I have a table that has three columns: Category, Timestamp and Value.
What I want is a SQL select that will give me the 5 most recent values of each category. How would I go about and do that?
I tried this:
select
a."Category",
b."Timestamp",
b."Value"
from
(select "Category" from "Table" group by "Category" order by "Category") a,
(select a."Category", c."Timestamp", c."Value" from "Table" c
where c."Category" = a."Category" limit 5) b
Unfortunately, it won't allow it because "subquery in FROM cannot refer to other relations of same query level".
I'm using PostGreSQL 8.3, by the way.
Any help will be appreciated.
SELECT t1.category, t1.timestamp, t1.value, COUNT(*) as latest
FROM foo t1
JOIN foo t2 ON t1.id = t2.id AND t1.timestamp <= t2.timestamp
GROUP BY t1.category, t1.timestamp
HAVING latest <= 5;
Note: Try this out and see if it performs suitably for your needs. It will not scale well for large groups.
Related
As others have commented, I'm now going to add some code:
Imported tables
table3
Case No. is the primary key. Each report date shows one patient. Depending on if the patient is import or local, the cumulative column increases. You can see some days there are no cases so the date like 25/01/2020 is skipped
table2
Report date has no duplicate.
Now, I want to join the tables. Example outcome here:
enter image description here
The maximum cumulative of each date is joined into the new table. So although 26/01/2020 of table3 shows the increase from 6, 7, to 8, I only want the highest cumulative number there.
Thanks for letting me know how my previous query could be improved. Your opinion helps me a lot.
I have tried Gordon Linoff's by substituting the actual names (which I initially omitted because I thought they were ambiguous).
His code is as follows (I've upvoted):
SELECT t3.`Report date`,
max(max(t3.cumulative_local)) over (order by t3.`Report date`),
max(max(t3.cumulative_import)) over (order by t3.`Report date`)
from table3 t3 left join
table2 t2
using (`Report date`)
group by t2.`Report date`;
But I got an error
Error Code: 1055. Expression #1 of SELECT list is not in GROUP BY clause and contains nonaggregated column 'new.t3.Report date' which is not functionally dependent on columns in GROUP BY clause; this is incompatible with sql_mode=only_full_group_by
Anyways I am now experimenting. Both answers helped. If you know how to fix 1055, let me know, or if you could propose another solution. Thanks
I think you just want aggregation and window functions:
select t1.date,
max(max(cumulativea)) over (order by t1.date),
max(max(cumulativeb)) over (order by t1.date)
from table1 t1 left join
table2 t2
on t1.date = t2.date
group by t1.date;
This returns the maximum values of the two columns up to each date, which is, I think, what you are trying to describe.
I don't understand why you have cumulA and cumulB on table1. I suppose it will be to store the Max cumulA and cumulB for each days.
You must first self-join table2 to find the Max for each date (with a GROUP BY date) :
SELECT t2.id, t2.date, cA
FROM t2
JOIN (
SELECT id, MAX(cumulA) AS cA, date AS d2
FROM t2
GROUP BY d2
) AS td
ON t2.id=td.id
AND t2.date=d2
ORDER BY t2.date
After, you join left table1 on result of self-join table2 to have each days.
SELECT * FROM `t1` LEFT JOIN t2 ON t1.date = t2.date ORDER BY t1.date
Here is the fusion of the 2 junctions :
SELECT * FROM `t1` LEFT JOIN (
SELECT t2.id, t2.date, cA
FROM t2
JOIN (
SELECT id, MAX(cumulA) AS cA, date AS d2
FROM t2
GROUP BY d2
) AS td
ON t2.id=td.id
AND t2.date=d2
ORDER BY t2.date
) AS tt
ON t1.date = tt.date ORDER BY t1.date
You do the same for cumulB.
And after (I suppose), you INSERT INTO the result into table1.
I hope I answered your question.
Good continuation.
_Teddy_
I have a DB table with columns ID, time, and text. ID is non-unique. I'm trying to construct a query that says "For each distinct IDs, give me the row whose time is the greatest value before T". The second part is easy enough with a ORDER BY DESC LIMIT 1 WHERE time < T, but I don't know how to ensure I get coverage of all the IDs.
As an example:
ID,time,text
1,10,"hello"
1,20,"world"
2,10,"foo"
3,10,"bar"
4,50,"blah"
If I searched with time 25, I'd want:
1,20,"world"
2,10,"foo"
3,10,"bar"
I could do this is multiple queries by searching for DISTINCT IDs and then doing a search for each, but I was hoping there was a more efficient compound query form.
SELECT t.ID, t.time, t.text
FROM YourTable t
INNER JOIN (SELECT ID, MAX(time) AS MaxTime
FROM YourTable
WHERE time < T
GROUP BY ID) q
ON t.ID = q.ID
AND t.time = q.MaxTime
You can use a sub-query:
select t1.id, t1.time, t1.test
from yourtable t1
inner join
(
select id, max(time) MaxTime
from yourtable
where time < T
group by id
) t2
on t1.id = t2.id
and t1.time = t2.time
I have a table like below
id name dependency
-----------------------
1 xxxx 0
2 yyyy 1
3 zzzz 2
4 aaaaaa 0
5 bbbbbb 4
6 cccccc 5
the list goes on. I want to select group of rows from this table , by giving the name of 0 dependency in where clause of SQL and till it reaches a condition where there is no more dependency. (For ex. rows 1,2, 3 forms a group, and rows 4,5,6 is another group) .please help
Since you did not specify a product, I'll go with features available in the SQL specification. In this case, I'm using a common-table expression which are supported by many database products including SQL Server 2005+ and Oracle (but not MySQL):
With MyDependents As
(
Select id, name, 0 As level
From MyTable
Where dependency = 0
And name = 'some value'
Union All
Select T.id, T.name, T.Level + 1
From MyDependents As D
Join MyTable As T
On T.id = D.dependency
)
Select id, name, level
From MyDependents
Another solution which does not rely on common-table expressions but does assume a maximum level of depth (in this case two levels below level 0) would something like
Select T1.id, T1.name, 0 As level
From MyTable As T1
Where T1.name = 'some value'
Union All
Select T2.id, T2.name, 1
From MyTable As T1
Join MyTable As T2
On T2.Id = T1.Dependency
Where T1.name = 'some value'
Union All
Select T3.id, T3.name, 2
From MyTable As T1
Join MyTable As T2
On T2.Id = T1.Dependency
Join MyTable As T3
On T3.Id = T2.Dependency
Where T1.name = 'some value'
Sounds like you want to recursively query your table, for which you will need a Common Table Expression (CTE)
This MSDN article explains CTEs very well. They are confusing at first but surprisingly easy to implement.
BTW this is obviously only for SQL Server, I'm not sure how you'd achieve that in MySQL.
This is the first thing that came to mind. It can be probably done more directly/succinctly, I'll try to dwell on it a little.
SELECT *
FROM table T1
WHERE T1.id >=
(SELECT T2.id FROM table T2 WHERE T2.name = '---NAME HERE---')
AND T1.id <
(SELECT MIN(id)
FROM table T3
WHERE T3.dependency = 0 AND T3.id > T2.id)
If you can estimate a max depth, this works out to something like:
SELECT
COALESCE(t4.field1, t3.field1, t2.field1, t1.field1, t.field1),
COALESCE(t4.field2, t3.field2, t2.field2, t1.field2, t.field2),
COALESCE(t4.field3, t3.field3, t2.field3, t1.field3, t.field3),
....
FROM table AS t
LEFT JOIN table AS t1 ON t.dependency = t1.id
LEFT JOIN table AS t2 ON t1.dependency = t2.id
LEFT JOIN table AS t3 ON t2.dependency = t3.id
LEFT JOIN table AS t4 ON t3.dependency = t4.id
....
This is a wild guess just to be different, but I think it's kind of pretty, anyway. And it's at least as portable as any of the others. But I don't want to look to closely; I'd want to use sensible data, start testing, and check for sensible results.
Hierarchical query will do:
SELECT *
FROM your_table
START WITH id = :id_of_group_header_row
CONNECT BY dependency = PRIOR id
Query works like this:
1. select all rows satisfying START WITH condition (this rows are roots now)
2. select all rows satisfying CONNECT BY condition,
keyword PRIOR means this column's value will be taken from the root row
3. consider rows selected on step 2 to be roots
4. go to step 2 until there are no more rows
There is a table with Columns as below:
Id : long autoincrement;
timestamp:long;
price:long
Timestamp is given as a unix_time in ms.
Question: what is the average time difference between the records ?
First thought is a sub-query grabbing the record immediately previous:
SELECT timestamp -
(select top 1 timestamp from Table T1 where T1.Id < Table.Id order by Id desc)
FROM Table
Then you can take the average of that:
SELECT AVG(delta)
from (SELECT timestamp -
(select top 1 timestamp from Table T1 where T1.Id < Table.Id order by Id desc) as delta
FROM Table) T
There will probably need to be some handling of the null that results for the first row, but I haven't tested to be sure.
In SQL Server, you could write something like that to get that information:
SELECT
t1.ID, t2.ID,
DATEDIFF(MILLISECOND, t2.PriceTime, test2.PriceTime)
FROM table t1
INNER JOIN table t2 ON t2.ID = t1.ID-1
WHERE t1.ID > (SELECT MIN(ID) FROM table)
and if you're only interested in the AVG across all entries, you could use:
SELECT
AVG(DATEDIFF(MILLISECOND, t2.PriceTime, test2.PriceTime))
FROM table t1
INNER JOIN table t2 ON t2.ID = t1.ID-1
WHERE t1.ID > (SELECT MIN(ID) FROM table)
Basically, you need to join the table with itself, and use "t1.ID = t2.ID-1" to associate item no. 2 in one table with item no. 1 in the other table and then calculate the time difference between the two. In order to avoid accessing item no. 0 which doesn't exist, use the "T1.ID > (SELECT MIN(ID) FROM table)" clause to start from the second item.
