MySQL查询，多个计数和总和

Question

我有一个输出到php表的MySQL查询，但是在连接两个都使用COUNT的表时遇到问题：

$query = "SELECT mqe.registration, 
        COUNT(*) AS numberofenqs,
        COUNT(DISTINCT ucv.ip) AS unique_views,
        SUM(ucv.views) AS total_views
        FROM main_quick_enquiries AS mqe
        LEFT OUTER JOIN used_car_views AS ucv
        ON ucv.numberplate = mqe.registration
        WHERE mqe.registration IS NOT NULL
        GROUP BY mqe.registration ORDER BY numberofenqs DESC";

查询运行，但是numberofenqs列中的数字始终是错误的，正如我从单独执行该查询所知道的那样，它具有正确的结果：

SELECT registration, COUNT(*) AS numberofenqs FROM main_quick_enquiries GROUP BY registration ORDER BY numberofenqs DESC

为什么COUNT（*）在顶部查询代码中无法正常工作，以及从何处获得数据？

Answer 1

可能是因为左外连接...

尝试运行此命令：

SELECT registration
, count(*)
FROM main_quick_enquiries
GROUP BY registration

并将其与此结果进行比较

SELECT mqe.registration
, count(*)
FROM main_quick_enquiries mqe
LEFT OUTER JOIN used_car_views ucv
ON ucv.numberplate = mqe.registration
GROUP BY mqe.registration

可能有问题:)在重复行中...尝试找到一个特定的注册号，并比较两个查询的详细信息

SELECT *
FROM main_quick_enquiries
WHERE registration = XXXX

+

SELECT *
FROM main_quick_enquiries mqe
LEFT OUTER JOIN used_car_views ucv
ON ucv.numberplate = mqe.registration
WHERE registration = XXXX

你应该看到差异

Answer 2

谢谢大家，但我想我用COUNT（DISTINCT mqe.id）代替了COUNT（*）。