加入statiment中的别名问题

Question

我有以下查询的问题:

SELECT 
    g_contac.contid, g_contac.name, g_contac.email, f_sync.foreign_key,
    (
        SELECT COUNT(g_cpers.cpersid) 
        FROM g_cpers 
        WHERE g_cpers.contid = g_contac.contid
    ) AS employee_count
FROM f_sync 
    FULL OUTER JOIN g_contac ON 
    (
        g_contac.contid = f_sync.external_id AND 
        model = case when f_sync.employee_count = 0 then 'PRIVATE' else 'COMPANY' end
    )
WHERE model = 'COMPANY' or model = 'PRIVATE' OR model IS null

当我执行它时,我收到错误:

列名称"employee_count"无效.

如何解决这个问题呢？

Answer 1

这是因为您f_sync.employee_count在查询中提到,但f_sync没有一个名为的列employee_count:您刚刚在查询中使用别名 创建了一个动态列employee_count.

简单的解决方法是重复计算:

SELECT 
g_contac.contid, g_contac.name, g_contac.email, f_sync.foreign_key,
(
    SELECT COUNT(g_cpers.cpersid) 
    FROM g_cpers 
    WHERE g_cpers.contid = g_contac.contid
) AS employee_count
FROM f_sync 
FULL OUTER JOIN g_contac ON 
(
    g_contac.contid = f_sync.external_id AND 
    model = case when (SELECT COUNT(g_cpers.cpersid) 
    FROM g_cpers 
    WHERE g_cpers.contid = g_contac.contid) = 0 then 'PRIVATE' else 'COMPANY' end
)
WHERE model = 'COMPANY' or model = 'PRIVATE' OR model IS null;

更好的解决方法是创建一个包含此列的视图,这意味着它只能计算一次;

编辑:改进的查询和合并的评论

您可以通过使用SQL not exists而不是count(*) = 0:来提高清晰度:

SELECT 
g_contac.contid, g_contac.name, g_contac.email, f_sync.foreign_key
FROM f_sync 
FULL OUTER JOIN g_contac ON 
(
    g_contac.contid = f_sync.external_id AND 
    model = case when not exists (SELECT * FROM g_cpers 
      WHERE g_cpers.contid = g_contac.contid) then 'PRIVATE' else 'COMPANY' end
)
WHERE model = 'COMPANY' or model = 'PRIVATE' OR model IS null;