Question

我有一张名为＆＃39; users＆＃39;我的用户拥有4个不同的登录选项：passwd，facebook_login，google_login和aprofiel_login。我想创建一个获得结果的查询：

用户总数
仅使用passwd设置的用户
仅设置了facebook_login的用户
仅设置了google_login的用户
仅设置了aprofiel_login的用户
具有上述4个选项集中的1个以上的用户

我提出了以下查询，它给出了前5个，但我无法得到计算倍数的那个。

我的查询是这样的：

select count(*) total,
    sum(case when passwd is not null and (facebook_login is null and google_login is null and aprofiel_login is null) then 1 else 0 end) gad,
    sum(case when facebook_login is not null and (passwd is null and google_login is null and aprofiel_login is null) then 1 else 0 end) facebook,
    sum(case when google_login is not null and (passwd is null and facebook_login is null and aprofiel_login is null) then 1 else 0 end) google,
    sum(case when aprofiel_login is not null and (passwd is null and facebook_login is null and google_login is null) then 1 else 0 end) aprofiel
from `users`
where auth_level = 1
and is_account_active = 1

然后我考虑过，我有我的总金额，我有所有其他单一的金额，如果我用所有其他的减去总数，我有多个登录用户的结果。但是我如何在MySQL中做到这一点？

Answer 1

将此查询粘贴到子查询中，然后对派生列进行数学运算：

SELECT
   t1.*,
   total - (gad + facebook + google + aprofiel) as otherfield
FROM
   (
      select count(*) total,
        sum(case when passwd is not null and (facebook_login is null and google_login is null and aprofiel_login is null) then 1 else 0 end) gad,
        sum(case when facebook_login is not null and (passwd is null and google_login is null and aprofiel_login is null) then 1 else 0 end) facebook,
        sum(case when google_login is not null and (passwd is null and facebook_login is null and aprofiel_login is null) then 1 else 0 end) google,
        sum(case when aprofiel_login is not null and (passwd is null and facebook_login is null and google_login is null) then 1 else 0 end) aprofiel
      from `users`
      where auth_level = 1
        and is_account_active = 1
    ) t1

我如何使用case中的值来在mysql中进行计算？

1 个答案: