Php “帮助获得”;“正确”;数据库中的mysql计数
嘿,伙计们,我有个问题Php “帮助获得”;“正确”;数据库中的mysql计数,php,mysql,database,count,Php,Mysql,Database,Count,嘿,伙计们,我有个问题 "SELECT COUNT(purchase_log.id), purchase_log.date_purchased, purchase_log.total_cost, purchase_log.payment_status, cart_contents.product_name, members.first_name, members.last_name, members.email FROM `purchase_log` LEFT JOIN `cart_conten
"SELECT COUNT(purchase_log.id), purchase_log.date_purchased, purchase_log.total_cost, purchase_log.payment_status, cart_contents.product_name, members.first_name, members.last_name, members.email FROM `purchase_log` LEFT JOIN `cart_contents` ON purchase_log.id = cart_contents.purchase_id LEFT JOIN `members` ON purchase_log.member_id = members.id";
如你所见,我有3个不同的表,我试图从中提取数据。运行此查询后,返回的计数为9。这实际上是表“cart\u contents”中的总行数,但我希望它返回5,因为表“purchase\u log”包含每行的总事务,并且每个事务在表“cart\u contents”中可以有1行或多行
那么如何形成查询以计算正确的金额呢?您可以尝试
选择计数(不同)
购买_log.id),
购买日,
采购日志。总成本,
采购\日志。付款\状态,
购物车内容。产品名称,
members.first_name,members.last_name,
members.email FROMpurchase\u log
LEFT
加入cart\u内容
ON
购买日志id=
购物车\u contents.purchase\u id左加入
会员
购买日志。会员id=
成员id
是的,这是有效的…我确实尝试过这个,但我很愚蠢,我把不同的计数(purchase_log.id)而不是你所拥有的…谢谢你的帮助!