How can I get the latest data of a specific day in MySQL?
Let's assume that I have a column of dates recorded on my database
dates | time | value
---------------------------------------------
2015-08-05 | 11:03:02 | 200
2015-08-05 | 23:04:22 | 2400
2015-08-07 | 8:00:22 | 500
2015-08-08 | 13:00:11 | 400
2015-08-08 | 13:23:11 | 200
2015-08-09 | 17:00:23 | 2200
2015-08-09 | 17:06:00 | 1290
2015-08-09 | 19:22:00 | 900
2015-08-13 | 01:01:22 | 1010
I want to get the latest data or transaction of a specific date, my desired result would be like this
dates | time | value
---------------------------------------------
2015-08-05 | 23:04:22 | 2400
2015-08-07 | 8:00:22 | 500
2015-08-08 | 13:23:11 | 200
2015-08-09 | 19:22:00 | 900
2015-08-13 | 01:01:22 | 1010
Only the latest data of a spefic or distinct date is chosen, what is the possible query with this?
You need to do this way to get that for each dates
select t1.dates,t1.time,t1.value from table as t1 inner join
(
select dates,max(time) as time from table group by dates
) as t2 on t1.dates=t2.dates and t1.time=t2.time
Use
SELECT * FROM TABLE_NAME GROUP BY dates ORDER BY time desc LIMIT 1;
Or
SELECT * FROM TABLE_NAME GROUP BY dates HAVING time <= '23:59:59' LIMIT 1;
Try this:
SELECT `dates`, MAX(`time`) AS `time`, MAX(`value`) AS `value`
FROM `tbl_name`
GROUP BY `dates`
ORDER BY `dates` ASC
Try this query:
SELECT * FROM `table` GROUP BY dates ORDER BY time DESC;
Related
I've this MySQL table my_table:
+-------+------------+-----------+
|Student| Date | Classroom |
+-------+------------+-----------+
| 1 | 2018-01-01 | 101 |
| 2 | 2018-01-01 | 102 |
| 3 | 2018-01-01 | 103 |
| 1 | 2018-03-01 | 104 |
| 2 | 2018-06-01 | 103 |
| 3 | 2018-09-01 | 104 |
| 1 | 2018-11-01 | 106 |
| 2 | 2018-12-01 | 101 |
+-------+------------+-----------+
The students stay in the assigned classroom till changed.
I'm trying to get which classroom they were in for a certain month.
For example in October(10), student 1 was in 104, 2 was in 103, and 3 was in 104.
I'm really unsure on how to proceed with this one so any help is appreciated.
Currently using this query based on Strawberry answer
SELECT x.*
FROM my_table x
LEFT OUTER JOIN my_table y
ON y.student = x.student
AND y.date < x.date
WHERE x.date <= LAST_DAY('2018-10-01')
GROUP BY student
DROP TABLE IF EXISTS my_table;
CREATE TABLE my_table
(Student INT NOT NULL, Date DATE NOT NULL, Classroom INT NOT NULL,PRIMARY KEY(student,classroom));
INSERT INTO my_table VALUES
(1,'2018-01-01',101),
(2,'2018-01-01',102),
(3,'2018-01-01',103),
(1,'2018-03-01',104),
(2,'2018-06-01',103),
(3,'2018-09-01',104),
(1,'2018-11-01',106),
(2,'2018-12-01',101);
SELECT x.*
FROM my_table x
JOIN
( SELECT student
, MAX(date) date
FROM my_table
WHERE date <= LAST_DAY('2018-10-01')
GROUP
BY student
) y
ON y.student = x.student
AND y.date = x.date;
+---------+------------+-----------+
| Student | Date | Classroom |
+---------+------------+-----------+
| 1 | 2018-03-01 | 104 |
| 2 | 2018-06-01 | 103 |
| 3 | 2018-09-01 | 104 |
+---------+------------+-----------+
Here's a go at it (snippet to go in a stored procedure; assumes table called example & output to table months). It produces a row per student for each month of the range.
drop table months;
create table months (month date, student integer, classroom integer);
set #month = (select min(date) from example);
start_loop: LOOP
insert into months select #month, s1.student, classroom from
(select student, max(date) as maxdate from example where date <= #month group by student) s1
join example s2 on s1.student = s2.student and maxdate = date;
if #month = (select max(date) from example) then
leave start_loop;
end if;
set #month = #month + interval 1 month;
END LOOP start_loop;
Let's break the problem into two parts. Firstly, find all the rooms which have been allocated to student A so far and sort them using the date. Next, find the record which is just before or equal to the required month.
For example:
Consider student 1. We get
+-------+------------+-----------+
|Student| Date | Classroom |
+-------+------------+-----------+
| 1 | 2018-01-01 | 101 |
| 1 | 2018-03-01 | 104 |
| 1 | 2018-11-01 | 106 |
+-------+------------+-----------+
Now, let's say for month June we try to find month just less than or equal to 2018-06-01 to get the required room number. I hope this will help.
