我有这个查询,在检查了各种教程后应该可以工作 - 但事实并非如此。
$query="SELECT week, year, COUNT(week) AS week_no
FROM archive_agent_booking
LEFT JOIN invoice_additions ON invoice_additions.week = archive_agent_booking.week
WHERE client_id='$account_no' GROUP BY week, year ORDER BY week DESC";
表格如下:
archive_agent_booking
+---------+----------+----------+----------+----------+---------+---------+
| job_id | week | year | desc | price | date | acc_no |
+---------+----------+----------+----------+----------+---------+---------+
invoice_additions
+---------+----------+----------+----------+----------+---------+
| acc_no | week | year | desc | am_price | am_date |
+---------+----------+----------+----------+----------+---------+
我基本上想计算两个表中的每个周元素并将它们显示为一个总数,即使其中一个表中没有显示周值之一。不知道这是否是最好的解决方案,因此我愿意接受替代方案。
最佳答案
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:我对你想看到的内容做了一些巨大的假设
关于php - mysql PHP查询包含连接计数,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/13399626/