Question

我无法理解这一个！所以我正在寻求帮助......

对于每个用户，我都试图对分配给他们的帖子类型求和。

用户表：crm_users 列：users_id，users_first，users_last

帖子表：crm_entities 列：crm_id，users_id，settype，post

我想COUNT（）用户的总帖子，其中settype = draft以及settype = published，例如：

Name------------Published---------Drafts
John Smith---------15---------------3
Nancy Grace--------11-------------  5
Jay Martin----------7--------------14

我相信我这让它变得比它可能更难......或者至少我认为！

感谢您的任何建议！

Answer 1

因为MySQL使用1代表true而0代表false，所以你可以使用：

SELECT
    u.users_first,
    u.users_last,
    SUM(p.settype='published') as Published,
    SUM(p.settype='draft') as Drafts
FROM
    crem_users u
LEFT JOIN
    crm_entities p
ON
    u.users_id = p.users_id
GROUP BY
    u.users_id
ORDER BY Published DESC;

Answer 2

尝试：

select u.users_first, (select count(*) as qty from crm_entities e where e.settype='draft' and e.users_id = u.users_id) drafts, (select count(*) as qty from crm_entities e where e.settype='published' and e.users_id = u.users_id) published from crm_users u

PHP / MySQL - 每个用户在1个表中共计2个不同的总计

2 个答案: