I am using two table named Transactions and Items. Refer below
Table 1 : Transactions
C_ID - State - Time
1 Start 2016-07-13 16:02:42
1 Passed 2016-07-13 20:28:21
2 Passed 2016-07-11 17:39:13
3 Passed 2016-07-07 20:23:00
4 Start 2016-07-01 13:19:54
4 Passed 2016-07-01 17:37:41
5 Start 2016-07-07 16:16:21
5 Passed 2016-07-07 21:04:01
6 Passed 2016-07-07 21:11:39
7 Passed 2016-07-08 20:30:46
Table 2 : Items
C_No - C_ID
C1 - 5
C2 - 3
C3 - 9
C4 - 7
C5 - 6
C6 - 8
C7 - 2
C8 - 4
C9 - 10
C10 - 1
I would like to join these tables and need output as mentioned below;
Output
C_No - State - Time
C10 - Start 2016-07-13 16:02:42
C10 - Passed 2016-07-13 20:28:21
C8 - Start 2016-07-01 13:19:54
C8 - Passed 2016-07-01 17:37:41
C1 - Start 2016-07-07 16:16:21
C1 - Passed 2016-07-07 20:00:01
In addition to join two tables, I want a filter on State and Time. The conditions are (State = 'Start' and Time <= 17:00) and (State = 'Passed' and Time <= 21:00)
I don't want any item doesn't have both Start and Passed.
I used the following query
{SELECT distinct(c.C_No), p.State, p.Time FROM Items c
inner join Transitions p on p.c_id = c.c_id and date(p.Time) between '2016-07-01' and CURRENT_DATE()
and ((p.State = 'Start' and time(p.Time) <= '17:00:00') or p.State = 'Passed' )
order by c.C_No, State;}
SQLFiddle added to question.
Try this statement
select c_no,state,time from transactions inner join items on transactions.c_id=items.c_id where (state='Start' and right(time,8)<='17:00:00') or (state='Passed' and right(time,8)<='21:00:00')
In your question, with your sample data, I don't think C1 should exist in the result, because when C_ID = 5, and its State is 'Passed' and its Time is '2016-07-07 21:04:01', it does not match your condition. So try this:
select t2.C_No, t.`State`, t.`Time`
from Transactions t
join (
select C_ID
from Transactions
where (State = 'Start' and time(`Time`) <= '17:00:00')
or (State = 'Passed' and time(`Time`) <= '21:00:00')
group by C_ID
having count(distinct State) > 1
) t1 on t.C_ID = t1.C_ID
left join Items t2 on t1.C_ID = t2.C_ID
order by t2.C_No, t.`State`;
SqlFiddle Demo
Ideas as follows:
with t_start as (
select t2.c_no, t1.state, t1.time
from transactions t1
inner join items t2 on t2.c_id=t1.c_id
where t1.state='State' and time(t1.time)<='17:00:00'
) t_passed as(
select t2.c_no, t1.state, t1.time
from transactions t1
inner join items t2 on t2.c_id=t1.c_id
where t1.state='Passed' and time(t1.time)<='21:00:00'
) t_total as(
select * from t_start
union
select * from t_passed
)
select * from t_total order by c_no, state;
Related
Below is my table let's call account
**ID accountID score tracking_date
1 1 3 2014-09-25 00:01:05
2 2 4 2014-09-26 01:05:18
3 1 6 2014-09-27 09:23:05
4 2 9 2014-09-28 20:01:05
5 1 1 2014-09-28 23:21:34
6 3 7 2014-09-21 00:01:00
7 2 1 2014-09-22 01:45:24
8 2 9 2014-09-27 14:01:43
9 3 1 2014-09-24 22:01:27
I want to select record with max date and also the difference of score with the records having tracking_date as minimum for that accountId. So I want output like below
ID accountID score_with_maxdate diff_score_with_mindate max_tracking_date
1 1 1 -2 2014-09-28 23:21:34
2 2 9 8 2014-09-28 20:01:05
3 3 1 -6 2014-09-24 22:01:27
Any help?
