我正在使用 MySQL 和 PHP。这是我收到错误的查询。
$query =
"SELECT days.day, count(myDataTable.appId) as countf, count(myDataTable.appId) as counts
FROM
(
select curdate() as day
union select curdate() - interval 1 day
union select curdate() - interval 2 day
union select curdate() - interval 3 day
union select curdate() - interval 4 day
union select curdate() - interval 5 day
union select curdate() - interval 6 day
union select curdate() - interval 7 day
union select curdate() - interval 8 day
union select curdate() - interval 9 day
) days
left join myDataTable as n1
on days.day = n1.date AND n1.appId = '$id' AND n1.status = 'ERROR'
group by days.day
left join myDataTable as n2
on days.day = n2.date AND n2.appId = '$id' AND n2.status = 'SUCCESS'
group by days.day";
错误日志是:
You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'left join myDataTable on days.day = myDataTable.date AND myDataTable.appId ' at line 17
最佳答案
正确缩进 SQL 有助于您发现错误。通过主要关键字(SELECT、FROM、WHERE、HAVING、GROUP BY 和 ORDER BY)缩进将帮助您快速发现它们:
SELECT
days.day,
count (myDataTable.appId) as countf,
count(myDataTable.appId) as counts
FROM
(
select curdate() as day
union select curdate() - interval 1 day
union select curdate() - interval 2 day
union select curdate() - interval 3 day
union select curdate() - interval 4 day
union select curdate() - interval 5 day
union select curdate() - interval 6 day
union select curdate() - interval 7 day
union select curdate() - interval 8 day
union select curdate() - interval 9 day
) days
left join myDataTable as n1
on days.day = n1.date AND n1.appId = '$id' AND n1.status = 'ERROR'
group by
days.day
left join myDataTable as n2
on days.day = n2.date AND n2.appId = '$id' AND n2.status = 'SUCCESS'
group by
days.day
您可以看到有两个 GROUP BY 不起作用。此外,您在第一个 GROUP BY
子句中有一个 LEFT JOIN
,它也不起作用。删除第一个 GROUP BY
将使您更接近:
SELECT
days.day,
count (myDataTable.appId) as countf,
count(myDataTable.appId) as counts
FROM
(
select curdate() as day
union select curdate() - interval 1 day
union select curdate() - interval 2 day
union select curdate() - interval 3 day
union select curdate() - interval 4 day
union select curdate() - interval 5 day
union select curdate() - interval 6 day
union select curdate() - interval 7 day
union select curdate() - interval 8 day
union select curdate() - interval 9 day
) days
left join myDataTable as n1
on days.day = n1.date AND n1.appId = '$id' AND n1.status = 'ERROR'
left join myDataTable as n2
on days.day = n2.date AND n2.appId = '$id' AND n2.status = 'SUCCESS'
group by
days.day
不,你有一个正确的 FROM 子句。这是数据库查看的第一部分,因此它知道从哪里获取数据以及如何将数据连接在一起。您的表别名在此处设置,然后在查询中的其他任何地方使用。这导致了第二个问题。
您在 SELECT 子句中引用了 myDataTable
,但当数据库查看您的 SELECT
时,myDataTable
已不在上下文中。不过,别名 n1
和 n2
是,因此请更改这些别名以引用您的表别名:
SELECT
days.day,
count (n1.appId) as countf,
count(n2.appId) as counts
FROM
(
select curdate() as day
union select curdate() - interval 1 day
union select curdate() - interval 2 day
union select curdate() - interval 3 day
union select curdate() - interval 4 day
union select curdate() - interval 5 day
union select curdate() - interval 6 day
union select curdate() - interval 7 day
union select curdate() - interval 8 day
union select curdate() - interval 9 day
) days
left join myDataTable as n1
on days.day = n1.date AND n1.appId = '$id' AND n1.status = 'ERROR'
left join myDataTable as n2
on days.day = n2.date AND n2.appId = '$id' AND n2.status = 'SUCCESS'
group by
days.day
最后,您可以在 SELECT
中使用 CASE
语句,而不是为每种状态加入两次 myDataTable
:
SELECT
days.day,
SUM(CASE WHEN n1.status = 'ERROR' THEN 1 ELSE 0 END) as countf,
SUM(CASE WHEN n1.status = 'SUCCESS' THEN 1 ELSE 0 END) as counts
FROM
(
select curdate() as day
union select curdate() - interval 1 day
union select curdate() - interval 2 day
union select curdate() - interval 3 day
union select curdate() - interval 4 day
union select curdate() - interval 5 day
union select curdate() - interval 6 day
union select curdate() - interval 7 day
union select curdate() - interval 8 day
union select curdate() - interval 9 day
) days
left join myDataTable as n1
on days.day = n1.date AND n1.appId = '$id'
group by
days.day
关于mysql - 无法根据两个条件两次 JOIN 同一张表,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/44078578/