Here is my database structure
Table name :set_inventory, and columns are below
inventory_id | room_id | quantity_start_date | quantity_end_date | total_rooms
1 | 2 | 2015-10-10 | 2015-10-12 | 5
2 | 2 | 2015-10-13 | 2015-10-14 | 10
3 | 2 | 2015-10-15 | 2015-10-17 | 0
Another Table
Table name : rooms, amd columns are
room_id | room_type | room_picture | room_description
2 | standard | pic_link | demo description
Description: In inventory table admin able to set the inventory based on multiple date ranges.
My query:
SELECT rooms.room_id, rooms.room_pic1, rooms.room_type, rooms.maximum_adults,
rooms.maximum_children, rooms.room_amenities,set_inventory.room_id,
set_inventory.quantity_start_date, set_inventory.quantity_start_date,
set_inventory.total_rooms
from rooms, set_inventory
WHERE rooms.room_id = set_inventory.room_id
AND quantity_start_date <= '2015-10-11'
AND quantity_end_date > '2015-10-13'
It shows me the inventory only 5, based in above query,
What i am looking for :actually, i am looking for the result,
for example my check_in_date "2015-10-11" and check_out_date is "2015-10-17"
In this i will be able to notify the customer that
"room is only available for 4 days,
Plz help me , thanks
UPDATED
Try this WHERE condition:
WHERE rooms.room_id = set_inventory.room_id AND DATE(quantity_start_date) <= '2015-10-11' AND DATE(quantity_end_date) > '2015-10-13';
So, minding you want to have the amount of days between these to dates, your query should contain somwthing like this:
SELECT ABS(DATEDIFF(a1.start_date, a2.end_date)) FROM (
SELECT quantity_start_date as start_date FROM set_inventory WHERE quantity_start_date <= '2015-10-11' ORDER BY start_date DESC LIMIT 1) a1, (
SELECT quantity_end_date as end_date FROM set_inventory WHERE quantity_end_date > '2015-10-13' ORDER BY end_date ASC LIMIT 1) a2;
If you try it, you'll get 4 in result.
This may be a little confusing but please bear with me. Here's the thing:
I have a database that contains ~1000 records, as the following table illustrates:
+------+----------+----------+
| id | date | amount |
+------+----------+----------+
| 0001 | 14/01/15 | 100 |
+------+----------+----------+
| 0002 | 14/02/04 | 358 |
+------+----------+----------+
| 0003 | 14/05/08 | 1125 |
+------+----------+----------+
What I want to do is this:
Retrieve all the records beginning at 2014 and until yesterday:
WHERE `date` > '14-01-01' AND `date` < CURDATE()
But also get the sum of amount up to the current date, this is:
WHERE `date` < CURDATE()
I've already got this working by just selecting all the records based on the second condition, getting the sum, and then excluding those which don't match the first condition. Something like this:
SELECT `id`, `date`, `amount` FROM `table`
WHERE `date` < CURDATE()
And then:
$rows = fetchAll($PDOStatement);
foreach($rows as $row) {
$sum += $row->amount;
if (
strtotime($row->date) > strtotime('14-01-01') &&
strtotime($row->date) < strtotime(date('Y-m-d'))
) {
$valid_rows[] = $row;
}
}
unset $rows;
Is there a way to achieve this in a single query, efficiently? Would a transaction be more efficient than sorting out the records in PHP? This has to be SQL-standard compliant (I'll be doing this on MySQL and SQLite).
Update:
It doesn't matter if the result ends up being something like this:
+------+----------+----------+-----+
| id | date | amount | sum |
+------+----------+----------+-----+
| 0001 | 14/01/15 | 100 | 458 |
+------+----------+----------+-----+
| 0002 | 14/02/04 | 358 | 458 |
+------+----------+----------+-----+
| 0003 | 14/05/08 | 1125 | 458 |
+------+----------+----------+-----+
The worst case would be when the resulting set ends up being the same as the set that gives the sum (in this case appending the sum would be irrelevant and would cause an overhead), but for any other regular cases the bandwith save would be huge.
You can create a special record with your sum and add it at the end of your first query
SELECT * FROM `table` WHERE `date` > '14-01-01' AND `date` < CURDATE()
UNION
SELECT 9999, CURDATE(), SUM(`amount`) FROM `table` WHERE `date` < CURDATE()
Then you will have all your desired record and the record with id 9999 or whatever is your sum
This could be achieved by correlated subquery, something like below:
SELECT *, (SELECT SUM(amount) FROM t WHERE t.date < t1.date) AS PrevAmount
FROM t AS t1
WHERE `date` > '14-01-01' AND `date` < CURDATE()
However it is very unefficient if the number of records is large.
