MySQL 查询、多次计数和求和

Question

我有一个输出到 php 表的 MySQL 查询，但在连接两个都使用 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 numberofenqs 列中的数字总是错误的，因为我通过单独执行该查询知道它会带来正确的结果:

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

为什么 COUNT(*) 在顶级查询代码中无法正常工作？它从哪里获取数字？

Answer 1

这可能是因为 LEFT OUTER JOIN ...

尝试运行这个:

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

口是心非的行中可能存在问题...尝试查找一个特定的注册号，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