0

我有一个看起来像这样的查询：

 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;

如何计算返回的行数？

我尝试使用 SQL_CALC_FOUND_ROWS 但我不能完全适应这个查询。

任何帮助将不胜感激。

谢谢，斯特凡。

4

2 回答 2

1

Easiest is just add a subquery... 

 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;

If all you want is the count, then: 

 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
于 2012-11-20T15:56:58.983 回答
0

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
;

（请注意，我冒昧地将您LEFT JOIN的 s 更改为常规JOINs，因为该WHERE子句阻止它们实际上作为LEFT JOINs 运行。如果您想要真正LEFT JOIN的 s，您可以将条件从WHERE子句移动到ON子句中：

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
;

)

于 2012-11-20T15:58:42.163 回答