It's hackish, but:
> select * from foo;
+------+------+
| id | val |
+------+------+
| 1 | 1 |
| 2 | 2 |
| 3 | 3 |
| 4 | 4 |
| 5 | 5 |
+------+------+
5 rows in set (0.02 sec)
> select * from foo
left join (
select sum(val)
from foo
where id < 3
) AS bar ON 1=1
where id < 4;
+------+------+----------+
| id | val | sum(val) |
+------+------+----------+
| 1 | 1 | 3 |
| 2 | 2 | 3 |
| 3 | 3 | 3 |
+------+------+----------+
Basically, do your summing in a joined subquery. That'll attach the sum result to every row in the outer table's results. You'll waste a bit of bandwidth sending that duplicated value out with every row, but it does get you the results in a "single" query.
EDIT:
You can get the SUM using a LEFT OUTER JOIN.
SELECT t1.`id`, t1.`date`, t2.sum_amount
FROM
`table` t1
LEFT OUTER JOIN
(
SELECT SUM(`amount`) sum_amount
FROM `table`
WHERE `date` < CURDATE()
) t2
ON 1 = 1
WHERE t1.`date` > STR_TO_DATE('01,1,2014','%d,%m,%Y') AND t1.`date` < CURDATE();
This will do what you want it to do...optimizing the subquery is the real challenge:
SELECT id,date,amount,(SELECT SUM(amount) FROM table) AS total_amount
FROM table
WHERE date BETWEEN '14-01-01' AND DATE_ADD(CURDATE(), INTERVAL -1 DAY)
Sorry if my question makes no sense. Not sure if we can do this with mysql only. Lets say I have this query:
SELECT SUM(win) * 100 as win_profit, date, uid FROM `tips` WHERE uid = 60 AND placed = 1 GROUP by date
This would obviously get the sum of the win column each day that is in the database.
Lets say the database had:
|___win___|____date____|
| 10 | 2014-04-16 |
| 10 | 2014-04-16 |
| 10 | 2014-04-17 |
| 10 | 2014-04-18 |
| 10 | 2014-04-18 |
| 10 | 2014-04-18 |
| 10 | 2014-04-19 |
| 10 | 2014-04-19 |
| 10 | 2014-04-19 |
This would result:
20
10
30
30
How can I get it to result so each adds up, mysql query only. So the result would be:
20
30
60
90
You could get all distinct dates, and LEFT JOIN to find the sum of all values up to that date; I kept the 100 multiplier from your sample query, but you need to remove it to get a result matching your desired result.
SELECT 100 * SUM(b.win), a.date
FROM (SELECT DISTINCT date FROM `tips`) a
LEFT JOIN tips b ON a.date >= b.date
GROUP BY a.date
ORDER BY a.date
An SQLfiddle to test with.
This could be another way to do it...
SET #full_sum=0;
SELECT #full_sum+SUM(win) as win_profit, date as this_date, uid,
#full_sum:=(SELECT #full_sum+SUM(win)
FROM `testing` WHERE uid = 60
GROUP by date HAVING date=this_date)
FROM `testing` WHERE uid = 60 GROUP by date;
MYSQL Table trial_list structure as follows...
id | product_id | expiry_date(date) | by_user | curr_datentime(timestamp)
we are able to extend any trial, and if we do that it simply another row with new expiry_date.
Now we would like to get rows got expired yesterday, we are currently using following sql query.....
Sample MYSQL DATASET
+----+-------------+-------------+-------------+----------+---------------------+
| id | product_id | comment | expiry_date | by_user | dnt |
+----+-------------+-------------+-------------+----------+---------------------+
| 2 | 50 | testing | 2011-02-18 | tester | 2011-02-17 23:36:12 |
+----+-------------+-------------+-------------+----------+---------------------+
| 3 | 50 | again | 2011-02-20 | tester | 2011-02-19 20:36:12 |
+----+-------------+-------------+-------------+----------+---------------------+
| 4 | 50 | extend | 2011-02-23 | tester | 2011-02-21 22:36:12 |
+----+-------------+-------------+-------------+----------+---------------------+
$sql = 'SELECT id, product_id, expiry_date, by_user, curr_datentime FROM trial_list WHERE expiry_date < CURDATE() ORDER BY expiry_date DESC';
We believe this is not correct as its getting all rows which date is older than yesterday not updated expiry_date, suppose we have given some user expiry date 1st feb 2011 and then we change again with 12th feb 2011, so it selects 1st feb 2011 entry. I think it makes sense.
What you have to do first is get the latest item per product_id. After that you can further filter it down to those which are expired. Something like:
SELECT a.* FROM
trial_list AS a
LEFT JOIN trial_list AS b ON a.product_id = b.product_id AND a.id < b.id
WHERE b.product_id IS NULL
AND a.expiry_date < curdate()
See http://dev.mysql.com/doc/refman/5.5/en/example-maximum-row.html
Try using NOW() instead of CURDATE(), you are comparing a Date to a Timestamp, NOW() will compare timestamps.