I have a database that stores the temperature reading of each sensor in a along with the sensor ID, and date the reading was taken.
SELECT DISTINCT `date` FROM `temperatureData` ORDER BY `date` ASC LIMIT 10
this allows me to select the last 10 readings that are going to be plotted in a chart.
there are up to 40 sensor readings for each date.
I tried doing the following.
SELECT `date`, `sensor`, `temp`
FROM `temperatureData`
WHERE `date` = (
SELECT DISTINCT `date` FROM `temperatureData` ORDER BY `date` ASC LIMIT 10
)
Can anyone assist me as to how to select all the readings for the dates that are returned back from the last 10 dates?
Thanks in advance.
Boris
You just need in instead of =:
SELECT `date`, `sensor`, `temp`
FROM `temperatureData`
WHERE `date` IN (
SELECT DISTINCT `date` FROM `temperatureData` ORDER BY `date` DESC LIMIT 10
)
NB: if you want the readings of the last 10 dates, you probably want to ORDER BY date DESC instead of ASC. I changed that too.
In MySQL 8.0, this could also be rewritten with window function dense_rank():
SELECT `date`, `sensor`, `temp`
FROM (
SELECT
`date`,
`sensor`,
`temp`
DENSE_RANK() OVER(ORDER BY `date` DESC) rn
FROM `temperatureData`
) t
WHERE rn <= 10
Edit
To workaround the limitation of MySQL 5.7 not supporting LIMIT in subqueries with IN, you can use a join instead:
SELECT t.`date`, t.`sensor`, t.`temp`
FROM `temperatureData` t
INNER JOIN (
SELECT DISTINCT `date` FROM `temperatureData` ORDER BY `date` DESC LIMIT 10
) d ON d.`date` = t.`date`
Related
How can i find the employee id, date and minimum_time and maximum_time order by id,date,time from this table?
SELECT emp_id,
date,
MIN(time) AS in_time,
MAX(time) AS out_time
FROM yourTable
GROUP BY emp_id,
date
ORDER BY emp_id, date
The MIN() and MAX() functions will find the earliest and latest time of each employee, on each date.
It's simple your query syntax would be like
SELECT atr1, atr2,.. FROM table_name ORDER BY atr1, atr2,..
Try:
SELECT `emp_id` , `date` , MIN( `time` ) as in_time , MAX( `time` ) as out_time
FROM `Table`
WHERE 1
GROUP BY `date`
ORDER BY `time` DESC
LIMIT 0 , 30
As i can see there are same emp_id for many dates, so group by date will get all data for each date.
You can just use the SELECT Query
SELECT employee.id, employee.date, employee.minimum_time, employee.maximum_time
From table_name
ORDER BY employee.id
I need to select first value for every hour from my db. But I don't know how to reverse order on GROUP BY statement.
How can i rewrite my query (now it selects last value in hour)?
SELECT HOUR(`time`) as hour, mytable.*
FROM mytable
WHERE DATE(`time`) ="2015-09-12" GROUP BY HOUR(`time`) ORDER BY `time` ASC;
This query gave me expected result:
SELECT HOUR(`time`) as hour, sortedTable.* FROM
(SELECT electrolysis.* FROM electrolysis
WHERE DATE(`time`)='2015-09-12' ORDER BY `time`) as sortedTable
GROUP BY HOUR(`time`);
You can just select the MIN HOUR in sub query , try using the query:
SELECT * from mytable WHERE `time` IN (
SELECT MIN(HOUR(`time`)) as `hour`
FROM mytable
WHERE DATE(`time`) ="2015-09-12"
GROUP BY HOUR(`time`) ) ORDER BY `time` ASC;
You can do something like this:-
SELECT sub0.min_time,
mytable.*
FROM mytable
INNER JOIN
(
SELECT MIN(`time`) AS min_time
FROM mytable
GROUP BY HOUR(`time`)
) sub0
ON mytable.`time` = sub0.min_time
WHERE DATE(`time`) ="2015-09-12"
ORDER BY `time` ASC
This is using a sub query to get the smallest time in each hour. This is then joined back against your main table on this min time to get the record that has this time.
