Question

我有以下SQL语句：

 SELECT DISTINCT ON (watch.url)
        watch.url, watch.name, user_url.frequency, watch.selector, price.date 
   FROM watch 
        JOIN user_url on user_url.url = watch.url 
        LEFT JOIN price on price.url = watch.url 
  WHERE (1400575234 > price.date + user_url.frequency)
     OR price.date is null
  ORDER BY watch.url, price.date DESC

其中1400575234是当前的unix时间戳。表格价格以不同的时间间隔保存商品的价格，每行一个。

目前此查询将返回所有项目，因为在某些时候他们有一个小于当前日期的条目。虽然我只是想抓住其price.date小于当前时间的URL，但我只想匹配价格表中url的最高日期。

我想我需要在order by工作之前运行select distinct。有没有人有任何想法？

（希望这是有道理的）

Answer 1

您可以尝试使用窗口功能：

SELECT url, name, frequency, selector, date
FROM (
 SELECT watch.url, watch.name, user_url.frequency, watch.selector, price.date,
        row_number() over (partition by watch.url order by price.date desc) as rn
   FROM watch 
        JOIN user_url on user_url.url = watch.url 
        LEFT JOIN price on price.url = watch.url 
  WHERE (1400575234 > price.date + user_url.frequency)
     OR price.date is null
) t
WHERE rn = 1
ORDER BY url, date;

^{顺便说一下：date是列的可怕名称}

Answer 2

我似乎使用以下

来解决它

SELECT DISTINCT ON (watch.url)
watch.url, watch.name, user_url.frequency, watch.selector, p.date 
FROM watch 
JOIN user_url on user_url.url = watch.url
LEFT JOIN (
    SELECT max(price.date) as date, url FROM price GROUP BY url
) as p ON p.url = watch.url
WHERE 1400575234 > p.date + user_url.frequency
ORDER BY watch.url

感谢大家的帮助。如果你能看到更有效的查询请告诉我。

SELECT DISTINCT没有正确排序

2 个答案: