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我实际上在 virturt mart 表上有三个表，一个是订单，另一个是项目，一个是 order_user_info 以获取用户名我需要加入order_user_info表

但是当我加入时，它显示结果信息加倍，下面我提到了查询和结果，请指导如何避免双重结果

*对于加入名字，我正在使用下面提到的查询 *

LEFT JOIN `urbanite_virtuemart_order_userinfos` 作为 Uinfo ON Uinfo.virtuemart_order_id=i.virtuemart_order_id

*完成查询 *

SELECT SQL_CALC_FOUND_ROWS o.created_on AS 间隔，CAST(i.`created_on` AS DATE) AS created_on, Uinfo.`first_name`, o.`order_number`, SUM(DISTINCT i.product_item_price * product_quantity) as order_subtotal_netto, SUM(DISTINCT i. product_subtotal_with_tax) 作为 order_subtotal_brutto，COUNT(DISTINCT i.virtuemart_order_id) 作为 count_order_id，SUM(i.product_quantity) 作为 product_quantity FROM `urbanite_virtuemart_order_items` 作为 i

LEFT JOIN `urbanite_virtuemart_orders` as o ON o.virtuemart_order_id=i.virtuemart_order_id
LEFT JOIN `urbanite_virtuemart_order_userinfos` 作为 Uinfo ON Uinfo.virtuemart_order_id=i.virtuemart_order_id AND Uinfo.created_on = i.created_on AND Uinfo.virtuemart_user_id = o.virtuemart_user_id
WHERE (`i`.`order_status` = "S") AND i.virtuemart_vendor_id = "63" AND DATE(o.created_on) BETWEEN "2013-06-01 05:00:00" AND "2013-06-30 05" :00:00"

按间隔分组
ORDER BY created_on DESC LIMIT 0, 400

结果我得到了像下面这样的加入

intervals | Created_on | order_no | order_subtotalnetto | order_subtotalbruto | count_order_id | product_quantity

2013-06-12 09:47:16 |2013-06-12 | 43940624 | 200.00000 | 200.00000 | 1 | 2

结果我得到了加入名字，如下所示

intervals | Created_on | order_no | f_name | order_subtotalnetto | order_subtotalbruto | count_order_id | product_quantity

2013-06-12 09:47:16 |2013-06-12 | Fatin Bokhari | 43940624 | 200.00000 | 200.00000 | 1 | 4

在没有加入名字的情况下看到它显示 product_quantity = 2 但是当我加入时它显示值加倍，我尝试了不同但不能这样，因为它每次都显示产品数量 = 1

急需救援！

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1 回答 1

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哦，实际上这些行在urbanite_virtuemart_order_userinfos表中出现了两次，所以我使用了 where 子句并且它有效

WHERE (`i`.`order_status` = "S") AND i.virtuemart_vendor_id = "63" AND DATE(o.created_on) BETWEEN "2013-06-01 05:00:00" AND "2013-06-30 05" :00:00" AND Uinfo.`address_type` = 'BT'
于 2013-06-14T13:08:35.470 回答