Reputation: 249
I've been stumped with some SQL where I've got several rows of data, and I want to subtract a row from the previous row and have it repeat all the way down.
So here is the table:
CREATE TABLE foo ( id, length )
INSERT INTO foo (id,length) VALUES(1,1090) INSERT INTO foo (id,length) VALUES(2,888) INSERT INTO foo (id,length) VALUES(3,545) INSERT INTO foo (id,length) VALUES(4,434) INSERT INTO foo (id,length) VALUES(5,45)
I want the results to show a third column called difference which is one row subtracting from the one below with the final row subtracting from zero.
+------+------------------------+ | id |length | difference | +------+------------------------+ | 1 | 1090 | 202 | | 2 | 888 | 343 | | 3 | 545 | 111 | | 4 | 434 | 389 | | 5 | 45 | 45 |
I've tried a self join but I'm not exactly sure how to limit the results instead of having it cycle through itself. I can't depend that the id value will be sequential for a given result set so I'm not using that value. I could extend the schema to include some kind of sequential value.
This is what I've tried:
SELECT id, f.length, f2.length, (f.length - f2.length) AS difference FROM foo f, foo f2
Thank you for the assist.
Upvotes: 12
Views: 67624
Reputation: 57177
edit: fixed when re-read Q (misunderstood)
SELECT f.id,
f2.id,
f.length,
f2.length,
(f.length -f2.length) AS difference
FROM foo f,
foo f2
where f2.id = f.id+1
id was ambiguous
edit: note: tested in mysql 5.0
Upvotes: 1
Reputation: 39047
Select f1.id, f1.seqnum, f2.seqnum, f1.length, f2.length, f1.length-f2.length
From (
Select Id, length, row_number(order by length) 'seqnum'
From
foo
) f1
Inner join (
Select
Id, length, row_number(order by length) 'seqnum' from foo union select 0, 0, 0
) f2
On f1.seqnum = f2.seqnum + 1
Order by f1.length desc
Upvotes: 1
Reputation: 1
I had this problem and it was interesting to look at your solutions. I find it strange that such a normal-life problem is so complicated in SQL. As I need the values in a report only, I chose a completely different solution. I'm running Ruby on Rails as the front end of my sqlite3 database, and just did the subtraction in the view like this:
In your ruby controller, there is a object variable @foo that holds the rows returned by your query.
In the view, just do
<table border=1>
<tr>
<th>id</th>
<th>length</th>
<th>difference</th>
</tr>
<% for i in [email protected] do %>
<tr>
<td><%=h @foo[i].id %></td>
<td><%=h @foo[i].length %></td>
<td><% if ([email protected]) then %>
<%= @foo[i].length %>
<% else %>
<%= @foo[i+1].length.to_i - @foo[i].length.to_i %>
<% end %>
</td>
</tr>
<% end %>
</table>
Seems to be more robust than the SQL solutions.
Upvotes: 0
Reputation: 89711
So they are just ordered largest to smallest?
SELECT f.id, f.length, (f.length - ISNULL(t.length, 0)) AS difference
FROM foo AS f
LEFT JOIN (
SELECT f1.id
,MAX(f2.length) as length
FROM foo AS f1
INNER JOIN foo AS f2
ON f1.length > f2.length
GROUP BY f1.id
) AS t -- this is the triangle
ON t.id = f.id
You can use COALESCE
(or IFNULL
) instead of ISNULL
for MySQL.
Upvotes: 3
Reputation: 33474
This might help you (somewhat).
select a.id, a.length,
coalesce(a.length -
(select b.length from foo b where b.id = a.id + 1), a.length) as diff
from foo a
Upvotes: 13
Reputation: 40517
Yipee!!! this does the trick:
SELECT f.id, f.length,
(f.length - ISNULL(f2.length,0)) AS diff
FROM foo f
LEFT OUTER JOIN foo f2
ON f2.id = (f.id +1)
Please check for other cases also, it is working for the values you posted! Note this is for SQL Server 2005
Upvotes: 7
Reputation: 69262
What about something like this:
SELECT T2.ID, T2.[Length], T2.[Length]-T1.[Length] AS 'Difference'
FROM Foo AS T1 RIGHT OUTER JOIN Foo AS T2 ON ( T1.ID = (T2.ID-1) )
ORDER BY T1.ID
Upvotes: 1