mysql查询帮助答案

【问题标题】：help with a mysql querymysql查询帮助
【发布时间】：2011-11-07 14:16:23
【问题描述】：

我有一个数据库，其中包含我的在线网店的所有交易，我正在尝试进行查询以打印出一份简单的财务报表。

它将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>

查询失败：#1111 - 无效使用组函数

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");

谁能指出我正确的方向？

【问题讨论】：

你真的有一个叫myDB的表吗？

标签： mysql sql mysql-error-1111

【解决方案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;

使用聚合函数时需要指定列
year(timestamp) 不用于 mysql 索引（如果您在时间戳上定义了索引）
count(incoming_amount > '0') 上的聚合函数不正确
sum 看起来也不正确

【讨论】：

【解决方案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");

【讨论】：

【解决方案3】：

基于@ajreal 的回答，您可以通过重用之前计算的值来加快查询速度，如下所示：

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，您可以将 DESC 修饰符添加到 group by 以首先获取最新结果。

【讨论】：