Note that there is a potential problem here if there are multiple records that share the same time as the smallest one for an hour. There are ways around this, but that will depend on your data (eg, if you have a unique id field which is always ascending with time then you could select the min id for each hour and join based on that)
You can use below query, which is more optimized just make sure that time field should be indexed.
SELECT HOUR(m.time), m.*
FROM mytable AS m
JOIN
(
SELECT MIN(`time`) AS tm
FROM mytable
WHERE `time` >= '2015-09-12 00:00:00' AND `time` <= '2015-09-12 23:59:59'
GROUP BY HOUR(`time`)
) AS a ON m.time=a.tm
GROUP BY HOUR(m.time)
ORDER BY m.time;
I need to group some data by date, but I have a very special case where the name of the final field should be the same of the original field and I can't use an expression in the GROUP BY
I have created this sqlfiddle with some example data:
http://sqlfiddle.com/#!2/8771a/1
I need this result:
DATE PAGEVIEWS
2013-12 69
2013-11 70
Note 1: I can't change the group by, if I do this I get the result, but I need to group by date and date should be the formatted date, and not the real date in the table:
SELECT DATE_FORMAT(`date`, "%Y-%m") AS `date`, SUM(pageviews) AS pageviews
FROM `domains_data`
GROUP BY DATE_FORMAT(`date`, "%Y-%m")
ORDER BY `date` DESC
Note 2: I can't rename the field, it should have the name "date" at the end, this isn't possible for me:
SELECT DATE_FORMAT(`date`, "%Y-%m") AS `date2`, SUM(pageviews) AS pageviews
FROM `domains_data`
GROUP BY `date2`
ORDER BY `date` DESC
There is some way to do it with MySQL?
Is the use of nested query allowed?
SELECT `date`, SUM(pageviews) AS pageviews FROM
(SELECT DATE_FORMAT(`date`, "%Y-%m") AS `date`, SUM(pageviews) AS pageviews
FROM `domains_data`
GROUP BY `date`) AS Ref
GROUP BY `date`
ORDER BY `date` DESC
If you can change the from clause:
SELECT `date`, SUM(pageviews) AS pageviews
FROM (select DATE_FORMAT(`date`, "%Y-%m") AS `date`, pageviews
from `domains_data` dd
) dd
GROUP BY `date`
ORDER BY `date` DESC;
However, given the constraint that you cannot change the group by, you probably cannot change the from either. Can you explain why your query has these limitations?
I need to select 40 rows with date from today and 10 records with older date, ordered by date.
If MySQL supported negative offset, it would look like this:
SELECT * FROM `mytable` WHERE `date` >= '2013-10-29' ORDER BY date LIMIT -10, 40;
Negative offset is not supported. How can I solve the problem? Thanks!!!
Use UNION to combine two queries:
(
SELECT *
FROM mytable
WHERE date < '2013-10-29'
ORDER BY date DESC
LIMIT 10
) UNION ALL (
SELECT *
FROM mytable
WHERE date >= '2013-10-29'
ORDER BY date
LIMIT 40
)
ORDER BY date -- if results need to be sorted
I have this table:
ID(INT) DATE(DATETIME)
Under the DATE column there are a lot of different dates, and I want to figure out the most common hour between all the rows of the table, regardless of the day.
How can I do that with a MySQL query?
SELECT HOUR(date) AS hr, COUNT(*) AS cnt
FROM yourtable
GROUP BY hr
ORDER BY cnt DESC
LIMIT 1
relevant docs: http://dev.mysql.com/doc/refman/5.5/en/date-and-time-functions.html#function_hour
Try this -
SELECT HOUR(`DATE`) AS `hour`, COUNT(*)
FROM `table`
GROUP BY `hour`
You could do a query like:
SELECT COUNT(daterow) AS occurrences FROM table GROUP BY daterow ORDER BY occurrences DESC LIMIT 1;
SELECT COUNT( id ) , HOUR( date )
FROM test
GROUP BY HOUR( date )
ORDER BY COUNT( id ) DESC
LIMIT 1