Marc
At a guess:
SELECT AVG(timestamp)
I think you need to provide more information in your question for us to help.
If you mean difference between each-other row:
select AVG(x) from (
select a.timestamp - b.timestamp as x
from table a, table b -- this multiplies a*b ) sub
SELECT AVG(T2.Timestamp - T1.TimeStamp)
FROM Table T1
JOIN Table T2 ON T2.ID = T1.ID + 1
try this
Select Avg(E.Timestamp - B.Timestamp)
From Table B Join Table E
On E.Timestamp =
(Select Max(Timestamp)
From Table
Where Timestamp < R.Timestamp)
I have two tables with a 1:n relationship: "content" and "versioned-content-data" (for example, an article entity and all the versions created of that article). I would like to create a view that displays the top version of each "content".
Currently I use this query (with a simple subquery):
SELECT
t1.id,
t1.title,
t1.contenttext,
t1.fk_idothertable
t1.version
FROM mytable as t1
WHERE (version = (SELECT MAX(version) AS topversion
FROM mytable
WHERE (fk_idothertable = t1.fk_idothertable)))
The subquery is actually a query to the same table that extracts the highest version of a specific item. Notice that the versioned items will have the same fk_idothertable.
In SQL Server I tried to create an indexed view of this query but it seems I'm not able since subqueries are not allowed in indexed views. So... here's my question... Can you think of a way to convert this query to some sort of query with JOINs?
It seems like indexed views cannot contain:
subqueries
common table expressions
derived tables
HAVING clauses
I'm desperate. Any other ideas are welcome :-)
Thanks a lot!
This probably won't help if table is already in production but the right way to model this is to make version = 0 the permanent version and always increment the version of OLDER material. So when you insert a new version you would say:
UPDATE thetable SET version = version + 1 WHERE id = :id
INSERT INTO thetable (id, version, title, ...) VALUES (:id, 0, :title, ...)
Then this query would just be
SELECT id, title, ... FROM thetable WHERE version = 0
No subqueries, no MAX aggregation. You always know what the current version is. You never have to select max(version) in order to insert the new record.
Maybe something like this?
SELECT
t2.id,
t2.title,
t2.contenttext,
t2.fk_idothertable,
t2.version
FROM mytable t1, mytable t2
WHERE t1.fk_idothertable == t2.fk_idothertable
GROUP BY t2.fk_idothertable, t2.version
HAVING t2.version=MAX(t1.version)
Just a wild guess...
You Might be able to make the MAX a table alias that does group by.
It might look something like this:
SELECT
t1.id,
t1.title,
t1.contenttext,
t1.fk_idothertable
t1.version
FROM mytable as t1 JOIN
(SELECT fk_idothertable, MAX(version) AS topversion
FROM mytable
GROUP BY fk_idothertable) as t2
ON t1.version = t2.topversion
I think FerranB was close but didn't quite have the grouping right:
with
latest_versions as (
select
max(version) as latest_version,
fk_idothertable
from
mytable
group by
fk_idothertable
)
select
t1.id,
t1.title,
t1.contenttext,
t1.fk_idothertable,
t1.version
from
mytable as t1
join latest_versions on (t1.version = latest_versions.latest_version
and t1.fk_idothertable = latest_versions.fk_idothertable);
M
If SQL Server accepts LIMIT clause, I think the following should work:
SELECT
t1.id,
t1.title,
t1.contenttext,
t1.fk_idothertable
t1.version
FROM mytable as t1 ordery by t1.version DESC LIMIT 1;
(DESC - For descending sort; LIMIT 1 chooses only the first row and
DBMS usually does good optimization on seeing LIMIT).
I don't know how efficient this would be, but:
SELECT t1.*, t2.version
FROM mytable AS t1
JOIN (
SElECT mytable.fk_idothertable, MAX(mytable.version) AS version
FROM mytable
) t2 ON t1.fk_idothertable = t2.fk_idothertable
Like this...I assume that the 'mytable' in the subquery was a different actual table...so I called it mytable2. If it was the same table then this will still work, but then I imagine that fk_idothertable will just be 'id'.
SELECT
t1.id,
t1.title,
t1.contenttext,
t1.fk_idothertable
t1.version
FROM mytable as t1
INNER JOIN (SELECT MAX(Version) AS topversion,fk_idothertable FROM mytable2 GROUP BY fk_idothertable) t2
ON t1.id = t2.fk_idothertable AND t1.version = t2.topversion
Hope this helps