I have the following table (simplified) in an oracle db:
productId | modelDescription
1 | thing
2 | another thing
3 | not a thing
4 | thing
I want to select the modeldescription which has the highest appearence in this table. The problem is that there can be nearly infinite model descriptions.
So the resultset should look like something like this:
modelDescription | appearance
thing | 2
another thing | 1
... | ...
select modeldescription, count(modeldescription)
from products
group by modeldescription
order by 2 desc
In addition, if you only want the highest add the following:
Select Top 1 modeldescription......
Related
Is there a build in function in sql, to reverse the order in which the groupby works? I try to groupby a certain key but i would like to have the last inserted record returned and not the first inserted record.
Changing the order with orderby does not affect this behaviour.
Thanx in advance!
EDIT:
this is the sample data:
id|value
-----
1 | A
2 | B
3 | B
4 | C
as return i want
1 | A
3 | B
4 | C
not
1 | A
2 | B
4 | C
when using group by id don't get the result i want.
Question here is how are you identifying last inserted row. Based on your example, it looks like based on id. If id is auto generated, or a sequence then you can definitely do this.
select max(id),value
from your_table
group by value
Ideally in a table design, people uses a date column which holds the time a particular record was inserted, so it is easy to order by that.
Use Max() as your aggregate function for your id:
SELECT max(id), value FROM <table> GROUP BY value;
This will return:
1 | A
3 | B
4 | C
As for eloquent, I've not used it but I think it would look like:
$myData = DB::table('yourtable')
->select('value', DB::raw('max(id) as maxid'))
->groupBy('value')
->get();
For an assignment I have to write several SQL queries for a database stored in a PostgreSQL server running PostgreSQL 9.3.0. However, I find myself blocked with last query. The database models a reservation system for an opera house. The query is about associating the a spectator the other spectators that assist to the same events every time.
The model looks like this:
Reservations table
id_res | create_date | tickets_presented | id_show | id_spectator | price | category
-------+---------------------+---------------------+---------+--------------+-------+----------
1 | 2015-08-05 17:45:03 | | 1 | 1 | 195 | 1
2 | 2014-03-15 14:51:08 | 2014-11-30 14:17:00 | 11 | 1 | 150 | 2
Spectators table
id_spectator | last_name | first_name | email | create_time | age
---------------+------------+------------+----------------------------------------+---------------------+-----
1 | gonzalez | colin | colin.gonzalez#gmail.com | 2014-03-15 14:21:30 | 22
2 | bequet | camille | bequet.camille#gmail.com | 2014-12-10 15:22:31 | 22
Shows table
id_show | name | kind | presentation_date | start_time | end_time | id_season | capacity_cat1 | capacity_cat2 | capacity_cat3 | price_cat1 | price_cat2 | price_cat3
---------+------------------------+--------+-------------------+------------+----------+-----------+---------------+---------------+---------------+------------+------------+------------
1 | madama butterfly | opera | 2015-09-05 | 19:30:00 | 21:30:00 | 2 | 315 | 630 | 945 | 195 | 150 | 100
2 | don giovanni | opera | 2015-09-12 | 19:30:00 | 21:45:00 | 2 | 315 | 630 | 945 | 195 | 150 | 100
So far I've started by writing a query to get the id of the spectator and the date of the show he's attending to, the query looks like this.
SELECT Reservations.id_spectator, Shows.presentation_date
FROM Reservations
LEFT JOIN Shows ON Reservations.id_show = Shows.id_show;
Could someone help me understand better the problem and hint me towards finding a solution. Thanks in advance.
So the result I'm expecting should be something like this
id_spectator | other_id_spectators
-------------+--------------------
1| 2,3
Meaning that every time spectator with id 1 went to a show, spectators 2 and 3 did too.
Note based on comments: Wanted to make clear that this answer may be of limited use as it was answered in the context of SQL-Server (tag was present at the time)
There is probably a better way to do it, but you could do it with the 'stuff 'function. The only drawback here is that, since your ids are ints, placing a comma between values will involve a work around (would need to be a string). Below is the method I can think of using a work around.
SELECT [id_spectator], [id_show]
, STUFF((SELECT ',' + CAST(A.[id_spectator] as NVARCHAR(10))
FROM reservations A
Where A.[id_show]=B.[id_show] AND a.[id_spectator] != b.[id_spectator] FOR XML PATH('')),1,1,'') As [other_id_spectators]
From reservations B
Group By [id_spectator], [id_show]
This will show you all other spectators that attended the same shows.
