2012-11-15 68 views
0

我有这个查询后检查各种教程应该工作 - 但它没有。mysql php查询包含加入计数

$query="SELECT week, year, COUNT(week) AS week_no 
FROM archive_agent_booking 
LEFT JOIN invoice_additions ON invoice_additions.week = archive_agent_booking.week 
WHERE client_id='$account_no' GROUP BY week, year ORDER BY week DESC"; 

的表如下所示:

archive_agent_booking 

+---------+----------+----------+----------+----------+---------+---------+ 
| job_id | week | year | desc | price | date | acc_no | 
+---------+----------+----------+----------+----------+---------+---------+ 


invoice_additions 

+---------+----------+----------+----------+----------+---------+ 
| acc_no | week | year | desc | am_price | am_date | 
+---------+----------+----------+----------+----------+---------+ 

我基本上是想从两个表中统计每个星期元素并将其显示为一个总即便一周值中的一个并不在一个显示表格。不知道这是否是最好的解决方案,所以我愿意接受替代品。

+0

什么是你查询的输出? – ethrbunny

+0

只是想出了set die()语句。 – Sideshow

回答

1
select 
    week, 
    sum(items) 
from 
    (
     (select week, count(*) as items from archive_agent_booking group by week) 
    union 
     (select week, count(*) from invoice_additions group by week) 
    ) 
group by 
    week 

编辑:我做了你希望看到什么一些巨大的假设

+0

谢谢:)虽然我没有时间去测试它,但似乎在表面上工作 - 接下来的半小时...... – Sideshow