Question

我有一张这样的表：

// mytable
+----+-----------------+
| id |    timestamp    |
+----+-----------------+
| 1  | 1452001861      |    -- yesterday
| 2  | 1452088272      |    -- today
| 3  | 1452088283      |    -- today
| 4  | 1451915461      |    -- last week
| 5  | 1452001861      |    -- yesterday
| 6  | 1452088263      |    -- today
| 7  | 1252388263      |    -- out of {today, yesterday, last week}
| 8  | 1452088312      |    -- today
| 9  | 1452001762      |    -- yesterday
| 10 | 1222388263      |    -- out of {today, yesterday, last week}
| 12 | 1451915459      |    -- last week
+----+-----------------+

这是我的疑问：

SELECT * FROM mytable 
   WHERE DATE(FROM_UNIXTIME(timestamp)) >= CURDATE() - INTERVAL 7 DAY
   ORDER BY timestamp DESC

上面的查询选择所有行，从现在到最后7天。

现在有时所有行都在前7天之前（我的意思是最新的行例如在最后8天添加）然后该查询的输出将是空的（选择0行）。

我想要的是，如果没有选择行（0行），则选择5个最后一行（之前已添加）。我怎么能这样做？

Answer 1

您可以尝试这种方法：

SET @row_count := (SELECT COUNT(*) 
                   FROM mytable 
                   WHERE DATE(FROM_UNIXTIME(timestamp)) >= CURDATE() - INTERVAL 7 DAY);

IF (@row_count > 0) THEN
    SELECT * 
    FROM mytable 
    WHERE DATE(FROM_UNIXTIME(timestamp)) >= CURDATE() - INTERVAL 7 DAY
    ORDER BY timestamp DESC;
ELSE
    SELECT * 
    FROM mytable 
    ORDER BY timestamp DESC
    LIMIT 5;
END IF;

选择了多少行？

1 个答案: