Question

我有一个数据库，包含我在线网上商店的所有交易，我试图查询打印出简单的财务报表。

它会打印在这样的表格中：

<th>month</th>
<th>number of sales</th>
<th>money in</th>
<th>money out</th>
<th>result</th>

失败的查询：＃1111 - 无效使用组功能

SELECT 
month(transaction_date) as month,
count(incoming_amount > '0') as number_of_sales,
sum(incoming_amount / 1.25) as money_in,
sum(outgoing_amount) as money_out,
sum((incoming_amount / 1.25) - sum(outgoing_amount)) as result
FROM myDB WHERE year(timestamp) = '2011' order by id desc");

有人能指出我正确的方向吗？

Answer 1

SELECT 
month(transaction_date) as month,
sum(if(incoming_amount>0,1,0)) as number_of_sales,
sum(incoming_amount)/1.25 as money_in,
sum(outgoing_amount) as money_out,
sum((incoming_amount/1.25)-outgoing_amount) as result
FROM myDB 
WHERE timestamp>='2011-01-01 00:00:00' AND timestamp<='2011-12-11 23:59:59'
GROUP BY month;

您需要在使用聚合函数时指定列
year(timestamp)没有使用mysql索引（如果你在时间戳上定义了一个索引）
count(incoming_amount > '0')上的汇总功能不正确
sum看起来也不正确

Answer 2

按语句添加分组：

SELECT 
month(transaction_date) as month,
count(incoming_amount > '0') as number_of_sales,
sum(incoming_amount / 1.25) as money_in,
sum(outgoing_amount) as money_out,
sum((incoming_amount / 1.25) - sum(outgoing_amount)) as result
FROM myDB WHERE year(timestamp) = '2011' GROUP BY month order by id desc");

Answer 3

在@ ajreal的答案的基础上，您可以通过重复使用以前计算的值来加快此查询：

SELECT s.*,
       (s.money_in - s.money_out) as result 
FROM
  (
  SELECT 
    month(transaction_date) as month,
    /*  year(transaction_date) as year   */  
    sum(incoming_amount>0) as number_of_sales, -- true = 1, false = 0.
    sum(incoming_amount)/1.25 as money_in,
    sum(outgoing_amount) as money_out,
  FROM myDB 
  WHERE transaction_date BETWEEN '2011-01-01 00:00:00' AND '2011-12-31 23:59:59'
  GROUP BY /*year,*/ month DESC;
  ) AS s

如果您选择超出年份，请取消注释相关部分请注意，您可以向DESC添加group by修饰符，以便首先获得最新结果。

帮助mysql查询

3 个答案: