当我在窗口系统中运行此查询时,行为正确UNNSET 但是当我运行这个查询时,Linux表现不同。不同行上的重复记录列表
SELECT DISTINCT
"billing_billmanagement"."creation_date",
"billing_billmanagement"."bill_number",
unnest(array_agg(DISTINCT "inventory_product"."product_name")) AS "product",
unnest(array_agg(DISTINCT "services_service"."name")) AS "service"
FROM "billing_billmanagement"
INNER JOIN "users_staffuser" ON ("billing_billmanagement"."staff_id" = "users_staffuser"."id")
INNER JOIN "auth_user" ON ("users_staffuser"."user_id" = "auth_user"."id")
LEFT OUTER JOIN "billing_customerproductbill" ON ("billing_billmanagement"."id" = "billing_customerproductbill"."bill_id")
LEFT OUTER JOIN "inventory_product" ON ("billing_customerproductbill"."product_id" = "inventory_product"."id")
LEFT OUTER JOIN "billing_customerservicebill" ON ("billing_billmanagement"."id" = "billing_customerservicebill"."bill_id")
LEFT OUTER JOIN "services_service" ON ("billing_customerservicebill"."service_id" = "services_service"."id")
WHERE "billing_billmanagement"."creation_date" BETWEEN '2017-12-04' AND '2017-12-06'
GROUP BY billing_billmanagement.creation_date,
billing_billmanagement.bill_number
ORDER BY "billing_billmanagement"."creation_date" ASC
答案 0 :(得分:0)
如果出现重复行问题,请尝试使用
SELECT billing_billmanagement.creation_date,
billing_billmanagement.bill_number,
inventory_product.product_name AS product,
services_service.name AS service
FROM billing_billmanagement
INNER JOIN users_staffuser ON (billing_billmanagement.staff_id = users_staffuser.id)
INNER JOIN auth_user ON (users_staffuser.user_id = auth_user.id)
LEFT OUTER JOIN billing_customerproductbill ON (billing_billmanagement.id = billing_customerproductbill.bill_id)
LEFT OUTER JOIN inventory_product ON (billing_customerproductbill.product_id = inventory_product.id)
LEFT OUTER JOIN billing_customerservicebill ON (billing_billmanagement.id = billing_customerservicebill.bill_id)
LEFT OUTER JOIN services_service ON (billing_customerservicebill.service_id = services_service.id)
WHERE billing_billmanagement.creation_date BETWEEN '2017-12-04' AND '2017-12-06'
GROUP BY 1,
2,
3,
4
ORDER BY 1 ASC;