A column consists of dates only. I need to extract the latest 2 dates from the table. What is the best way to do it?
Example: Values in a table having date column as 01-01-2021, 01-02-2021, 01-03-2021
I would need 01-02-2021,01-03-2021 as my output.
You can use rank() (or row_number() if there are no dupicliates):
select t.*
from t
qualify rank() over (order by datecol desc) <= 2;
Note: If you want distinct values, you could also use:
select distinct datecol
from t
order by datecol desc
limit 2;
You can order the column and then select the top X elements that you require
Select Top 2 col_Date
from MyTable
Order by col_Date DESC
The returned result would be a column
If you need to have a single string, there is a function that can help : STRING_SPLIT
Related
I have one table where multiple records inserted for each group of product. Now, I want to extract (SELECT) only the last entries. For more, see the screenshot. The yellow highlighted records should be return with select query.
The HAVING MAX and HAVING MIN clause for the ANY_VALUE function is now in preview
HAVING MAX and HAVING MIN were just introduced for some aggregate functions - https://cloud.google.com/bigquery/docs/release-notes#February_06_2023
with them query can be very simple - consider below approach
select any_value(t having max datetime).*
from your_table t
group by t.id, t.product
if applied to sample data in your question - output is
You might consider below as well
SELECT *
FROM sample_table
QUALIFY DateTime = MAX(DateTime) OVER (PARTITION BY ID, Product);
If you're more familiar with an aggregate function than a window function, below might be an another option.
SELECT ARRAY_AGG(t ORDER BY DateTime DESC LIMIT 1)[SAFE_OFFSET(0)].*
FROM sample_table t
GROUP BY t.ID, t.Product
Query results
You can use window function to do partition based on key and selecting required based on defining order by field.
For Example:
select * from (
select *,
rank() over (partition by product, order by DateTime Desc) as rank
from `project.dataset.table`)
where rank = 1
You can use this query to select last record of each group:
Select Top(1) * from Tablename group by ID order by DateTime Desc
I have table as below:
I want write a sql query to get output as below:
the query should select all the records from the table but, when multiple records have same Id column value then it should take only one record having latest Date.
E.g., Here Rudolf id 1211 is present three times in input---in output only one Rudolf record having date 06-12-2010 is selected. same thing with James.
I tried to write a query but it was not succssful. So, please help me to form a query string in sql.
Thanks in advance
You can partition your data over Date Desc and get the first row of each partition
SELECT A.Id, A.Name, A.Place, A.Date FROM (
SELECT
*,
ROW_NUMBER() OVER (PARTITION BY Id ORDER BY Date DESC) AS rn
FROM [Table]
) A WHERE A.rn = 1
you can use WITH TIES
select top 1 PERCENT WITH TIES * from t
order by (row_number() over(partition by id order by date desc))
https://dbfiddle.uk/?rdbms=sqlserver_2017&fiddle=280b7412b5c0c04c208f2914b44c7ce3
As i can see from your example, duplicate rows differ only in Date. If it's a case, then simple GROUP BY with MAX aggregate function will do the job for you.
SELECT Id, Name, Place, MAX(Date)
FROM [TABLE_NAME]
GROUP BY Id, Name, Place
Here is working example: http://sqlfiddle.com/#!18/7025e/2
I have a table with data where in Column A I have groups of repeating Data (one after another).
I want to select only first row of each group based on values in column A only (no other criteria). Mind you, I want all corresponding columns selected also for the mentioned new found row (I don't want to exclude them).
Can someone help me with a proper query.
Here is a sample:
SAMPLE
Thanks!
#standardSQL
SELECT row.*
FROM (
SELECT ARRAY_AGG(t LIMIT 1)[OFFSET(0)] row
FROM `project.dataset.table` t
GROUP BY columnA
)
you can try smth like this:
#standardSQL
SELECT
* EXCEPT(rn)
FROM (
SELECT
*,
ROW_NUMBER() OVER(PARTITION BY columnA ORDER BY columnA) AS rn
FROM
your_dataset.your_table)
WHERE rn = 1
that will return:
Row columnA col2 ...
1 AC1001 Z_Creation
2 ACO112BISPIC QN
...
Add LIMIT 1 at the end of the query
something like
SELECT name, year FROM person_table ORDER BY year LIMIT 1
You can now use qualify for a more concise solution:
select
*
from
your_dataset.your_table
where true
qualify ROW_NUMBER() OVER(PARTITION BY columnA ORDER BY columnA) = 1
In BigQuery the physical sequence of rows is not significant. “BigQuery does not guarantee a stable ordering of rows in a table. Only the result of a query with an explicit ORDER BY clause has well-defined ordering.”[1].
First, you need to define which property will determine the first row of your group, then you can run Vasily Bronsky’s query by changing ORDER BY with that property. Which means either you should add another column to the table to store the order of the rows or select one from the columns you have.
I have an oracle DB.
My table has ID and DATE columns (and more).
I would like to select for every ID the next available record after a certain date. For only one ID the query would be:
SELECT * FROM my_table
WHERE id = 1 AND date >= '01.01.2018'
(just ignoring the to_date() function)
How would that look like for multiple IDs? And I do want to SELECT *.
Thanks!
We can use ROW_NUMBER here:
SELECT ID, date -- and maybe other columns
FROM
(
SELECT *,
ROW_NUMBER() OVER (PARTITION BY ID ORDER BY date) rn
FROM my_table
WHERE date >= date '2018-01-01'
) t
WHERE rn = 1
The idea here is to assign a row number to each ID partition, starting with the earliest date which occurs after the cutoff you specify. The first record from each partition would then be the immediate next date, assuming it exists.
I would like to select queries a table with group by only 2 value in specific field. Like example bellow.
I have a table that normally having 2 value each minute. But some time having more than 2 value in each minute and I have selected by descending order.
Avoid that, I want to field of minute value [TimeStamp] can be group by 2 item based descending order and skip when have more than 2 value.
SELECT TOP 10
dbo.ConRadVacuum.[TimeStamp],
dbo.ConRadVacuum.Tag,
dbo.ConRadVacuum.[Value]
FROM
ConRadVacuum
ORDER BY
[TimeStamp] DESC
I want "selected row" (32 minute) can be selected only two like the other rows
I think you can use row_number() function by partitioning over minute part of your timestamp field, something like this:
select *
from (
select *,
row_number() over (
partition by datepart(minute, cast([TimeStamp] as datetime))
--^^ I use `datepart()` over cast of your timestamp field
--If your field is `datetime` you don't need to `cast`
order by [TimeStamp] desc) seq
from t) tt
where seq <= 2;
SQL Fiddle Demo
I didn't get your question very well, but i think you can put your query in CTE and add ROW_Number to it, then select your cte where your row number field is 1 or 2