Meaning that every time spectator with id 1 went to a show, spectators 2 and 3 did too.
In other words, you want a list of ...
all spectators that have seen all the shows that a given spectator has seen (and possibly more than the given one)
This is a special case of relational division. We have assembled an arsenal of basic techniques here:
How to filter SQL results in a has-many-through relation
It is special because the list of shows each spectator has to have attended is dynamically determined by the given prime spectator.
Assuming that (d_spectator, id_show) is unique in reservations, which has not been clarified.
A UNIQUE constraint on those two columns (in that order) also provides the most important index.
For best performance in query 2 and 3 below also create an index with leading id_show.
1. Brute force
The primitive approach would be to form a sorted array of shows the given user has seen and compare the same array of others:
SELECT 1 AS id_spectator, array_agg(sub.id_spectator) AS id_other_spectators
FROM (
SELECT id_spectator
FROM reservations r
WHERE id_spectator <> 1
GROUP BY 1
HAVING array_agg(id_show ORDER BY id_show)
#> (SELECT array_agg(id_show ORDER BY id_show)
FROM reservations
WHERE id_spectator = 1)
) sub;
But this is potentially very expensive for big tables. The whole table hast to be processes, and in a rather expensive way, too.
2. Smarter
Use a CTE to determine relevant shows, then only consider those
WITH shows AS ( -- all shows of id 1; 1 row per show
SELECT id_spectator, id_show
FROM reservations
WHERE id_spectator = 1 -- your prime spectator here
)
SELECT sub.id_spectator, array_agg(sub.other) AS id_other_spectators
FROM (
SELECT s.id_spectator, r.id_spectator AS other
FROM shows s
JOIN reservations r USING (id_show)
WHERE r.id_spectator <> s.id_spectator
GROUP BY 1,2
HAVING count(*) = (SELECT count(*) FROM shows)
) sub
GROUP BY 1;
#> is the "contains2 operator for arrays - so we get all spectators that have at least seen the same shows.
Faster than 1. because only relevant shows are considered.
3. Real smart
To also exclude spectators that are not going to qualify early from the query, use a recursive CTE:
WITH RECURSIVE shows AS ( -- produces exactly 1 row
SELECT id_spectator, array_agg(id_show) AS shows, count(*) AS ct
FROM reservations
WHERE id_spectator = 1 -- your prime spectator here
GROUP BY 1
)
, cte AS (
SELECT r.id_spectator, 1 AS idx
FROM shows s
JOIN reservations r ON r.id_show = s.shows[1]
WHERE r.id_spectator <> s.id_spectator
UNION ALL
SELECT r.id_spectator, idx + 1
FROM cte c
JOIN reservations r USING (id_spectator)
JOIN shows s ON s.shows[c.idx + 1] = r.id_show
)
SELECT s.id_spectator, array_agg(c.id_spectator) AS id_other_spectators
FROM shows s
JOIN cte c ON c.idx = s.ct -- has an entry for every show
GROUP BY 1;
Note that the first CTE is non-recursive. Only the second part is recursive (iterative really).
This should be fastest for small selections from big tables. Row that don't qualify are excluded early. the two indices I mentioned are essential.
SQL Fiddle demonstrating all three.
It sounds like you have one half of the total question--determining which id_shows a particular id_spectator attended.
What you want to ask yourself is how you can determine which id_spectators attended an id_show, given an id_show. Once you have that, combine the two answers to get the full result.
So the final answer I got, looks like this :
SELECT id_spectator, id_show,(
SELECT string_agg(to_char(A.id_spectator, '999'), ',')
FROM Reservations A
WHERE A.id_show=B.id_show
) AS other_id_spectators
FROM Reservations B
GROUP By id_spectator, id_show
ORDER BY id_spectator ASC;
Which prints something like this:
id_spectator | id_show | other_id_spectators
-------------+---------+---------------------
1 | 1 | 1, 2, 9
1 | 14 | 1, 2
Which suits my needs, however if you have any improvements to offer, please share :) Thanks again everybody!
SELECT TOP 1 Col1,col2
FROM table ... JOIN table2
...Some stuff...
ORDER BY DESC
gives different result. compared to
SELECT Col1,col2
FROM table ... JOIN table2
...Some stuff...
ORDER BY DESC
2nd query gives me some rows , When I want the Top 1 of this result I write the 1st query with TOP 1 clause. These both give different results.
why is this behavior different
This isn't very clear, but I guess you mean the row returned by the first query isn't the same as the first row returned by the second query. This could be because your order by has duplicate values in it.
Say, for example, you had a table called Test
+-----+------+
| Seq | Name |
+-----+------+
| 1 | A |
| 1 | B |
| 2 | C |
+-----+------+
If you did Select * From Test Order By Seq, either of these is valid
+-----+------+
| Seq | Name |
+-----+------+
| 1 | A |
| 1 | B |
| 2 | C |
+-----+------+
+-----+------+
| Seq | Name |
+-----+------+
| 1 | B |
| 1 | A |
| 2 | C |
+-----+------+
With the top, you could get either row.
Having the top 1 clause could mean the query optimizer uses a completely different approach to generate the results.
I'm going to assume that you're working in SQL Server, so Laurence's answer is probably accurate. But for completeness, this also depends on what database technology you are using.
Typically, index-based databases, like SQL Server, will return results that are sorted by the index, depending on how the execution plan is created. But not all databases utilize indices.
Netezza, for example, keeps track of where data lives in the system without the concept of an index (Netezza's system architecture is quite a bit different). As a result, selecting the 1st record of a query will result in a random record from the result set floating to the top. Executing the same query multiple times will likely result in a different order each time.
If you have a requirement to order data, then it is in your best interest to enforce the ordering yourself instead of relying on the arbitrary ordering that the database will use when creating its execution plan. This will make your results more predictable.
Your 1st query will get one table's top row and compare with another table with condition. So it will return different values compare to normal join.
I have a table for the link/relationship between two other tables, a table of customers and a table of groups. a group is made up of one or more customers. The link table is like
APP_ID | GROUP_ID | CUSTOMER_ID
1 | 1 | 123
1 | 1 | 124
1 | 1 | 125
1 | 2 | 123
1 | 2 | 125
2 | 3 | 123
3 | 1 | 123
3 | 1 | 124
3 | 1 | 125
I now have a need, given a list of customer IDs to be able to get the group ID for that list of customer IDs. Group ID may not be unique, the same group ID will contain the same list of customer IDs but this group may exist in more than one app_id.
I'm thinking that
SELECT APP_ID, GROUP_ID, COUNT(CUSTOMER_ID) AS COUNT
FROM GROUP_CUST_REL
WHERE CUSTOMER_ID IN ( <list of ids> )
GROUP BY APP_ID, GROUP_ID
HAVING COUNT(CUSTOMER_ID) = <number of ids in list>
will return me all of the group IDs that contain all of the customer ids in the given list and only those group ids. So for a list of (123,125) only group id 2 would be returned from the above example
I will then have to link with the app table to use its created timestamp to identify the most recent application that the group existed in so that I can then pull the correct/most up to date info from the group table.
Does anyone have any thoughts on whether this is the most efficient way to do this? If there is another quicker/cleaner way I'd appreciate your thoughts.
This smells like a division:
Division sample
Other related stack overflow question
Taking a look at the provided links you'll see the solution to similar issues from relational alegebra's point of view, doesn't seem to be quicker and arguably cleaner.
I didn't look at your solution at first, and when I solved this I turned out to have solved this the same way you did.
Actually, I thought this:
<number of ids in list>
Could be turned into something like this (so that you don't need the extra parameter):
select count(*) from (<list of ids>) as t
But clearly, I was wrong. I'd stay with your current solution if I were you.
Ok this one is realy tricky :D
i have a this table
bills_products:
- bill_id - product_id - action -
| 1 | 4 | add |
| 1 | 5 | add |
| 2 | 4 | remove |
| 2 | 1 | add |
| 3 | 4 | add |
as you can see product with the id 4 was added at bill 1 then removed in bill 2 and added again in bill 3
All Bills belong to a bill_group. But for the simplicity sake let's assume all the bills are in the same group.
Now i need a SQL Query that shows all the products that are currently added at this group.
In this example that would be 5, 1 and 4. If we would remove the bill with id 3 that would be 5 and 1
I've tried to do this with DISTINCT but it's not powerful enough or maybe I'm doing it wrong.
This seems to work in SQL Server at least:
select product_id
from (
select product_id,
sum((case when action='add' then 1 else -1 end)) as number
from bills_products
group by product_id
) as counts
where number > 0
SELECT DISTINCT product_id FROM bills_products WHERE action = 'add';
GSto almost had it, but you have to ORDER BY bill_id DESC to ensure you get the latest records.
SELECT DISTINCT product_id FROM bills_products
WHERE action = 'add'
ORDER BY bill_id DESC;
(P.S. I think most people would say it's a best practice to have a timestamp column on tables like this where you need to be able to know what the "newest" row is. You can't always rely on ids only ascending.)