Here is one option. We can self-join a subquery which finds both the min and max tracking dates, for each account, twice to your original table. This will bring in all metadata for those max tracking date records, including the scores.
SELECT
t1.accountID,
t2.score AS score_with_maxdate,
t2.score - t3.score AS diff_score_with_mindate,
t1.max_tracking_date
FROM
(
SELECT
accountID,
MAX(tracking_date) AS max_tracking_date,
MIN(tracking_date) AS min_tracking_date
FROM yourTable
GROUP BY accountID
) t1
INNER JOIN yourTable t2
ON t1.accountId = t2.accountID AND t2.tracking_date = t1.max_tracking_date
INNER JOIN yourTable t3
ON t1.accountId = t3.accountID AND t3.tracking_date = t1.min_tracking_date
ORDER BY
t1.accountID;
Demo
This is a somewhat tricky question. I think conditional aggregation is a convenient way to solve the problem:
select min(t.id) as id, t.accountId,
max(case when t.tracking_date = t2.max_td then t.score end) as score_with_maxdate,
max(case when t.tracking_date = t2.max_td then t.score
when t.tracking_date = t2.min_td then - t.score
end) as diff_score_with_mindate,
max(t.tracking_date) as max_tracking_date
from t join
(select t2.accountId, min(t2.tracking_date) as min_td, max(t2.tracking_date) as max_td
from t t2
group by t2.accountId
) t2
on t.accountId = t2.accountId
group by t.accountId;
Another hackish way of getting same results by using aggregate and string fucntion
select t.accountID,
t.score_with_maxdate,
t.score_with_maxdate - t.score_with_mindate score_with_maxdate,
t.max_tracking_date
from(
select accountID,
substring_index(group_concat(score order by tracking_date desc),',', 1) + 0 score_with_maxdate,
substring_index(group_concat(score order by tracking_date asc),',', 1) + 0 score_with_mindate,
max(tracking_date) max_tracking_date
from demo
group by accountID
) t
Demo
But i would suggest you to go with other solutions mentioned by Tim & Gordon
I just get confused. Already tried to search this whole site or google but didn't find the 'nearest' solution.
Ok let's say I have this table structure.
id date finger_id finger_time is_enter
1 2017-03-30 2 09:00 1
2 2017-03-30 2 17:13 0
3 2017-03-31 4 09:10 1
4 2017-03-31 3 09:01 1
5. 2017-03-31 3 17:00 0
I want to make the table to be like below.
date finger_id enter_time exit_time
2017-03-30 2 09:00 17:13
2017-03-30 4 09:10
2017-03-31 3 09:10 17:00
I have made sql statement but it turns like this.
date finger_id enter_time exit_time
2017-03-30 2 09:00
2017-03-30 2 17:13
2017-03-31 4 09:10
2017-03-31 3 09:01
2017-03-31 3 17:00
I just want to know how to merge the is_enter 1 with is_enter 0 on the same date by the finger_id column.
Here's my sql query for the reference.
SELECT *
FROM `tbl_fingerprint`
LEFT JOIN `tbl_employee` ON `tbl_employee`.`fingerprint_id`=`tbl_fingerprint`.`fingerprint_id`
LEFT JOIN `tbl_position` ON `tbl_position`.`position_id`=`tbl_employee`.`position_id`
WHERE `fingerprint_date` >= '2017-03-01'
AND `fingerprint_date` <= '2017-04-01'
GROUP BY `tbl_fingerprint`.`fingerprint_id`,
`tbl_fingerprint`.`fingerprint_date`,
`tbl_fingerprint`.`is_enter`
ORDER BY `fingerprint_date` ASC LIMIT 30
Thanks for your help guys.
