[英]Yii2 - Query count per day to ActiveRecord
我具有以下表结构,并使用Yii2 ActiveRecord
方法来提取供应商下周每天的预订数量( OrderLine
)(也需要输入0)。 因此,某种方式可以使每个供应商每天排成一行, num_bookings
或可能为0
具体取决于供应商。
/--------------------\ /------------\
| OrderLine |------------------|Availability|
|--------------------| 0..n 1 |------------|
|ID {PK} | |ID {PK} |
|availabilityID {FK} | |start |
|line_status | \------------/
|supplierID {FK} |
\--------------------/
| 1
|
|
| 1
/----------\
| Supplier |
|----------|
|ID {PK} |
\----------/
使用DAO,通过以下SQL直接查询数据库,可以(几乎)获得所需的结果,
select count(ol.ID) as num_bookings,
day(from_unixtime(a.start)) as order_day,
ol.supplierID
from order_line ol left join
availability a on ol.availabilityID = a.ID
where ol.line_status = "booked"
and a.start >= 1451952000 //magic number for midnight today
and a.start <= 1452556800 //magic number for seven days from now
group by order_day, ol.supplierID;
类似于
------------------------------------
| num_bookings|order_day|supplierID|
------------------------------------
| 1 | 5 | 3 |
| 2 | 5 | 7 |
| 1 | 6 | 7 |
| 1 | 7 | 7 |
------------------------------------
因此,对于给定的Supplier
没有预订的日期,应输入0
,就像这样
------------------------------------
| num_bookings|order_day|supplierID|
------------------------------------
| 1 | 5 | 3 |
| 0 | 6 | 3 |
| 0 | 7 | 3 |
| 2 | 5 | 7 |
| 1 | 6 | 7 |
| 1 | 7 | 7 |
------------------------------------
[days 8+ omitted for brevity...]
我有一些php / Yii代码,它们最终会给我类似的东西,但是涉及多个查询和数据库连接,如下所示,
$suppliers = Supplier::find()->all(); // get all suppliers
$start = strtotime('tomorrow');
$end = strtotime('+7 days', $start); // init times
// create empty assoc array with key for each of next 7 days
$booking_counts[date('D j', $start)] = 0;
for ($i=1; $i<7; ++$i) {
$next = strtotime('+'.$i." days", $start);
$booking_counts[date('D j', $next)] = 0;
}
foreach ($suppliers as $supplier) {
$bookings = OrderLine::find()
->joinWith('availability')
->where(['order_line.supplierID' => $supplier->ID])
->andWhere(['>=', 'availability.start', $start])
->andWhere(['<=', 'availability.start', $end])
->andWhere(['order_line.line_status' => 'booked'])
->orderBy(['availability.start' => SORT_ASC])
->all();
$booking_count = $booking_counts;
foreach ($bookings as $booking) {
$booking_count[date('D j', $booking->availability->start)] += 1;
}
}
这给了我一个供每个供应商使用的数组,其计数存储在适当的当日索引下,但感觉效率很低。
我可以重构该代码以更少的数据库调用和更少的“脚手架”代码返回所需的数据吗?
这可能是您选择的第一个问题
$results = OrderLine::find()
->select('count(order_line.ID) as num_bookings, day(from_unixtime(availability.start)) as order_day', order_line.supplierID )
->from('order_line')
->leftjoin('availability', 'order_line.availabilityID = availability.ID')
->where( 'order_line.line_status = "booked"
and a.start >= 1451952000
and a.start <= 1452556800')
->groupBy(order_day, order_line.supplierID)
->orderBy(['availability.start' => SORT_ASC])
->all();
通过这种方式,您应该获得供应商ID(和order_day)行,从而避免了供应商上的foreach
为了获取$ results-> num_bookings和order_day中的数据,您需要添加
public $num_bookings;
public $order_day;
在您的OrderLine模型中
我希望这是您要寻找的。
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