Mysql 从同一个表中选择总计作为三个字段作为一个查询？

我有一张桌子，里面有各种各样的订单：

ID | Date       | etc...
1  | 2013-01-01 | etc
2  | 2013-02-01 | etc
3  | 2013-03-01 | etc
4  | 2013-04-01 | etc
5  | 2013-05-01 | etc
6  | 2013-06-01 | etc
7  | 2013-06-01 | etc
8  | 2013-03-01 | etc
9  | 2013-04-01 | etc
10 | 2013-05-01 | etc

我想要一个以结果结尾的查询：

overallTotal | totalThisMonth | totalLastMonth
10           | 2              | 1

但我想在一个查询中实现这一点！我正试图找到一种使用子查询来实现这一点的方法。到目前为止，我已经：

SELECT * from (
  SELECT count(*) as overallTotal from ORDERS
)

如何将其与其他子查询相结合，以便在一个查询中获得总数

更新

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最初的问题是针对MySQL的，但我现在需要针对Firebird的。

使用IF语句结果的总和，IF检查它是否是相关日期

SELECT COUNT(*), 
    SUM(IF(YEAR(Date) = YEAR(CURDATE()) AND MONTH(Date) = MONTH(CURDATE()), 1, 0))
    SUM(IF((YEAR(Date) = YEAR(CURDATE()) AND MONTH(Date) = MONTH(CURDATE()) - 1) OR (YEAR(Date) = YEAR(CURDATE()) - 1 AND MONTH(Date) = 12 AND MONTH(CURDATE()) = 1), 1, 0))
from ORDERS

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复杂性是由于在查找前一个月时应对年份的变化而造成的。使用条件和，您可以这样做：

select 
    count(*) as overallTotal, 
    sum(if(Month(Date)+12*Year(Date)=Month(GetDate())+12*Year(GetDate()), 1, 0)) 
       as totalThisMonth 
    sum(if(Month(Date)+12*Year(Date)=Month(GetDate())+12*Year(GetDate())-1, 1, 0)) 
       as totalThisMonth;
from mytable

使用“月+12*年”公式可以避免年变化的问题

更新

对于Firebird，同样适用，您只需将Monthx替换为从x开始的提取月份，将Yearx替换为从x开始的提取年份，将Getdate替换为当前时间。这看起来很难看，所以我不会把它放在这里，但是你可以自己做

SELECT
  (SELECT COUNT(*) FROM C AS total) AS T,
  (SELECT COUNT(*) AS thisMonth FROM C WHERE MONTH(d) = 6) AS A,
  (SELECT COUNT(*) AS lastMonth FROM C WHERE MONTH(d) = 5) AS B;

请注意，月份是硬编码的，但在应用程序级别提取当前月份应该不会太困难。如果您真的需要，可以在SQL级别使用以下方法：

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我看到您已经在使用子查询了，那么为什么不执行以下操作呢

SELECT
(SELECT count(*) from ORDERS) as overallTotal,
(SELECT count(*) from ORDERS where Date between ... and ...) as totalThisMonth,
(SELECT count(*) from ORDERS where Date between ... and ...) as overallTotal

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选择*的必要性是什么？您真的需要以列而不是行的形式显示结果吗？我想行也可以？

SELECT
(SELECT count(*) from ORDERS) as overallTotal,
(SELECT count(*) from ORDERS where Date between ... and ...) as totalThisMonth,
(SELECT count(*) from ORDERS where Date between ... and ...) as overallTotal