我只想选择当天的记录。这样我每天就可以看到有多少访客(“bezoekers”)。
这是我使用的代码:
Select FROM_UnixTime((sensordata1.time), '%Y-%c-%d-%H') "Datum (Descending)",
COUNT(DISTINCT address) "Bezoekers"
FROM sensordata1
ORDER BY 1
FROM_UnixTime((sensordata1.time), '%Y-%c-%d-%H') DESC
我尝试过:
Select FROM_UnixTime((sensordata1.time), '%Y-%c-%d-%H') "Datum (Descending)",
COUNT(DISTINCT address) "Bezoekers"
FROM sensordata1
WHERE 1, DATE('Datum') = DATE(CURDATE()
ORDER BY 1
FROM_UnixTime((sensordata1.time), '%Y-%c-%d-%H') DESC
我想要什么:
Datum | Bezoekers
2017-4-20-15 | 31
2017-4-20-14 | 34
2017-4-20-13 | 20
2017-4-20-12 | 26
我不想要其中包含例如 2017-04-21 的内容。 如果我第二天回来,它必须是从那天开始,所以你不能设定日期。
最佳答案
您缺少GROUP BY
。如果您想要每个 FROM_UnixTime(sensordata1.time, '%Y-%c-%d-%H')
一个结果行,则按其分组。
Select
FROM_UnixTime(sensordata1.time, '%Y-%c-%d-%H') AS "Datum (Descending)",
COUNT(DISTINCT address) AS "Bezoekers"
FROM sensordata1
GROUP BY FROM_UnixTime(sensordata1.time, '%Y-%c-%d-%H')
ORDER BY FROM_UnixTime(sensordata1.time, '%Y-%c-%d-%H') DESC;
仅当天相同(添加了相应的 WHERE
子句):
Select
FROM_UnixTime(sensordata1.time, '%Y-%c-%d-%H') AS "Datum (Descending)",
COUNT(DISTINCT address) AS "Bezoekers"
FROM sensordata1
WHERE FROM_UnixTime(sensordata1.time, '%Y-%m-%d') = curdate()
GROUP BY FROM_UnixTime(sensordata1.time, '%Y-%c-%d-%H')
ORDER BY FROM_UnixTime(sensordata1.time, '%Y-%c-%d-%H') DESC;
关于mysql - 如何只显示当前的记录,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/43539533/