SQL获取值，然后这些值的总数

Question

我正在编写一个拍卖脚本，其中包含一个看起来像所附图像的表格。

在拍卖中，两个用户可以出价相同，例如，两个玩家可以出价“1”，这将使该出价总数为 2。

我需要一个查询来获取所有用户的单个出价金额以及拍卖 ID 范围内该出价金额的总数。

就目前而言，我首先得到用户的出价：

SELECT bid_amount FROM all_bids WHERE auction_id = '129' AND user_id = '9'

然后我在 PHP 中遍历每个数量并执行以下操作

SELECT COUNT(*) FROM all_bids WHERE auction_id = '129' AND bid_amount = 'xxx'

但这当然是非常耗电的，我相信它可以在一个查询中完成。

我已经做到了

SELECT bid_amount,COUNT(*) FROM (SELECT bid_amount FROM all_bids WHERE auction_id = '129' AND user_id ='9') as foo GROUP BY bid_amount

返回正确的出价金额但计数都是 1 这是错误的，因为我猜我的错误查询仅计算子查询内的值，而不是仅计算该用户放置的出价之外的值

Answer 1

 SELECT  a.*, b.cnt from all_bids a 
 join (select bid_amount,COUNT(*) cnt from all_bids group by bid_amount) b 
 on a.bid_amount=b.bid_amount
 WHERE a.auction_id = '123' AND a.player_id = '456'

Answer 2

我想你想要group by:

select bid_amount, count(*)
from all_bids 
where auction_id = 123 and player_id = 456
group by bid_amount;

请注意，数字常量不需要单引号。

嗯。您可能希望将其表述为：

select ab.bid_amount, count(*)
from all_bids ab
where ab.auction_id = 123 and
      ab.bid_amount in (select ab2.bid_amount
                        from all_bids ab2
                        where ab2.auction_id = ab.auction_id and ab2.player_id = 456
                       )
group by ab.bid_amount;