MySQL的计数行加入

Question

我有一个查询，看起来像这样：

Select x.date, x.id, x.phone, 
    x.product, xy.policy,xy.date 
from (y left join z 
      on y.customer_id=z.customer_id) 
    left join x 
     on x.id=z.id 
    left join xy 
     on xy.id=x.id 
where x.date > '2000-01-01' 
    and y.detail =foo 
    and xy.policy like 'foo_____' 
    and xy.policy_type = foo; 

我怎么能算的行这个回报率是多少？

我试过使用SQL_CALC_FOUND_ROWS，但我不能完全适合这个查询。

任何帮助将不胜感激。

谢谢， Stefan。

Answer 1

你可以写：

SELECT COUNT(1) 
    FROM y 
    JOIN z 
    ON y.customer_id = z.customer_id 
    JOIN x 
    ON x.id = z.id 
    JOIN xy 
    ON xy.id = x.id 
WHERE x.date > '2000-01-01' 
    AND y.detail = foo 
    AND xy.policy LIKE 'foo_____' 
    AND xy.policy_type = foo 
; 

（注我冒昧地将LEFT JOIN改为JOIN，因为WHERE条款无论如何阻止它们实际上起到LEFT JOIN的作用。如果你想真正LEFT JOIN S，你可以移动从WHERE子句条件为ON条款：

SELECT COUNT(1) 
    FROM y 
    LEFT 
    JOIN z 
    ON z.customer_id = y.customer_id 
    LEFT 
    JOIN x 
    ON x.id = z.id 
    AND x.date > '2000-01-01' 
    LEFT 
    JOIN xy 
    ON xy.id = x.id 
    AND xy.policy LIKE 'foo_____' 
    AND xy.policy_type = foo 
WHERE y.detail = foo 
; 

）

Answer 2

最简单的就是只需添加一个子查询...

Select x.date, x.id, x.phone, 
    x.product, xy.policy,xy.date, 
    (Select Count(*) 
    From (y left join z on y.customer_id=z.customer_id) 
     left join x on x.id=z.id 
     left join xy on xy.id=x.id 
    where x.date > '2000-01-01' 
     and y.detail =foo 
     and xy.policy like 'foo_____' 
     and xy.policy_type = foo) RecordCount 
from (y left join z 
      on y.customer_id=z.customer_id) 
    left join x 
     on x.id=z.id 
    left join xy 
     on xy.id=x.id 
where x.date > '2000-01-01' 
    and y.detail =foo 
    and xy.policy like 'foo_____' 
    and xy.policy_type = foo; 

如果你想要的是计数，那么：

Select Count(*) 
From (y left join z on y.customer_id=z.customer_id) 
    left join x on x.id=z.id 
    left join xy on xy.id=x.id 
where x.date > '2000-01-01' 
    and y.detail =foo 
    and xy.policy like 'foo_____' 
    and xy.policy_type = foo