依此类推。。。然后左键连接到我的事务表并暂时使用@Crazy,如果您可以使用Elounting进行自己的查询,那么就开始吧。在较小的数据集中,性能不会成为问题;一旦您开始遇到性能瓶颈,到那时,您就有望升级您的mariadb版本。在这种情况下,您只需使用窗口函数。此查询将比使用union和left join更有效。但当数据变大时,差异就会变得明显。是的,表格本身已经达到了6位数的记录。我认为演出会很慢。我认为需要找到其他方法来实现这一目标。
SELECT
@roll_qty := CASE WHEN @id_itm = dt.id_item
THEN @roll_qty + dt.qty_change
ELSE dt.qty_change
END AS qty,
@id_itm := dt.id_item AS id_item,
dt.year,
dt.month
FROM
(
SELECT
t.id_item,
SUM(t.qty * t.t_in - t.qty * t.t_out) AS qty_change,
YEAR(t.created_at) AS `year`,
LPAD(MONTH(t.created_at), 2, '0') AS `month`
FROM your_table AS t
GROUP BY t.id_item, `year`, `month`
ORDER BY t.id_item, `year`, `month`
) AS dt
CROSS JOIN (SELECT @roll_qty := 0,
@id_itm := 0
) AS user_init_vars;
| id_item | year | month | qty |
| ------- | ---- | ----- | --- |
| 1 | 2018 | 07 | 5 |
| 1 | 2018 | 08 | 5 |
| 1 | 2018 | 09 | 15 |
| 1 | 2018 | 10 | 10 |
select yyyy, mm, total,
(@t := if(@ym = concat_ws('-', yyyy, mm), @t + total,
@ym := concat_ws('-', yyyy, mm), total
)
) as running_total
from (select year(created_at) as yyyy, month(created_at) as mm,
id_item,
sum(case T_IN when 1 then qty else - qty end) as total
from transactions
where created_at < '2018-11-01'
group by id_item
order by id_item, min(created_at)
) i cross join
(select @ym := '', @n := 0);