I have a table with the fields id, group, left, level and createdAt.
Every row belongs to a group. The row with level = 0 is the group "leader".
I want to sort the table by the leaders' date, and within each group sort the rows by left. For example, in this table:
Id - Group - Left - Level - CreatedAt
1 1 1 0 00:10
2 1 2 1 00:20
3 2 1 0 00:00
4 1 3 1 00:30
5 2 2 1 00:40
The order should be:
Id - Group - Left - Level - CreatedAt
3 2 1 0 00:00
5 2 2 1 00:40
1 1 1 0 00:10
2 1 2 1 00:20
4 1 3 1 00:30
Because row 3 is the newest group leader, it should be first and followed by all it's group ordered by left. After that is row 1 which is the second most new leader, followed by it's group ordered by left.
Etc..
I hope I explained it clear enough.
Thanks!
Essentially, you need to join your table with the leader's time:
SELECT my_table.*
FROM my_table NATURAL JOIN (
SELECT my_table.Group, MIN(my_table.CreatedAt) AS LeaderTime
FROM my_table
WHERE my_table.Level = 0
GROUP BY my_table.Group
) t
ORDER BY t.LeaderTime, my_table.Left
See it on sqlfiddle.
If you can guarantee that there is an unambiguous leader for every group—e.g. because you have defined a UNIQUE constraint on (Group, Level), which you cannot have because your example contains two records in Group = 1 with Level = 1—then you can avoid the grouping operation:
SELECT my_table.*
FROM my_table JOIN my_table AS leader
ON leader.Group = my_table.Group AND leader.Level = 0
ORDER BY leader.CreatedAt, my_table.Left
In order to sort on the group leader's createdAt, you can join each row to the row describing the group leader, then do the sorting:
SELECT t1.*
FROM my_table t1
INNER JOIN my_table t2
ON t1.Group = t2.Group AND t2.Level = 0
ORDER BY t2.CreatedAt, t1.Left;
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'm very inexperienced. I've prepared a select statement which gives the information I need to populate a matches table. However it is not suitable because it contains a where clause. Is there a different way to use it, or how can I change it so that it is suitable for INSERT INTO.
The tables are as follows:-
match_order
match_order_id||match_descrip||first_player||second_player
1 1v2 1 2
2 1v3 1 3
3 2v3 2 3
4 1v4 1 4
5 2v4 2 4
6 3v4 3 4
entries
entry_id||round_id||league_id||box_id||box_position
1 1 1 1 1
2 1 1 1 2
3 1 1 1 3
4 1 2 1 4
5 1 2 1 2
6 1 2 1 1
7 1 2 1 1
matches
match_id||round_id||league_id||box_id||match_order_id||player1||player2
I need to insert new rows every month for a new round of matches. League size, box size & positions change each month.
This is the statement which gives the correct rows.
SELECT e.round_id, e.league_id, e.box_id, mo.match_order_id, e.entry_id as player1, e1.entry_id as player2
FROM match_order mo
LEFT JOIN entries e ON mo.first_player = e.box_position
LEFT JOIN entries e1 ON mo.second_player = e1.box_position
WHERE e.round_id = e1.round_id AND e.league_id = e1.league_id AND e.box_id = e1.box_id
ORDER BY round_id, league_id, box_id, match_order_id
Any help & advise would be greatly appreciated.
Thank you
Assuming match_id is an auto-increment column, you have the data for the other columns. You can just add the INSERT statement before your SELECT.
INSERT INTO matches(round_id, leage_id, box_id, match_order_id, player1, player2)
SELECT e.round_id, e.league_id, e.box_id, mo.match_order_id, e.entry_id as player1, e1.entry_id as player2
FROM match_order mo
LEFT JOIN entries e ON mo.first_player = e.box_position
LEFT JOIN entries e1 ON mo.second_player = e1.box_position
WHERE e.round_id = e1.round_id AND e.league_id = e1.league_id AND e.box_id = e1.box_id
I'm trying to make an update on a table so that it can increment the values on 1 column depending on another's order.
Here's how it'd go
ID GROUP_ID ORDER(Desired) ORDER(NOW)
1 1 1 2
2 1 2 3
3 1 3 1
4 2 1 2
5 2 2 1
6 3 1 1
7 3 2 1
8 3 3 2
So what I need is for each ID, to update the ORDER column so it can be consecutive, starting from 1, within each GROUP_ID.
I have found some solutions to similar problems regarding the updates and orders, but none that uses multiple orders for groups within the same table.
