本文介绍了从一个表中选择，从另一个 id 链接的表中计数的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着跟版网的小编来一起学习吧！
问题描述
这是我的代码:
$sql = mysql_query("select c.name, c.address, c.postcode, c.dob, c.mobile, c.email,
count(select * from bookings where b.id_customer = c.id) as purchased, count(select * from bookings where b.the_date > $now) as remaining,
from customers as c, bookings as b
where b.id_customer = c.id
order by c.name asc");
您可以看到我正在尝试做什么，但我不确定如何正确编写此查询.
you can see what i am trying to do, but im not sure how to write this query properly.
这是我得到的错误:
警告:mysql_fetch_assoc():已提供参数不是有效的 MySQL 结果资源
Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource
这是我的 mysql_fetch_assoc:
heres my mysql_fetch_assoc:
<?php
while ($row = mysql_fetch_assoc($sql))
{
?>
<tr>
<td><?php echo $row['name']; ?></td>
<td><?php echo $row['mobile']; ?></td>
<td><?php echo $row['email']; ?></td>
<td><?php echo $row['purchased']; ?></td>
<td><?php echo $row['remaining']; ?></td>
</tr>
<?php
}
?>
推荐答案
尝试改变...
count(select * from bookings where b.id_customer = c.id)
...到...
(select count(*) from bookings where b.id_customer = c.id)
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