当我使用 left join 和 group by 时,sum 不起作用
sum doesn't work while I use left join and group by
我有两个table:进口和订单:我已经分别附上了。
我想要以下内容:
1. 进口总额product_idtable
2. 订单 table 中相同 product_id 件的总和作为状态明智。
我的查询是:
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`
此查询的结果:
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
)
)
)
我的预期结果是:
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
)
)
)
我该怎么做?我尝试了不同的方式,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`
但运气不好:'(
我猜您需要有关每种产品的信息。如果是这样,您可以通过几种方式做到这一点。这是一个 union
all 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
我有两个table:进口和订单:我已经分别附上了。
我想要以下内容: 1. 进口总额product_idtable 2. 订单 table 中相同 product_id 件的总和作为状态明智。
我的查询是:
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`
此查询的结果:
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
)
)
)
我的预期结果是:
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
)
)
)
我该怎么做?我尝试了不同的方式,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`
但运气不好:'(
我猜您需要有关每种产品的信息。如果是这样,您可以通过几种方式做到这一点。这是一个 union
all 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