Question

如何将这些从我的数据库中提取统计信息的SQL查询结合起来以获得更好的性能？

$total= mysql_query("SELECT COUNT(*) as number, SUM(order_total) as sum FROM history");
$month = mysql_query("SELECT COUNT(*) as number, SUM(order_total) as sum FROM history WHERE date >= UNIX_TIMESTAMP(DATE_ADD(CURDATE(),INTERVAL -30 DAY))");
$day = mysql_query("SELECT COUNT(*) as number, SUM(order_total) as sum FROM history WHERE date >= UNIX_TIMESTAMP(CURDATE())");

Answer 1

如果您想要查询单个查询中的所有数据，您有两种选择：

使用UNION查询（as sugested by bishop in his answer）
调整查询以获得单行所需内容

我会显示选项2（已经涵盖了选项1）。

注意：我使用用户变量（init子查询中的内容）来避免一次又一次地写表达式。另外，要过滤汇总数据，请使用case ... end表达式。

select 
    -- Your first query:
    count(*) as number, sum(order_total) as `sum`
    -- Your second query:
  , sum(case when `date` <= @prev_date then 1 else 0 end) as number_prev
  , sum(case when `date` <= @prev_date then order_total else 0 end) as sum_prev
    -- Your third query:
  , sum(case when `date` <= @cur_date then 1 else 0 end) as number_cur
  , sum(case when `date` <= @cur_date then order_total else 0 end) as sum_cur
from (
         select @cur_date := unix_timestamp(curdate())
              , @prev_date := unix_timestamp(date_add(curdate(), interval -30 day))
     ) as init
   , history;

希望这有帮助

Answer 2

由于查询具有相同的列结构，您可以要求MySQL将它们与UNION操作结合起来：

(SELECT 'total' AS kind, COUNT(*) as number, SUM(order_total) as sum FROM history~
UNION
(SELECT 'by-month' AS kind, COUNT(*) as number, SUM(order_total) as sum FROM history WHERE date <= UNIX_TIMESTAMP(DATE_ADD(CURDATE(),INTERVAL -30 DAY)))
UNION
(SELECT 'by-day' AS kind, COUNT(*) as number, SUM(order_total) as sum FROM history WHERE date <= UNIX_TIMESTAMP(CURDATE()))

结合数据库查询

2 个答案: