I got a question in my homework for SQL about selecting the maximum values from the same table that have different class "Letters"
For example:
ID Student Group Avg(value)
-------------------------------------
1 stud1 A 9
2 stud2 A 9.5
3 stud3 B 8
4 stud4 B 8.5
What my query should do, is to show stud2 and stud4.The maximum from their respective groups.
I managed to do it in the end, but it took a lot of characters so I thought that maybe there's a shorter way to do. Any ideas? I used to first search the id or the stud that has max avg(value) from group A, intersecting with the id of the stud that has max avg(value) from B and then putting everything into one big select and then using those intersected IDs into another query that requested to show some different things about those IDs. But as I said, it looked far too long and thought that maybe there's an shorter way.
Try this (I renamed group to grp and avg to avg_val as those are reserved keywords):
select t1.*
from your_table t1
inner join (
select grp, max(avg_val) avg_val
from your_table
group by grp
) t2 on t1.grp = t2.grp
and t1.avg_val = t2.avg_val;
It finds maximum avg value per group and joins it with original table to get the corresponding students.
Please note that if there are multiple students with same avg as the max value of the that group, all of those students will be returned.
Related
Sorry if the title is misleading, I don't really know the terminology for what I want to accomplish. But let's consider this table:
CREATE TABLE entries (
id INT NOT NULL,
number INT NOT NULL
);
Let's say it contains four numbers associated with each id, like this:
id number
1 0
1 9
1 17
1 11
2 5
2 8
2 9
2 0
.
.
.
Is it possible, with a SQL-query only, to count the numbers of matches for any two given numbers (tuples) associated with a id?
Let's say I want to count the number of occurrences of number 0 and 9 that is associated with a unique id. In the sample data above 0 and 9 does occur two times (one time where id=1 and one time where id=2). I can't think of how to write a SQL-query that solves this. Is it possible? Maybe my table structure is wrong, but that's how my data is organized right now.
I have tried sub-queries, unions, joins and everything else, but haven't found a way yet.
You can use GROUP BY and HAVING clauses:
SELECT COUNT(s.id)
FROM(
SELECT t.id
FROM YourTable t
WHERE t.number in(0,9)
GROUP BY t.id
HAVING COUNT(distinct t.number) = 2) s
Or with EXISTS():
SELECT COUNT(distinct t.id)
FROM YourTable t
WHERE EXISTS(SELECT 1 FROM YourTable s
WHERE t.id = s.id and s.id IN(0,9)
HAVING COUNT(distinct s.number) = 2)
I have 2 different tables called observations and intervals.
observations:
id,
type,
date
1 recess 03.05.2011 17:00
2 recess 03.06.2011 12:00
intervals:
id,
observation id,
value
1 1 5
2 1 8
3 2 4
4 2 4
I want a view that will display:
observation_id
percent_positive ((count where value = 5)/(total number of observations))
1 .5
2 0
I know
Select observation_id, Count(*) from intervals where value = 5 Group by
observation_id
will give me:
1 1
1 0
and
Select observation_id, Count(*) from intervals Group by
observation_id
will give me:
1 2
2 2
So how do I combine these to create a view with the percent_positive variable I'm looking for?
You can use joins to fetch data from two tables having a common column field .
For more ,please read it in detail Multiple values from multiple Tables
This gave me your desired result. Not proficient enough in SQL to determine if this is the optimal way of solving the issue though.
SELECT
observation_id as obs,
(SELECT COUNT(*) FROM intervals WHERE observation_id = obs AND value = 5)/(SELECT COUNT(*) FROM INTERVALS WHERE observation_id = obs) as percent
FROM observation
JOIN intervals ON observation.id = intervals.observation_id
GROUP BY observation_id;
SELECT
i.observation_id,
SUM(IF(i.value=5,1,0)) / counts.num as 'percent_positive'
FROM intervals i
inner join (
select observation_id, count(1) as num from intervals group by observation_id
) counts on counts.observation_id = i.observation_id
group by i.observation_id
order by i.observation_id
;
That oughta get you close, can't actually run to test at the moment. I'm not sure about the significance of the value 5 meaning positive, so the i.value=5 test might need to be modified to do what you want. This will only include observation IDs with intervals that refer to them; if you want ALL observations then you'll need to join that table (select from that table and left join the others, to be precise). Of course the percentage for those IDs will be 0 divided by 0 anyway...
I have table that looks like this:
id rank
a 2
a 1
b 4
b 3
c 7
d 1
d 1
e 9
I need to get all the distinct rank values on one column and count of all the unique id's that have reached equal or higher rank than in the first column.
So the result I need would be something like this:
rank count
1 5
2 4
3 3
4 3
7 2
9 1
I've been able to make a table with all the unique id's with their max rank:
SELECT
MAX(rank) AS 'TopRank',
id
FROM myTable
GROUP BY id
I'm also able to get all the distinct rank values and count how many id's have reached exactly that rank:
SELECT
DISTINCT TopRank AS 'rank',
COUNT(id) AS 'count of id'
FROM
(SELECT
MAX(rank) AS 'TopRank',
id
FROM myTable
GROUP BY id) tableDerp
GROUP BY TopRank
ORDER BY TopRank ASC
But I don't know how to get count of id's where the rank is equal OR HIGHER than the rank in column 1. Trying SUM(CASE WHEN TopRank > TopRank THEN 1 END) naturally gives me nothing. So how can I get the count of id's where the TopRank is higher or equal to each distinct rank value? Or am I looking in the wrong way and should try something like running totals instead? I tried to look for similar questions but I think I'm completely on a wrong trail here since I couldn't find any and this seems a pretty simple problem that I'm just overthinking somehow. Any help much appreciated.
One approach is to use a correlated subquery. Just get the list of ranks and then use a correlated subquery to get the count you are looking for:
SELECT r.rank,
(SELECT COUNT(DISTINCT t2.id)
FROM myTable t2
WHERE t2.rank >= r.rank
) as cnt
FROM (SELECT DISTINCT rank FROM myTable) r;
Here is a simplified version of my table:
group price spec
a 1 .
a 2 ..
b 1 ...
b 2
c .
. .
. .
I'd like to produce a result like this: (I'll refer to this as result_table)
price_a |spec_a |price_b |spec_b |price_c ...|total_cost
1 |. |1 |.. |... |
(min) (min) =1+1+...
Basically I want to:
select the rows containing the min price within each group
combine columns into a single row
I know this can be done using several queries and/or combined with some non-sql processing on the results, but I suspect that there maybe better solutions.
The reason that I want to do task 2 (combine columns into a single row)
is because I want to do something like the following with the result_table:
select *,
(result_table.total_cost + table1.price + table.2.price) as total_combined_cost
from result_table
right join table1
right join table2
This may be too much to ask for, so here is some other thoughts on the problem:
Instead of trying to combine multiple rows(task 2), store them in a temporary table
(which would be easier to calculate the total_cost using sum)
Feel free to drop any thoughts, don't have to be complete answer, I feel it's brilliant enough if you have an elegant way to do task 1 !
==Edited/Added 6 Feb 2012==
The goal of my program is to identify best combinations of items with minimal cost (and preferably possess higher utilitarian value at the same time).
Consider #ypercube's comment about large number of groups, temporary table seems to be the only feasible solution. And it is also pointed out there is no pivoting function in MySQL (although it can be implemented, it's not necessary to perform such operation).
Okay, after study #Johan's answer, I'm thinking about something like this for task 1:
select * from
(
select * from
result_table
order by price asc
) as ordered_table
group by group
;
Although looks dodgy, it seems to work.
==Edited/Added 7 Feb 2012==
Since there could be more than one combination may produce the same min value, I have modified my answer :
select result_table.* from
(
select * from
(
select * from
result_table
order by price asc
) as ordered_table
group by group
) as single_min_table
inner join result_table
on result_table.group = single_min_table.group
and result_table.price = single_min_table.price
;
However, I have just realised that there is another problem I need to deal with:
I can not ignore all the spec, since there is a provider property, items from different providers may or may not be able to be assembled together, so to be safe (and to simplify my problem) I decide to combine items from the same provider only, so the problem becomes:
For example if I have an initial table like this(with only 2 groups and 2 providers):
id group price spec provider
1 a 1 . x
2 a 2 .. y
3 a 3 ... y
4 b 1 ... y
5 b 2 x
6 b 3 z
I need to combine
id group price spec provider
1 a 1 . x
5 b 2 x
and
2 a 2 .. y
4 b 1 ... y
record (id 6) can be eliminated from the choices since it dose not have all the groups available.
So it's not necessarily to select only the min of each group, rather it's to select one from each group so that for each provider I have a minimal combined cost.
You cannot pivot in MySQL, but you can group results together.
The GROUP_CONCAT function will give you a result like this:
column A column B column c column d
groups specs prices sum(price)
a,b,c some,list,xyz 1,5,7 13
Here's a sample query:
(The query assumes you have a primary (or unique) key called id defined on the target table).
SELECT
GROUP_CONCAT(a.`group`) as groups
,GROUP_CONCAT(a.spec) as specs
,GROUP_CONCAT(a.min_price) as prices
,SUM(a.min_prices) as total_of_min_prices
FROM
( SELECT price, spec, `group` FROM table1
WHERE id IN
(SELECT MIN(id) as id FROM table1 GROUP BY `group` HAVING price = MIN(price))
) AS a
See: http://dev.mysql.com/doc/refman/5.0/en/group-by-functions.html
Producing the total_cost only:
SELECT SUM(min_price) AS total_cost
FROM
( SELECT MIN(price) AS min_price
FROM TableX
GROUP BY `group`
) AS grp
If a result set with the minimum prices returned in row (not in column) per group is fine, then your problem is of the gretaest-n-per-group type. There are various methods to solve it. Here's one:
SELECT tg.grp
tm.price AS min_price
tm.spec
FROM
( SELECT DISTINCT `group` AS grp
FROM TableX
) AS tg
JOIN
TableX AS tm
ON
tm.PK = --- the Primary Key of the table
( SELECT tmin.PK
FROM TableX AS tmin
WHERE tmin.`group` = tg.grp
ORDER BY tmin.price ASC
LIMIT 1
)
OK so I have looked theough the other solutions an no help. So here is what I am trying to do.
I need to select the row with multiple columns where the value in one column is the max value.
here is sample data
orderfileid item number item cost warehouse
1 1234 3.45 ATL
1 2345 1.67 DFW
3 2345 2.45 NYY
3 678 2.4 ORD
2 1234 1.67 DFW
I need to select the entire row where the orderfileid is the max for each unique item number
the returned dataset should look like
orderfileid item number item cost warehouse
2 1234 1.67 DFW
3 2345 2.45 NYY
3 6789 2.4 ORD
I think i tried every combination of select max(orderfileid) i can think of
Any help would be appriciated.
thanks
You need to find your MAX values in a subquery, then use those results to join to your main table to retrieve the columns.
SELECT t.OrderFileId, t.ItemNumber, t.ItemCost, t.Warehouse
FROM YourTable t
INNER JOIN (SELECT ItemNumber, MAX(OrderFileId) AS MaxOrderId
FROM YourTable
GROUP BY ItemNumber) q
ON t.ItemNumber = q.ItemNumber
AND t.OrderFileId = q.MaxOrderId
select
t.*
from
table t
inner join (
select itemnumber, max(orderfileid) maxof
from table
group by itemnumber
) m on t.itemnumber = m.itemnumber
and t.orderfileid = m.maxof
I wouldn't even use Max. Just combine GROUP BY and ORDER BY
SELECT * FROM orders GROUP BY item_number ORDER BY orderfileid DESC
then for minimum just change to ASC
Try
SELECT * FROM `TABLE` WHERE orderfileid=(select max(orderfileid) from TABLE)
you can refer to a similar problem on how to group things using partitioning and picking one per partition in mysql
Deleting Rows: No Single Member Has More Than x Records
this is something similar to doing rank over in Oracle. my previous post was for oracle. my bad..
I think what you are looking for is the "Having" clause. Take a look at this.
select orderfileid, max(itemnumber), itemcost, warehouse from MyTable group by orderfileid having max(itemnumber) ;