1

我必须在哪里写一个查询，我需要获取上周、上个月以及所有的记录。对于这个问题，我写了 3 个不同的查询（上周，上个月和所有）

对于每周信息：-

SELECT bu.brand_name AS 'Brand_Name',COUNT(s.unique) AS '# Item Sold',SUM(s.price) AS 'Total_Price'
FROM item_details s
LEFT JOIN sales_order o ON s.fk_sales_order = o.id_sales_order 
LEFT JOIN customer_info AS c ON o.fk_customer_id = c.id_customer
LEFT JOIN simple_details cc ON s.unique = cc.unique
LEFT JOIN config_details  cf ON cc.fk_config_id = cf.config_id 
LEFT JOIN brand_details  cb ON cf.fk_brand_id = cb.brand_id 
LEFT JOIN category_details  ctc ON cf.fk_category_id = ctc.category_id 
LEFT JOIN gender_details  g ON cf.fk_gender_id = g.gender_id
LEFT JOIN buyers AS bu  ON bu.brand_name =  cb.name AND bu.category_name = ctc.name AND bu.gender = g.name
WHERE  bu.buyers = 'xyz'  AND DATE_FORMAT(o.created_date,'%Y-%m-%d') >= @weekstartdate AND  DATE_FORMAT(o.created_date,'%Y-%m-%d') <= @weekenddate
GROUP BY bu.brand_name

每月信息：-

SELECT bu.brand_name AS 'Brand_Name',COUNT(s.unique) AS '# Item Sold',SUM(s.price) AS 'Total_Price'
FROM item_details s
LEFT JOIN sales_order o ON s.fk_sales_order = o.id_sales_order 
LEFT JOIN customer_info AS c ON o.fk_customer_id = c.id_customer
LEFT JOIN simple_details cc ON s.unique = cc.unique
LEFT JOIN config_details  cf ON cc.fk_config_id = cf.config_id 
LEFT JOIN brand_details  cb ON cf.fk_brand_id = cb.brand_id 
LEFT JOIN category_details  ctc ON cf.fk_category_id = ctc.category_id 
LEFT JOIN gender_details  g ON cf.fk_gender_id = g.gender_id
LEFT JOIN buyers AS bu  ON bu.brand_name =  cb.name AND bu.category_name = ctc.name AND bu.gender = g.name
WHERE  bu.buyers = 'xyz'  AND DATE_FORMAT(o.created_date,'%Y-%m-%d') >= @monthstartdate AND  DATE_FORMAT(o.created_date,'%Y-%m-%d') <= @monthenddate
GROUP BY bu.brand_name

对于所有记录：-

SELECT bu.brand_name AS 'Brand_Name',COUNT(s.unique) AS '# Item Sold',SUM(s.price) AS 'Total_Price'
FROM item_details s
LEFT JOIN sales_order o ON s.fk_sales_order = o.id_sales_order 
LEFT JOIN customer_info AS c ON o.fk_customer_id = c.id_customer
LEFT JOIN simple_details cc ON s.unique = cc.unique
LEFT JOIN config_details  cf ON cc.fk_config_id = cf.config_id 
LEFT JOIN brand_details  cb ON cf.fk_brand_id = cb.brand_id 
LEFT JOIN category_details  ctc ON cf.fk_category_id = ctc.category_id 
LEFT JOIN gender_details  g ON cf.fk_gender_id = g.gender_id
LEFT JOIN buyers AS bu  ON bu.brand_name =  cb.name AND bu.category_name = ctc.name AND bu.gender = g.name
WHERE  bu.buyers = 'xyz'  
GROUP BY bu.brand_name

这些工作正常（提供正确的输出）。但问题是，我必须将这三个查询合并为一个。输出应该是品牌名称，item_sold(week)，total_price(week)，item_sold(month)，total_price(month)，item_sold(all)，total_price(all) 我该如何编写这个查询？

4

4 回答 4

3

无需深入研究您的代码，显而易见的解决方案是

SELECT
    all.brand_name
    pw.items_sold items_sold_week
    pw.total_price total_price_week
    pm.items_sold items_sold_month
    pm.total_price total_price_month
    all.items_sold items_sold_all
    all.total_price total_price_all
FROM
    (your all-time select) all
    JOIN (your per-month select) pm ON all.brand_name = pm.brand_name
    JOIN (your per-week select) pw ON all.brand_name = pw.brand_name

尽管您可能应该重新考虑您的整个方法，并确定您是否真的想要在 DB 层中使用这种逻辑，或者最好是在您的应用程序中。

于 2012-06-28T07:31:39.660 回答
2

您可以使用case将聚合限制为行的子集：

select  bu.brand_name
,       count(case when date_format(o.created_date,'%Y-%m-%d') >= @weekstartdate
             and date_format(o.created_date,'%Y-%m-%d') <= @weekenddate 
             then 1 end) as '# Item Sold Week'
,       sum(case when date_format(o.created_date,'%Y-%m-%d') >= @weekstartdate
             and date_format(o.created_date,'%Y-%m-%d') <= @weekenddate 
             then s.price end) as 'Total_Price Week'
,       count(case when date_format(o.created_date,'%Y-%m-%d') >= @monthstartdate
             and date_format(o.created_date,'%Y-%m-%d') <= @monthstartdate
             then 1 end) as '# Item Sold Month'
,       ...
于 2012-06-28T07:33:28.253 回答
1

如果所有三个选择在结果中使用相同的字段，您可以将它们联合起来：

SELECT * 
   FROM  (SELECT 1) AS a
   UNION (SELECT 2) AS b
   UNION (SELECT 3) AS c

如果您需要相互区分周/周一/所有记录 - 只需添加包含“周”或“周一”的常量字段

于 2012-06-28T07:30:20.560 回答
0

您可以在查询之间使用 UNION.keyword 将它们捆绑在一起。但是所有查询中的列类型和序列必须相同。您可以为每个集合添加一个标识符

于 2012-06-28T07:31:05.190 回答