从一个表中选择，从另一个 id 链接的表中计数

Question

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。

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

继承我的 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   
}

?>

Answer 1

尝试改变喜欢...

count(select * from bookings where b.id_customer = c.id)

...到...

(select count(*) from bookings where b.id_customer = c.id)

Answer 2

您的查询错误地使用了 COUNT，这已被 @Will A 的回答。

我还想建议一个可能更好的构造替代方案，我认为它反映了相同的逻辑：

SELECT
  c.name,
  c.address,
  c.postcode,
  c.dob,
  c.mobile,
  c.email,
  COUNT(*) AS purchased,
  COUNT(b.the_date > $now OR NULL) AS remaining
FROM customers AS c
  INNER JOIN bookings AS b ON b.id_customer = c.id
GROUP BY c.id
ORDER BY c.name ASC

注意：通常您应该将所有非聚合 SELECT 表达式包含到 GROUP BY 中。然而MySQL支持缩短的GROUP BY列表，所以指定唯一标识您正在提取的所有非聚合数据的键表达式就足够了。 请避免随意使用该功能。如果未包含在 GROUP BY 中的列每组有多个值，则您无法控制在不进行聚合拉出该列时实际返回哪个值。