[英]sum doesn't work while I use left join and group by
我有兩個表:進口和訂單:我分別附上了這些。
我想要以下內容:1。導入表中相同product_id的總和2.作為狀態明智的訂單表中相同product_id的總和。
我的查詢是:
SELECT `Import`.*,
SUM( case when orders.status = "sold" THEN orders.pieces else 0 end) as total_sell,
SUM( case when orders.status = "No contact" THEN orders.pieces else 0 end) as no_contact,
SUM( case when orders.status = "confirmed" THEN orders.pieces else 0 end) as confirmed,
SUM( case when orders.status = "canceled" THEN orders.pieces else 0 end) as canceled
FROM `amrajegeachi`.`imports` AS `Import`
LEFT JOIN `orders`
ON `Import`.`product_id` = `orders`.`product_id`
WHERE 1 = 1
GROUP BY `Import`.`id`
和此查詢的結果:
Array
(
[0] => Array
(
[Import] => Array
(
[id] => 1
[category_id] => 2
[product_id] => 2
[amount] => 50
[cost] => 8320
[comment] => transportation and others cost: 100
[created] => 2015-06-23 19:21:10
)
[0] => Array
(
[total_sell] => 10
[no_contact] => 1
[confirmed] => 2
[canceled] => 0
)
)
[1] => Array
(
[Import] => Array
(
[id] => 2
[category_id] => 2
[product_id] => 2
[amount] => 15
[cost] => 3000
[comment] =>
[created] => 2015-06-22 18:10:36
)
[0] => Array
(
[total_sell] => 10
[no_contact] => 1
[confirmed] => 2
[canceled] => 0
)
)
[2] => Array
(
[Import] => Array
(
[id] => 3
[category_id] => 2
[product_id] => 1
[amount] => 15
[cost] => 2000
[comment] =>
[created] => 2015-06-23 19:20:15
)
[0] => Array
(
[total_sell] => 10
[no_contact] => 0
[confirmed] => 0
[canceled] => 0
)
)
)
我的預期結果是:
Array
(
[0] => Array
(
[Import] => Array
(
[id] => 1
[category_id] => 2
[product_id] => 2
[amount] => 65
[cost] => 8320
[comment] => transportation and others cost: 100
[created] => 2015-06-23 19:21:10
)
[0] => Array
(
[total_sell] => 10
[no_contact] => 1
[confirmed] => 2
[canceled] => 0
)
)
[2] => Array
(
[Import] => Array
(
[id] => 3
[category_id] => 2
[product_id] => 1
[amount] => 15
[cost] => 2000
[comment] =>
[created] => 2015-06-23 19:20:15
)
[0] => Array
(
[total_sell] => 10
[no_contact] => 0
[confirmed] => 0
[canceled] => 0
)
)
)
我怎樣才能做到這一點? 我嘗試過不同的方式,例如:
SELECT `Import`.*, SUM(`Import`.`amount`) as total_import,
SUM( case when orders.status = "sold" THEN orders.pieces else 0 end) as total_sell,
SUM( case when orders.status = "No contact" THEN orders.pieces else 0 end) as no_contact,
SUM( case when orders.status = "confirmed" THEN orders.pieces else 0 end) as confirmed,
SUM( case when orders.status = "canceled" THEN orders.pieces else 0 end) as canceled
FROM `amrajegeachi`.`imports` AS `Import`
LEFT JOIN `orders`
ON `Import`.`product_id` = `orders`.`product_id`
WHERE 1 = 1
GROUP BY `Import`.`id`
但沒有運氣:'(
我猜您想要了解每種產品的信息。 如果是這樣,您可以通過幾種方式實現此目的。 這是一個union
所有方法:
SELECT product_id, sum(amount) as total_import,
sum( case when status = 'sold' THEN pieces else 0 end) as total_sell,
sum( case when status = 'No contact' THEN pieces else 0 end) as no_contact,
sum( case when status = 'confirmed' THEN pieces else 0 end) as confirmed,
sum( case when status = 'canceled' THEN pieces else 0 end) as canceled
from ((select i.product_id, amount, NULL as status, NULL as pieces
from `amrajegeachi`.`imports` i
) union all
(select o.product_id, NULL, o.status, o.pieces
from `orders` o
)
) io
group by product_id;
SELECT `Import`.*, SUM(`Import`.`amount`) as total_import,
SUM( case when orders.status = "sold" THEN orders.pieces else 0 end) as total_sell,
SUM( case when orders.status = "No contact" THEN orders.pieces else 0 end) as no_contact,
SUM( case when orders.status = "confirmed" THEN orders.pieces else 0 end) as confirmed,
SUM( case when orders.status = "canceled" THEN orders.pieces else 0 end) as canceled
FROM `amrajegeachi`.`imports` AS `Import`
LEFT JOIN `orders`
ON `Import`.`product_id` = `orders`.`product_id`
WHERE 1 = 1
GROUP BY `Import`.`id
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