You can do a group by date and finger_id fields and use conditional expression (case or if()) within an aggregate function to get the expected outcome. The conditional statements within the aggregate function make sure that they return value only if the right value is set in is_enter field. I leave out the employee details, since those do not form part of your question:
SELECT date, fingerprint_id, max(if(is_enter=1,finger_time,null) as enter_time, max(if(is_enter=0,finger_time,null) as exit_time
FROM `tbl_fingerprint`
WHERE `fingerprint_date` >= '2017-03-01'
AND `fingerprint_date` <= '2017-04-01'
GROUP BY `tbl_fingerprint`.`fingerprint_id`,
`tbl_fingerprint`.`fingerprint_date`,
ORDER BY `fingerprint_date` ASC LIMIT 30
SELECT * FROM `tbl_fingerprint`
LEFT JOIN `tbl_employee` ON `tbl_employee`.`fingerprint_id`=`tbl_fingerprint`.`fingerprint_id`
LEFT JOIN `tbl_position` ON `tbl_position`.`position_id`=`tbl_employee`.`position_id`
LEFT JOIN (SELECT * FROM tbl_fingerprint WHERE is_enter = 0) a
ON a.finger_id = tbl_fingerprint.finger_id AND a.date = tbl_fingerprint.date
WHERE `fingerprint_date` >= '2017-03-01' AND `fingerprint_date` <= '2017-04-01' AND tbl_fingerprint.is_enter = 1
GROUP BY `tbl_fingerprint`.`fingerprint_id`, `tbl_fingerprint`.`fingerprint_date`, `tbl_fingerprint`.`is_enter`
ORDER BY `fingerprint_date` ASC LIMIT 30
Try This (This will work if finger_time is of time type):-
SELECT date, finger_id, min(finger_time) enter_time, if (min(finger_time) = max(finger_time), null, max(finger_time)) exit_time FROM xyz group by finger_id, date
SELECT a1.*, a3.time as time_out FROM attendance as a1
INNER JOIN (SELECT MIN(id) as id FROM attendance where is_enter = '1' group by date, f_id ) as a2
ON a2.id = a1.id
LEFT JOIN attendance as a3 ON a3.date = a1.date AND a1.f_id = a3.f_id and a3.is_enter = '0'
you may need to cast the date to not include the time portion or to char with the yyyy-mm-dd format
For Example, I have table like this:
Date | Id | Total
-----------------------
2014-01-08 1 15
2014-01-09 3 24
2014-02-04 3 24
2014-03-15 1 15
2015-01-03 1 20
2015-02-24 2 10
2015-03-02 2 16
2015-03-03 5 28
2015-03-09 5 28
I want the output to be:
Date | Id | Total
---------------------
2015-01-03 1 20
2014-02-04 3 24
2015-03-02 2 16
2015-03-09 5 28
Here the distinct values are Id. I need latest Total for each Id.
You can use left join as
select
t1.* from table_name t1
left join table_name t2
on t1.Id = t2.Id and t1.Date >t2.Date
where t2.Id is null
http://dev.mysql.com/doc/refman/5.0/en/example-maximum-column-group-row.html
You can also use Max() in sql:
SELECT date, id, total
FROM table as a WHERE date = (SELECT MAX(date)
FROM table as b
WHERE a.id = b.id
)
You can do it as below
SELECT *
FROM YourTable D
WHERE date = (SELECT MAX(date) FROM YourTable WHERE ID = D.ID)
Another way is by using INNER JOIN
Find the latest date per ID then join result back to the table to get the value
select A.ID,A.Date,A.value
from yourtable A
INNER JOIN
(
select MAX(date) as Date,ID
from yourtable
group by ID
) B
ON A.ID =B.ID and A.Date = B.Date
The other answers didn't work for me. I found the following code, which worked great for me:
SELECT * FROM TABLE WHERE DATE IN (SELECT MAX(DATE) FROM TABLE)
I am using SSMS 2014, SQLServer
I have a table like following columns:
id date price pid
---------------------------
1 2015-02-01 34 2
2 2015-02-02 34 2
3 2015-02-03 34 2
4 2015-02-04 78 2
5 2015-02-05 78 2
6 2015-02-06 78 2
7 2015-02-07 52 2
8 2015-02-08 52 2
9 2015-02-09 52 2
10 2015-02-10 34 2
11 2015-02-11 34 2
12 2015-02-12 34 2
Now I want following result:
date_from date_to price pid
-------------------------------------
2015-02-01 2015-02-03 34 2
2015-02-04 2015-02-06 78 2
2015-02-07 2015-02-09 52 2
2015-02-10 2015-02-12 34 2
IMPORTANT:
I don't want to group the price "34" in this case.
One solution i can think of using user defined variables also i assume the id part is set to auto_increment
select min(t1.date) date_from,
max(t1.date) date_to,
t1.price,
t1.pid
from (
select t.*,
#r:= case when #g = price then #r else #r + 1 end r,
#g:= price g
from test t
cross join (select #g:=null,#r:=0) t
order by id
) t1
group by t1.r
DEMO
Borrowing M Khalid Junaid's fiddle...
SELECT a.date date_from
, MIN(c.date) date_to
, price
, pid
FROM test a
LEFT
JOIN test b
ON b.pid = a.pid
AND b.price = a.price
AND b.id = a.id - 1
LEFT
JOIN test c
ON c.pid = a.pid
AND c.price = a.price
AND c.id >= a.id
LEFT
JOIN test d
ON d.pid = a.pid
AND d.price = a.price
AND d.id = c.id + 1
WHERE b.id IS NULL
AND c.id IS NOT NULL
AND d.id IS NULL
GROUP
BY a.id, a.price, a.pid;
http://sqlfiddle.com/#!2/478f9/6
Try below query, it will group by three records on basis of date and price
SELECT min(created) AS date_from, max(created) AS date_to, price, pid
FROM t1
GROUP BY price, floor(DATEDIFF("2015-02-01", created) / 3);
Try this in MySQL. Here app_date is Date in your question:
set #rownum=1;
select min(app_date),max(app_date),price,pid from
(select t.id1 id,t.app_date app_date,t.price1 price,t.pid pid,
case when (t.price1=t.price2 or t.price2 is null) then #rownum else #rownum:=#rownum+1 end temp_num from
(select a.id id1,b.id id2,a.app_date app_date,a.price price1,b.price price2,a.pid pid from
test a left outer join test b on a.id=b.id+1) t
order by id1) temp
group by price,pid,temp_num
order by min(app_date);
I was trying to find a solution but did not succeed even if it seems simple. So this might be a newbie question...
I have a table userscores with 3 columns:
date userid points
2012-05-01 1 23
2012-06-01 1 34
2012-07-01 1 44
2012-05-01 2 78
2012-06-01 2 94
2012-07-01 2 99
2012-06-01 3 2
2012-07-01 3 9
Now I need to get the difference of the points between 2012-05-01 and 2012-06-01 for each user.
Users' points that are not existing (example userid 3) have to be calculated as 2 - 0... for this I guess I can use COALESCE(qa_points,0).
I read about combining two subqueries for the calculation but failed implementing it.
Any help appreciated.
PS: This is not working:
SELECT t1.userid, t1.points - t2.points AS mpoints FROM (
SELECT userid,points FROM `userscores`
WHERE YEAR(date) = YEAR('2012-05-01')
AND MONTH(date) = MONTH('2012-05-01') )
AS t1
JOIN (
SELECT userid,points FROM `userscores`
WHERE YEAR(date) = YEAR('2012-04-01')
AND MONTH(date) = MONTH('2012-04-01') )
AS t2
ORDER BY mpoints DESC, t1.userid DESC;
I suppose your query will look like this:
SELECT ul.userid,
ul.points - COALESCE(uf.points, 0) AS points_difference
FROM userscores ul
LEFT JOIN
(SELECT userid, points FROM userscores WHERE `date` = '2012-05-01') AS uf
ON uf.userid = ul.userid
WHERE ul.date = '2012-06-01'
LEFT JOIN is used because you told that there may be no records for this user/former date combination.
Use this query:
SELECT t1.userid,
( t1.points - (case t2.date when '2012-05-01' then t2.points else 0 end))
AS mpoints FROM userscores as t1
INNER JOIN userscores as t2
ON t1.date = '2012-06-01' AND t1.userid=t2.userid