mysql查询帮助

我有一个数据库，里面有我在线网店的所有交易记录，我正试图通过查询打印出一份简单的财务报表

它将打印在如下表格中：

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

有人能给我指出正确的方向吗？

选择
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;

月份（交易日）作为月份，
金额（如果（收入金额>0,1,0））作为销售数量，
金额（传入金额）/1.25作为货币输入，
金额（流出金额）为流出金额，
作为结果的金额（（传入金额/1.25）-传出金额）
来自myDB
其中时间戳>='2011-01-01 00:00:00'和时间戳'0'）

不正确

sum

看起来也不正确

选择
月份（交易日）作为月份，
金额（如果（收入金额>0,1,0））作为销售数量，
金额（传入金额）/1.25作为货币输入，
金额（流出金额）为流出金额，
作为结果的金额（（传入金额/1.25）-传出金额）
来自myDB
其中时间戳>='2011-01-01 00:00:00'和时间戳'0'）

不正确

sum

看起来也不正确

---

添加group by语句：

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");

---

添加group by语句：

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");

---

基于@ajreal的答案，您可以通过重用以前计算的值来加速此查询，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

如果选择“年度以外”，请取消对相关章节的注释。

---

注意：您可以将

DESC

修饰符添加到

groupby

以首先获得最新的结果。

基于@ajreal的答案，您可以通过重用以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

如果选择“年度以外”，请取消对相关章节的注释。

---

注：您可以将

DESC

修饰符添加到

groupby

以首先获得最新的结果。

您真的有一个名为

myDB

的表吗？您真的有一个名为

myDB

的表吗？