Hope I illustrated the problem right. Thanks in advance
You can do it by "ranking" the rows over again. Mysql doesn't support window functions but you can achieve the same results with join and count like this:
UPDATE YourTable t
INNER JOIN(SELECT s.id,s.group_id,count(*) as cnt
FROM YourTable s
INNER JOIN YourTable ss
ON(s.group_id = ss.group_id and s.id >= ss.id)
GROUP BY s.id,s.group_id) tt
ON (t.id = tt.id and t.group_id = tt.group_id)
SET t.order = tt.cnt
I have example data
ID DAY ORDER TIME PRODUCT
1 1 1 1 1
2 1 1 1 2
3 1 1 1 3
4 1 2 2 1
5 1 2 2 2
6 1 2 2 3
7 1 2 *3* 1
8 1 2 *3* 2
9 1 2 *3* 3
I want to prevent to having mltiple orders in different time at same day. if I set unique index on DAY,ORDER,TIME I will not be able to insert multiple time anyway, but I want to disable multiple different TIME. Is this possible with mysql?
Have can I find all records where there multiple different TIME value in same DAY and ORDER and delete them?
in this case I would like to delete records 7,8 ad 9 with SQL query because it is duplicate ORDER inserted.
I don't want to normalize table I will stick with this database structure.
Thank you very much
You can use delete with a join clause to find the duplicates and delete them:
delete
from t join
(select day, "order", min(time) as tokeeptime
from t
group by day, "order"
) tokeep
on t.day = tokeep.day and t."order" = tokeep."order" and t.time <> tokeeptime;
DELETE a
FROM tableName a
INNER JOIN
(
SELECT a.DAY, a.ORDER, MAX(a.TIME) Time
FROM tableName a
GROUP BY a.DAY, a.ORDER
HAVING COUNT(DISTINCT TIME) > 1
) b ON a.DAY = b.DAY AND
a.Order = b.Order AND
a.Time = b.Time
SQLFiddle Demo
I have three tables and am trying to get info from two and then perform a calculation on the third and display all the results in one query.
The (simplified) tables are:
table: employee_work
employee_id name
1 Joe
2 Bob
3 Jane
4 Michelle
table: carryover
employee_id days
1 5
2 10
3 3
table: timeoff
employee_id time_off_type days
1 Carryover 2
1 Leave 3
1 Carryover 1
2 Sick 4
2 Carryover 4
3 Leave 1
4 Sickness 4
The results I would like are:
employee_id, carryover.days, timeoff.days
1 5 3
2 10 4
3 3 0
However when I run the query, whilst I get the correct values in columns 1 and 2, I get the same number repeated in the third column for all entries.
Here is my query:
Select
employee_work.employee_id,
carryover.carryover,
(SELECT SUM(days) FROM timeoff WHERE timeoff.time_off_type = 'Carryover'
AND timeoff.start_date>='2013-01-01') AS taken
From
carryover Left Join
employee_work On employee_work.employee_id = carryover.employee_id Left Join
timeoff On employee_work.employee_id = timeoff.employee_id Left Join
Where
carryover.carryover > 0
Group By
employee_work.employee_id
I have tried to group by in the sub query but I then get told "Subquery returns more than one row" - how can I ensure that the sub query is respecting the join so it only looks at each employee at a time so I get my desired results?
The answer to your question is to use a correlated subquery. You don't need to mention the timeoff table twice in this case:
Select
employee_work.employee_id,
carryover.carryover,
(SELECT SUM(days)
FROM timeoff
WHERE timeoff.time_off_type = 'Carryover' and
timeoff.start_date>='2013-01-01' and
timeoff.employee_id = employee_work.employee_id
) AS taken
From
carryover Left Join
employee_work On employee_work.employee_id = carryover.employee_id
Where
carryover.carryover > 0
Group By
employee_work.employee_id;
An alternative structure is to do the grouping for all employees in the from clause. You can also remove the employee_work table, because it does not seem to be being used. (You can use carryover.employee_id for the id.)
Select co.employee_id, co.carryover, et.taken
From carryover c Left Join
(SELECT employee_id, SUM(days) as taken
FROM timeoff
WHERE timeoff.time_off_type = 'Carryover' and
timeoff.start_date>='2013-01-01'
) et
on co.employee_id = et.employee_id
Where c.carryover > 0;
I don't think the group by is necessary. If it is, then you should probably have an aggregation function in the original query.