Question

我的mysql数据库上有这个表：

date       | client | total
2015-01-01 | john   | 85.00
2015-01-01 | alfred | 35.00
2015-01-01 | georgy | 125.00
2015-01-01 | linda  | 42.00
2015-01-01 | alex   | 76.00
2015-01-01 | john   | 94.00
2015-01-01 | john   | 75.30

我想按客户名称分组计算总金额和百分比，所以我提出了这个问题：

SELECT client, total, round(total*100/t.th,1) as percent 
FROM ( 
    SELECT client, sum(total) as total 
    FROM mytable 
    WHERE date='2015-01-01' 
    GROUP BY `client`) c 
JOIN ( select sum(total) as th from mytable 
    WHERE date='2015-01-01') t 
order by percent desc

它运作良好，但现在我只想列出3个第一个结果，所以我在最后添加了LIMIT 3但是它不起作用，它仍然显示了相同数量的结果。

更新

当我执行此查询时，我获得：

client | total | percent
john   | 254.3 | 47.8
georgy | 125   | 23.5
alex   | 76    | 14.3
linda  | 42    | 7.9
alfred | 35    | 6.6

直到这里是正确的，但现在我只想要3个第一行，所以我修改：

SELECT client, total, round(total*100/t.th,1) as percent 
FROM ( 
    SELECT client, sum(total) as total 
    FROM mytable 
    WHERE date='2015-01-01' 
    GROUP BY `client`) c 
JOIN ( select sum(total) as th from mytable 
    WHERE date='2015-01-01') t 
order by percent desc LIMIT 3

但它不起作用，它仍然显示相同的结果，我想要这个：

client | total | percent
john   | 254.3 | 47.8
georgy | 125   | 23.5
alex   | 76    | 14.3

我该如何解决？

Answer 1

尝试以下查询，在变量中存储特定日期的行数，以便您不需要交叉加入：

select @rows := sum(total) from myTable where date = '2015-01-01';

SELECT client, total, round(total*100/@rows,1) as percent 
FROM ( 
    SELECT client, sum(total) as total 
    FROM mytable 
    WHERE date='2015-01-01' 
    GROUP BY `client`
     )
order by percent desc
LIMIT 3;

或者，您可以执行以下操作，而不是编写2个不同的查询：

SELECT client, total, round(total*100/total_sum,1) as percent 
    FROM ( 
        SELECT client, @total_sum:= @total_sum + sum(total) as total_sum, sum(total) as total 
        FROM mytable 
        WHERE date='2015-01-01' 
        GROUP BY `client`
         ), (select @total_sum:=0) as t
    order by percent desc
    LIMIT 3;

MySQL LIMIT不起作用

1 个答案: