Question

我有以下查询：

SELECT floor(datediff(u.created_at, curdate()) / 1) AS days_ago,
       count(DISTINCT u.id) AS "New Users in Cohort", 
       count(DISTINCT i.user_id) AS "Uniq Users who invited Cohort",
count(DISTINCT u.id) / count(DISTINCT i.user_id) AS "% who invite"
FROM users u 
LEFT JOIN invitations i
     ON u.id = i.user_id
WHERE u.onboarding_started_at IS NOT NULL
GROUP BY days_ago;

我正在尝试获取查询以返回以下内容：

+----------+-----------+--------------------------+--------------+
| days_ago | New Users | Unique Users who Invited | % who invite |
+----------+-----------+--------------------------+--------------+
|        1 |        20 |                       10 | 50%          |
+----------+-----------+--------------------------+--------------+

新用户将位于u.created_at是查询使用的日期

我在做什么错？谢谢！

Answer 1

您还没说够我判断是否需要DISTINCT。

LEFT似乎不必要。

“昨天”是`...> = CURDATE（）-间隔1天和...

SELECT count(DISTINCT u.id) AS "New Users in Cohort", 
       count(DISTINCT i.user_id) AS "Uniq Users who invited Cohort",
       count(DISTINCT u.id) / count(DISTINCT i.user_id) AS "% who invite"
FROM users u 
JOIN invitations i
     ON u.id = i.user_id
WHERE u.onboarding_started_at IS NOT NULL
  AND u.created_at >= CURDATE() - INTERVAL 1 DAY
  AND u.created_at  < CURDATE()

如果您希望连续数天计数，那么我将使用DATE(u.created_at)作为DATE的显示值和GROUPing BY。（除非您真的想要“ days_ago”。）

SQL查询以显示为YESTERDAY创建的所有用户，已发送多少邀请？

1 个答案: