使用 whereDate 从包含离今天最近的日期的行中检索数据
Retrieving data from the row containing the closest date to today with whereDate
我想在按日期筛选和分组后按价格排序。但是,因为关系不止一个,我得不到我想要的结果。
我想要的结果是得到与end_date最接近的关系的价格,并据此排序。
为此,查询、sql输出、表格和演示页面如下。
提前致谢..
$query->join('tableB', 'tableA.id', '=', 'tableB.pro_id')
->select('tableA.*', 'tableB.start_date', 'tableB.end_date', 'tableB.old_daily')
->where(function($sq) {
$today = Carbon::now()->format('Y-m-d');
$sq->whereDate('end_date', '>=', $today);
})
->groupBy('tableA.id')
->orderBy('price', desc);
查询:
select `tableA`.*, `tableB`.`start_date`, `tableB`.`end_date`, `tableB`.`price`
from `tableA`
inner join `tableB` on `tableA`.`id` = `tableB`.`pro_id`
where (date(`end_date`) >= 2021-03-07)
group by `tableA`.`id`
order by `price` desc
表A
| id | title |
|----|-------|
| 1 | pro1 |
| 2 | pro2 |
| 3 | pro3 |
表B
| id | start_date | end_date | price | pro_id |
|----|------------|------------|-------|--------|
| 1 | 2021-06-01 | 2021-06-05 | 750 | 2 |
| 2 | 2021-05-01 | 2021-05-05 | 850 | 2 |
| 3 | 2021-04-01 | 2021-04-05 | 650 | 2 |
| 4 | 2021-06-01 | 2021-06-05 | 2750 | 1 |
| 5 | 2021-05-01 | 2021-05-05 | 2850 | 1 |
| 6 | 2021-04-01 | 2021-04-05 | 2650 | 1 |
| 7 | 2021-06-01 | 2021-06-05 | 1750 | 3 |
| 8 | 2021-05-01 | 2021-05-05 | 1850 | 3 |
| 9 | 2021-04-01 | 2021-04-05 | 1650 | 3 |
the closest to the end_date and sort it accordingly.
您应该找到给定日期和结束日期之间的差异,然后按升序排序。
ORDER BY DATEDIFF(end_date, '2021-03-07') ASC
这个查询给出了你想要的结果。
这一步使用“右连接”会是一个不错的选择。
select `tableA`.*, `tableB`.`start_date`, `tableB`.`end_date`, `tableB`.`price`
from `tableA`
right join(
SELECT id, start_date, end_date, pro_id, price, DATEDIFF(`tableB`.`end_date`, '2021-03-07') diff
FROM `tableB`
GROUP BY id order by diff asc
) `tableB` on `tableA`.`id` = `tableB`.`pro_id`
where (date(`end_date`) >= '2021-03-07')
group by `tableA`.`id`
order by `price` desc
我想在按日期筛选和分组后按价格排序。但是,因为关系不止一个,我得不到我想要的结果。
我想要的结果是得到与end_date最接近的关系的价格,并据此排序。
为此,查询、sql输出、表格和演示页面如下。 提前致谢..
$query->join('tableB', 'tableA.id', '=', 'tableB.pro_id')
->select('tableA.*', 'tableB.start_date', 'tableB.end_date', 'tableB.old_daily')
->where(function($sq) {
$today = Carbon::now()->format('Y-m-d');
$sq->whereDate('end_date', '>=', $today);
})
->groupBy('tableA.id')
->orderBy('price', desc);
查询:
select `tableA`.*, `tableB`.`start_date`, `tableB`.`end_date`, `tableB`.`price`
from `tableA`
inner join `tableB` on `tableA`.`id` = `tableB`.`pro_id`
where (date(`end_date`) >= 2021-03-07)
group by `tableA`.`id`
order by `price` desc
表A
| id | title |
|----|-------|
| 1 | pro1 |
| 2 | pro2 |
| 3 | pro3 |
表B
| id | start_date | end_date | price | pro_id |
|----|------------|------------|-------|--------|
| 1 | 2021-06-01 | 2021-06-05 | 750 | 2 |
| 2 | 2021-05-01 | 2021-05-05 | 850 | 2 |
| 3 | 2021-04-01 | 2021-04-05 | 650 | 2 |
| 4 | 2021-06-01 | 2021-06-05 | 2750 | 1 |
| 5 | 2021-05-01 | 2021-05-05 | 2850 | 1 |
| 6 | 2021-04-01 | 2021-04-05 | 2650 | 1 |
| 7 | 2021-06-01 | 2021-06-05 | 1750 | 3 |
| 8 | 2021-05-01 | 2021-05-05 | 1850 | 3 |
| 9 | 2021-04-01 | 2021-04-05 | 1650 | 3 |
the closest to the end_date and sort it accordingly.
您应该找到给定日期和结束日期之间的差异,然后按升序排序。
ORDER BY DATEDIFF(end_date, '2021-03-07') ASC
这个查询给出了你想要的结果。 这一步使用“右连接”会是一个不错的选择。
select `tableA`.*, `tableB`.`start_date`, `tableB`.`end_date`, `tableB`.`price`
from `tableA`
right join(
SELECT id, start_date, end_date, pro_id, price, DATEDIFF(`tableB`.`end_date`, '2021-03-07') diff
FROM `tableB`
GROUP BY id order by diff asc
) `tableB` on `tableA`.`id` = `tableB`.`pro_id`
where (date(`end_date`) >= '2021-03-07')
group by `tableA`.`id`
order by `price` desc