Question

我有一张这样的表：

user_name   sent_at     data
---------------------------------------------
alpha       2015-11-01  { "status" : "pending", … }
beta        2015-11-01  { "status" : "pending", …}
gamma       2015-11-01  { "status" : "pending", … }
alpha       2015-11-02  { "status" : "complete", … }
alpha       2015-11-03  { "status" : "complete", … }

对于每个用户，我想找到他达到当前状态的行。所以对于2015-11-02的用户alpha。请注意，对于同一用户，可以多次重复相同的状态。

我最终得到了：

SELECT *
FROM
(
    SELECT 
        user_name,
        sent_at,
        status,
        row_number() OVER (PARTITION BY user_name ORDER BY sent_at DESC) as row
    FROM
    (
        SELECT
            user_name,
            sent_at,
            data->>'status' AS status,
            lag(data->>'status') OVER (PARTITION BY user_name ORDER BY sent_at) AS prev_status
        FROM log
    ) AS t
    WHERE status IS DISTINCT FROM prev_status
) AS u 
WHERE row = 1

有更简单的方法吗？

Answer 1

您的查询看起来不错imho。如果您想避免使用窗口函数，可以使用$ mvn -Dexec.mainClass=... -Dexec.classpathScope=test clean exec:java两次distinct on向相反方向排序：

sent_at

按列分组，然后在另一列中查找更改，然后获取最后一行

1 个答案: