I have this SQL query for pagination:
SELECT * FROM
(
SELECT T1.*,T2.*, ROW_NUMBER() over(ORDER BY ID DESC) row
FROM
table1 t1
LEFT JOIN
table2 t2 on t1.id = t2.pid
) tbl
WHERE row>= #start and row<#end
Now the problem is that the select result can be thousands of records, that will be executed for each page of each users.
Any suggestion that I can part the select (select less records?)
the ROW_NUMBER could be over order by ID or DATE.
and by the way, selecting * is just for simplicity of the sample code.
If you have SQL Server 2012 or above you can use the Offset and Fetch keywords as stated here
Related
I have code where I'm sampling 50,000 random records. I.e.,
SELECT * FROM Table1
SAMPLE 50000;
That works. However, what I really want to do is sample the number of records that are in a different table. I.e.,
SELECT * FROM Table1
SAMPLE count(*) FROM Table2;
I get an error. What am I doing wrong?
This is not randomized like sample, so bear that in mind. But there also won't be an obvious pattern, I believe it's determined by disk location (don't quote me on that).
SELECT *
FROM Table1
QUALIFY ROW_NUMBER() OVER
( PARTITION BY 1
ORDER BY 1
) <=
( SELECT COUNT(*)
FROM Table2
);
Better way
SELECT TMP.* -- Or list the columns you want with "rnd"
FROM ( SELECT RANDOM(-10000000,10000000) rnd,
T1.*
FROM Table1 T1
) TMP
QUALIFY ROW_NUMBER() OVER
( ORDER BY rnd
) <=
( SELECT COUNT(*)
FROM Table2
);
SELECT TOP 50000 * FROM Table1 ORDER BY NEWID()
I have a table with schema (id, date, value, source, ticker). I wanted to get record having highest ID group by date in sql server
Example Data
ID|date|value|source|ticker
3|10-Dec-2017|10|a|b
1|10-Dec-2017|11|p|q
Below query works in Sqlite. Do we know if I can do same with SqlServer
select max(id), date, value, source, ticker from table group by date
Expected return:-
ID|date|value|source|ticker
3|10-Dec-2017|10|a|b
Also how I can do same operation on UNION of 2 tables with same schema.
You can use subquery :
select t.*
from table t
where id = (select max(t1.id) from table t1 where t1.date = t.date);
However, you can also use row_number() function :
select top (1) with ties *
from table t
order by row_number() over (partition by [date] order by id desc);
You can also do it like below :
select t1.* from table1 t1
join (
select max(id) as id, [date] from table1
group by [date]
) as t2 on t1.id = t2.id
SQL HERE
I have a table with 2 fields
CREATE TABLE Temp_tab
(
id int identity primary key,
value float
);
INSERT INTO Temp_tab(value)
VALUES (65.09),(17.09);
I want to select all the records that are greater than Avg(Value).
Say... Select * from temp_tab where value > (select avg(value) from temp_tab);
This above query(using subquery) gives me the expected output
1 65.09
I want to achieve this without using Sub Query, CTE and Prodedure, since i am using Spark DB. Spark Db does not support Sub Queries, CTE and Prodedures
You can do this quite painfully with a cross join and aggregation:
Select t1.id, t1.value
from temp_tab t1 cross join
temp_tab t2
group by t1.id, t1.value
having t1.value > avg(t2.value);
As a note: Spark SQL claims to support subqueries (see here). So, your original query should work. If it only supports subqueries in the from clause, then you can do:
Select t.*
from temp_tab t join
(select avg(value) as avgvalue from temp_tab) a
on t.value > a.avgvalue;
spark-sql accept this query under version of 1.6.x
select * from (select * from tenmin_history order by TS_TIME DESC limit 144) a order by TS_TIME
This query solved my problem.
Table Essentially looks like:
Serial-ID, ID, Date, Data, Data, Data, etc.
There can be Multiple Rows for the Same ID. I'd like to create a view of this table to be used in Reports that only shows the most recent entry for each ID. It should show all of the columns.
Can someone help me with the SQL select? thanks.
There's about 5 different ways to do this, but here's one:
SELECT *
FROM yourTable AS T1
WHERE NOT EXISTS(
SELECT *
FROM yourTable AS T2
WHERE T2.ID = T1.ID AND T2.Date > T1.Date
)
And here's another:
SELECT T1.*
FROM yourTable AS T1
LEFT JOIN yourTable AS T2 ON
(
T2.ID = T1.ID
AND T2.Date > T1.Date
)
WHERE T2.ID IS NULL
One more:
WITH T AS (
SELECT *, ROW_NUMBER() OVER(PARTITION BY ID ORDER BY Date DESC) AS rn
FROM yourTable
)
SELECT * FROM T WHERE rn = 1
Ok, i'm getting carried away, here's the last one I'll post(for now):
WITH T AS (
SELECT ID, MAX(Date) AS latest_date
FROM yourTable
GROUP BY ID
)
SELECT yourTable.*
FROM yourTable
JOIN T ON T.ID = yourTable.ID AND T.latest_date = yourTable.Date
I would use DISTINCT ON
CREATE VIEW your_view AS
SELECT DISTINCT ON (id) *
FROM your_table a
ORDER BY id, date DESC;
This works because distinct on suppresses rows with duplicates of the expression in parentheses. DESC in order by means the one that normally sorts last will be first, and therefor be the one that shows in the result.
https://www.postgresql.org/docs/10/static/sql-select.html#SQL-DISTINCT
This seems like a good use for correlated subqueries:
CREATE VIEW your_view AS
SELECT *
FROM your_table a
WHERE date = (
SELECT MAX(date)
FROM your_table b
WHERE b.id = a.id
)
Your date column would need to uniquely identify each row (like a TIMESTAMP type).
Ok, got a tricky one here... If my data looks like this:
Table1
ID Date_Created
1 1/1/2009
2 1/3/2009
3 1/5/2009
4 1/10/2009
5 1/15/2009
6 1/16/2009
How do I get the records that are 2 days apart from each other? My end result set should be rows 1-3, and 5-6. Thanks!
SELECT l.*
FROM Table1 l
INNER JOIN Table1 r ON DATEDIFF(d, l.Date_Created, r.Date_Created) = 2
AND r.Date_Created = (SELECT TOP 1 * FROM Table1 WHERE Date_Created > l.Date_Created ORDER BY Date_Create)
select distinct t1.*
from Table1 t1
inner join Table1 t2
on abs(cast(t1.Date_Created - t2.Date_Created as float)) between 1 and 2
-- what does this give you?
select DISTINCT t1.id, t1.date_created, t2.id, t2.date_created from table1 t1, table1 t2 where datediff(dd,t1.date_created,t2.date_created) = 2 AND t1.id != t2.id ORDER BY t1.id;
Would this work?
select t1.id, t2.id
from table1 t1
join table1 t2
on t2.date_created - t1.date_created <= 2
I might suggest using programming code to do it. You want to collect groups of rows (separate groups). I don't think you can solve this using a single query (which would give you just one set of rows back).
If you want to get the rows which are WITHIN 'N' days apart, you can try this:
select t1.date_created, t2.date_created
from table1 t1, table1 t2
where t1.id <> t2.id and
t2.date_created-t1.date_created between 0 and N;
for exmaple, as you said, if you want to get the rows which are WITHIN 2 days a part,
you can use the below:
select t1.date_created,t2.date_created
from table1 t1, table1.t2
where t1.id <> t2.id and
t2.date_created-t1.date_created between 0 and 2;
I hope this helps....
Regards,
Srikrishna.
A cursor will be fastest, but here is a SELECT query that will do it. Note that for "up to N" days apart instead of 2 you'll have to replace the table Two with a table of integers from 0 to N-1 (and the efficiency will get worse).
I'll admit it's not entirely clear what you want, but I'm guess you want the ranges of rows that contain at least two rows in all and within which the successive rows are at most 2 days apart. If dates increase along with IDs, this should work.
with Two as (
select 0 as offset union all select 1
), r2(ID, Date_Created_o, dr) as (
select
ID, Date_Created+offset,
Date_Created + offset - dense_rank() over (
order by Date_Created+offset
) from r cross join Two
)
select
min(ID) as start, max(ID) as finish
from r2
group by dr
having min(ID) < max(ID)
order by dr;