I have table with names with eomonth history.
I need to join another table that has information about "tests" performed on names:
The result table should show the latest ID and date of performed test but the date of test cannot be greater than the data month:
Any ideas how to perform this?
I am guessing that you want the most recent EOMONTH from the second table for each row in the first table. If that is the correct interpretation, then you can simply use apply:
select t1.*, t2.*
from table1 t1 outer apply
(select top (1) t2.*
from table2 t2
where t2.test_id = t1.test_id and t2.eomonth <= t1.test_date
order by t2.eomonth desc
) t2;
This join works perfectly:
from table1 a
left join table2 b on a.name = b.name and a.eomonth >= b.id_date and a.eomonth < LAG(b.id_date,1,'3000-01-01') OVER (PARTITION BY name ORDER BY b.id_date desc)
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 am trying to build a SQL query to recover only the most young record of a table (it has a Timestamp column already) where the item by which I want to filter appears several times, as shown in my table example:
.
Basically, I have a table1 with Id, Millis, fkName and Price, and a table2 with Id and Name.
In table1, items can appear several times with the same fkName.
What I need to achieve is building up a single query where I can list the last record for every fkName, so that I can get the most actual price for every item.
What I have tried so far is a query with
SELECT DISTINCT [table1].[Millis], [table2].[Name], [table1].[Price]
FROM [table1]
JOIN [table2] ON [table2].[Id] = [table1].[fkName]
ORDER BY [table2].[Name]
But I don't get the correct listing.
Any advice on this? Thanks in advance,
A simple and portable approach to this greatest-n-per-group problem is to filter with a subquery:
select t1.millis, t2.name, t1.price
from table1 t1
inner join table2 t2 on t2.id = t1.fkName
where t1.millis = (select max(t11.millis) from table1 t11 where t11.fkName = t1.fkName)
order by t1.millis desc
using Common Table Expression:
;with [LastPrice] as (
select [Millis], [Price], ROW_NUMBER() over (Partition by [fkName] order by [Millis] desc) rn
from [table1]
)
SELECT DISTINCT [LastPrice].[Millis],[table2].[Name],[LastPrice].[Price]
FROM [LastPrice]
JOIN [table2] ON [table2].[Id] = [LastPrice].[fkName]
WHERE [LastPrice].rn = 1
ORDER BY [table2].[Name]
I have two tables.
Table1 is 1591 rows. Table2 is 270 rows.
I want to fetch specific column data from Table2 based on some condition between them and also exclude duplicates which are in Table2. Which I mean to join the tables but get only one value from Table2 even if the condition has occurred more than time. The result should be exactly 1591 rows.
I tried to make Left,Right, Inner joins but the data comes more than or less 1591.
Example
Table1
type,address,name
40,blabla,Adam
20,blablabla,Joe
Table2
type,currency
40,usd
40,gbp
40,omr
Joining on 'type'
Result
type,address,name,currency
40,blabla,name,usd
20,blblbla,Joe,null
try this it has to work
select *
from
Table1 h
inner join
(select type,currency,ROW_NUMBER()over (partition by type order by
currency) as rn
from
Table2
) sr on
sr.type=h.type
and rn=1
Try this. It's standard SQL, therefore, it should work on your rdbms system.
select * from Table1 AS t
LEFT OUTER JOIN Table2 AS y ON t.[type] = y.[type] and y.currency IN (SELECT MAX(currency) FROM Table2 GROUP BY [type])
If you want to control which currency is joined, consider altering Table2 by adding a new column active/non active and modifying accordingly the JOIN clause.
You can use outer apply if it's supported.
select a.type, a.address, a.name, b.currency
from Table1 a
outer apply (
select top 1 currency
from Table2
where Table2.type = a.type
) b
I typical way to do this uses a correlated subquery. This guarantees that all rows in the first table are kept. And it generates an error if more than one row is returned from the second.
So:
select t1.*,
(select t2.currency
from table2 t2
where t2.type = t1.type
fetch first 1 row only
) as currency
from table1 t1;
You don't specify what database you are using, so this uses standard syntax for returning one row. Some databases use limit or top instead.
I have two tables and i want to use some records from first table and get last related record from another one.
You can see my tables
I want to join table 1 with last record of table 2. (creationDate = 2018-07-20)
If you simply want to get the latest record in table 2 for every ID in table one then this will work:
select t1.ID, t1.Name, q.ID, q.CreationDate
from table1 t1
outer apply
(
select top 1 t2.ID, t2.CreationDate
from table2 t2
where t2.tbl_1_Id = t1.ID
order by t2.CreationDate desc
)q
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)