获取最后一个可用条目

Question

我正在使用 10.1.39-MariaDB - mariadb.org binary 并且我有以下 table:

| id    | date                | ticker | close     | createdAt           | updatedAt           | CompanyId |
|-------|---------------------|--------|-----------|---------------------|---------------------|-----------|
| 39869 | 2019-09-18 00:00:00 | AAPL   | 221.96160 | 2019-09-18 19:25:30 | 2019-09-18 19:25:30 | 238       |
| 39870 | 2019-09-17 00:00:00 | AAPL   | 220.70000 | 2019-09-18 19:25:30 | 2019-09-18 19:25:30 | 238       |
| 39871 | 2019-09-16 00:00:00 | AAPL   | 219.90000 | 2019-09-18 19:25:30 | 2019-09-18 19:25:30 | 238       |
| 39872 | 2019-09-13 00:00:00 | AAPL   | 218.75000 | 2019-09-18 19:25:30 | 2019-09-18 19:25:30 | 238       |
| 39873 | 2019-09-12 00:00:00 | AAPL   | 223.09000 | 2019-09-18 19:25:30 | 2019-09-18 19:25:30 | 238       |

此外，我有以下查询以获得最后一天的前 5 daily_returns:

SELECT *
FROM (SELECT prices.*,
             CAST((`close` - @old_close) / @old_close AS DECIMAL(20, 10)) AS daily_return,
             @old_close := `close`
      FROM prices,
           (SELECT @old_close := 0 AS) AS t
      ORDER BY ticker,
              `date` ASC) AS tt
WHERE DATE >= DATE(NOW()) - INTERVAL 1 DAY
ORDER BY `date` DESC,
         daily_return DESC
LIMIT 5

但是，周一我只得到空结果，因为市场停滞不前，没有任何价格输入数据库。

如何克服这个周末的问题并采用数据库中发布的最后价格？

Answer 1

1 个方法使用 CASE 语句 -

SELECT *
FROM (SELECT prices.*,
             CAST((`close` - @old_close) / @old_close AS DECIMAL(20, 10)) AS daily_return,
             @old_close := `close`
      FROM prices,
           (SELECT @old_close := 0 AS) AS t
      ORDER BY ticker,
              `date` ASC) AS tt
WHERE DATE >= CASE WHEN DAYNAME(NOW()) = 'Monday' THEN DATE(NOW()) - INTERVAL 3 DAY  -- OR 2 DAY, DEPENDS UPON YOUR REQUIREMENT
                   ELSE DATE(NOW()) - INTERVAL 1 DAY END
ORDER BY `date` DESC,
         daily_return DESC
LIMIT 5

获取最后一个可用条目

Get last available entry

sql

mariadb

groupwise-maximum