I have a table like this
and i want my output to look like this
I need to look at the ID and then take max created date and max completed date for that ID. There is also some cases where completed date is still empty so in that case i just need to look at the max created date. Im not sure how to tackle this, doing a group by doesnt account for my multiple scenarios
Use ROW_NUMBER:
WITH cte AS (
SELECT *, ROW_NUMBER() OVER (PARTITION BY QUOTE_NUMBER
ORDER BY WORKBOOK_CREATED_DATE DESC) rn
FROM yourTable
)
SELECT *
FROM yourTable
WHERE rn = 1;
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
How to retrieve data from the previous entry/row on a select statement?
Here is some data:
I want to retrieve facID, read_date, and previous date for each row based on facID.
I think using window function can help here like below code:
select factory_id,read_date from (
select factory_id,read_date,
rank() over (partition by factory_id order by read_date) as RN
from Table order by factory_id,read_date
) a
But not sure how to write a code to retrieve the previous date.
Thank you in advance
Use lag:
select
factory_id,
read_date,
lag(read_date) over (partition by factory_id order by read_date) as last_read_date
from Table
order by factory_id,read_date
In SQL Server, I am attempting to pull the second latest NOTE_ENTRY_DT_TIME (items highlighted in screenshot). With the query written below it still pulls the latest date (I believe it's because of the grouping but the grouping is required to join later). What is the best method to achieve this?
SELECT
hop.ACCOUNT_ID,
MAX(hop.NOTE_ENTRY_DT_TIME) AS latest_noteid
FROM
NOTES hop
WHERE
hop.GEN_YN IS NULL
AND hop.NOTE_ENTRY_DT_TIME < (SELECT MAX(hope.NOTE_ENTRY_DT_TIME)
FROM NOTES hope
WHERE hop.GEN_YN IS NULL)
GROUP BY
hop.ACCOUNT_ID
Data sample in the table:
One of the "easier" ways to get the Nth row in a group is to use a CTE and ROW_NUMBER:
WITH CTE AS(
SELECT Account_ID,
Note_Entry_Dt_Time,
ROW_NUMBER() OVER (PARTITION BY AccountID ORDER BY Note_Entry_Dt_Time DESC) AS RN
FROM dbo.YourTable)
SELECT Account_ID,
Note_Entry_Dt_Time
FROM CTE
WHERE RN = 2;
Of course, if an ACCOUNT_ID only has 1 row, then it will not be returned in the result set.
The OP's statement "The row will not always be 2." from the comments conflicts with their statement "I am attempting to pull the second latest NOTE_ENTRY_DT_TIME" in the question. At a best guess, this means that the OP has rows with the same date, that could be the "latest" date. If so, then would simply need to replace ROW_NUMBER with DENSE_RANK. Their sampple data, however, doesn't suggest this is the case.
You can use window functions:
select *
from (
select
n.*,
row_number() over(partition by account_id order by note_entry_dt_time desc) rn
from notes n
) t
where rn = 2
Situation:
I have three columns:
id
date
tx_id
The primary id column is tx_id and is unique in the table. Each tx_id is tied to an id and it has a record date. I would like to test whether or not the tx_id is incremental.
Objective:
I need to extract the first tx_id by id but I want to prevent using ROW_NUMBER
i.e
select id, date, tx_id, row_number() over(partition by id order by date asc) as First_transaction_id from table
and simply use
select id, date, MIN(tx_id) as First_transaction_id from table
So how can i make sure since i have more than 50 millions of ids that by using MINtx_id will yield the earliest transaction for each id?
How can i add a flag column to segment those that don't satisfy the condition?
how can i make sure since i have more than 50 millions of ids that by using MINtx_id will yield the earliest transaction for each id?
Simply do the comparison:
You can get the exceptions with logic like this:
select t.*
from (select t.*,
min(tx_id) over (partition by id) as min_tx_id,
rank() over (partition by id order by date) as seqnum
from t
) t
where tx_id = min_tx_id and seqnum > 1;
Note: this uses rank(). It seems possible that there could be two transactions for an id on the same date.
use corelated sunquery
select t.* from table_name t
where t.date= ( select min(date) from table_name
t1 where t1.id=t.id)
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