我收到的结果:
id sku name GROUP_CONCAT(quantity_received) GROUP_CONCAT(item_cost)
4 00004 Antibacterial Wipes 50,14,25,309,50,14,25,309,50,14,25,309,50,14,25,309,50,14,25,309,50,14,25,309,50,14,25,309,50,14,25,309,50,14,25,309 3.29,3.29,3.29,3.49,3.29,3.29,3.29,3.49,3.29,3.29,3.29,3.49,3.29,3.29,3.29,3.49,3.29,3.29,3.29,3.49,3.29,3.29,3.29,3.49,3.29,3.29,3.29,3.49,3.29,3.29,3.29,3.49,3.29,3.29,3.29,3.49
我想要的结果:
id sku name GROUP_CONCAT(DISTINCT quantity_received) GROUP_CONCAT(DISTINCT item_cost)
4 00004 Antibacterial Wipes 50,14,25,309 3.29,3.49
我解决此问题的方法是将 DISTINCT 放在 quantity_recieved 选择中。问题是,如果数量有两个相同的值,例如 50、50、14、25。结果将是 50、14、25。我只想摆脱重复的数字,只得到一次值.
这是查询:
SELECT `product`.`id`,`product`.`sku`,`product`.`name`,
case when coalesce(stock1.`quantity`, '') = ''
then '0'
else stock1.`quantity`
end as qty_warehouse,
case when coalesce(sum(distinct stock2.`quantity`), '') = ''
then '0'
else sum(distinct stock2.`quantity`)
end as qty_events,
case when coalesce(stock1.`quantity`, '') = ''
then '0'
else stock1.`quantity`
end +
case when coalesce(sum(distinct stock2.`quantity`), '') = ''
then '0'
else sum(distinct stock2.`quantity`)
end as qty_total
GROUP_CONCAT(DISTINCT quantity_received) ,
GROUP_CONCAT(DISTINCT item_cost)
FROM (`product`)
LEFT JOIN`shipping_event`
ON `shipping_event`.`product_id` = `product`.`id`
LEFT JOIN `product_stock` as stock1
ON `product`.`id` = `stock1`.`product_id` and `stock1`.`location_id` = 112
LEFT JOIN `product_stock` as stock2
ON `product`.`id` = `stock2`.`product_id` and `stock2`.`location_id` != 112
LEFT JOIN `shipping_list`
ON `shipping_event`.`shipping_list_id` = `shipping_list`.`id`
WHERE `shipping_list`.`type` = 'incoming'
AND `shipping_event`.`end_date` > '2004-01-01 01:01:01'
GROUP BY `product`.`id`
ORDER BY `sku` asc LIMIT 20
在这种情况下,我使用 group concat 只是为了显示结果。实际上,我将 quantity_received 相加,然后将它们乘以项目成本。