BigQuery - Extract last entry of each group - sql

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

Related

SQL Max or empty value grouped by conditions

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;

Find the second largest value with Groupings

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

Get minimum without using row number/window function in Bigquery

I have a table like as shown below
What I would like to do is get the minimum of each subject. Though I am able to do this with row_number function, I would like to do this with groupby and min() approach. But it doesn't work.
row_number approach - works fine
SELECT * FROM (select subject_id,value,id,min_time,max_time,time_1,
row_number() OVER (PARTITION BY subject_id ORDER BY value) AS rank
from table A) WHERE RANK = 1
min() approach - doesn't work
select subject_id,id,min_time,max_time,time_1,min(value) from table A
GROUP BY SUBJECT_ID,id
As you can see just the two columns (subject_id and id) is enough to group the items together. They will help differentiate the group. But why am I not able to use the other columns in select clause. If I use the other columns, I may not get the expected output because time_1 has different values.
I expect my output to be like as shown below
In BigQuery you can use aggregation for this:
SELECT ARRAY_AGG(a ORDER BY value LIMIT 1)[SAFE_OFFSET(1)].*
FROM table A
GROUP BY SUBJECT_ID;
This uses ARRAY_AGG() to aggregate each record (the a in the argument list). ARRAY_AGG() allows you to order the result (by value) and to limit the size of the array. The latter is important for performance.
After you concatenate the arrays, you want the first element. The .* transforms the record referred to by a to the component columns.
I'm not sure why you don't want to use ROW_NUMBER(). If the problem is the lingering rank column, you an easily remove it:
SELECT a.* EXCEPT (rank)
FROM (SELECT a.*,
ROW_NUMBER() OVER (PARTITION BY subject_id ORDER BY value) AS rank
FROM A
) a
WHERE RANK = 1;
Are you looking for something like below-
SELECT
A.subject_id,
A.id,
A.min_time,
A.max_time,
A.time_1,
A.value
FROM table A
INNER JOIN(
SELECT subject_id, MIN(value) Value
FROM table
GROUP BY subject_id
) B ON A.subject_id = B.subject_id
AND A.Value = B.Value
If you do not required to select Time_1 column's value, this following query will work (As I can see values in column min_time and max_time is same for the same group)-
SELECT
A.subject_id,A.id,A.min_time,A.max_time,
--A.time_1,
MIN(A.value)
FROM table A
GROUP BY
A.subject_id,A.id,A.min_time,A.max_time
Finally, the best approach is if you can apply something like CAST(Time_1 AS DATE) on your time column. This will consider only the date part regardless of the time part. The query will be
SELECT
A.subject_id,A.id,A.min_time,A.max_time,
CAST(A.time_1 AS DATE) Time_1,
MIN(A.value)
FROM table A
GROUP BY
A.subject_id,A.id,A.min_time,A.max_time,
CAST(A.time_1 AS DATE)
-- Make sure the syntax of CAST AS DATE
-- in BigQuery is as I written here or bit different.
Below is for BigQuery Standard SQL and is most efficient way for such cases like in your question
#standardSQL
SELECT AS VALUE ARRAY_AGG(t ORDER BY value LIMIT 1)[OFFSET(0)]
FROM `project.dataset.table` t
GROUP BY subject_id
Using ROW_NUMBER is not efficient and in many cases lead to Resources exceeded error.
Note: self join is also very ineffective way of achieving your objective
A bit late to the party, but here is a cte-based approach which made sense to me:
with mins as (
select subject_id, id, min(value) as min_value
from table
group by subject_id, id
)
select distinct t.subject_id, t.id, t.time_1, t.min_time, t.max_time, m.min_value
from table t
join mins m on m.subject_id = t.subject_id and m.id = t.id

how to get latest date column records when result should be filtered with unique column name in sql?

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

How do I use ROW_NUMBER()?

I want to use the ROW_NUMBER() to get...
To get the max(ROW_NUMBER()) --> Or i guess this would also be the count of all rows
I tried doing:
SELECT max(ROW_NUMBER() OVER(ORDER BY UserId)) FROM Users
but it didn't seem to work...
To get ROW_NUMBER() using a given piece of information, ie. if I have a name and I want to know what row the name came from.
I assume it would be something similar to what I tried for #1
SELECT ROW_NUMBER() OVER(ORDER BY UserId) From Users WHERE UserName='Joe'
but this didn't work either...
Any Ideas?
For the first question, why not just use?
SELECT COUNT(*) FROM myTable
to get the count.
And for the second question, the primary key of the row is what should be used to identify a particular row. Don't try and use the row number for that.
If you returned Row_Number() in your main query,
SELECT ROW_NUMBER() OVER (Order by Id) AS RowNumber, Field1, Field2, Field3
FROM User
Then when you want to go 5 rows back then you can take the current row number and use the following query to determine the row with currentrow -5
SELECT us.Id
FROM (SELECT ROW_NUMBER() OVER (ORDER BY id) AS Row, Id
FROM User ) us
WHERE Row = CurrentRow - 5
Though I agree with others that you could use count() to get the total number of rows, here is how you can use the row_count():
To get the total no of rows:
with temp as (
select row_number() over (order by id) as rownum
from table_name
)
select max(rownum) from temp
To get the row numbers where name is Matt:
with temp as (
select name, row_number() over (order by id) as rownum
from table_name
)
select rownum from temp where name like 'Matt'
You can further use min(rownum) or max(rownum) to get the first or last row for Matt respectively.
These were very simple implementations of row_number(). You can use it for more complex grouping. Check out my response on Advanced grouping without using a sub query
If you need to return the table's total row count, you can use an alternative way to the SELECT COUNT(*) statement.
Because SELECT COUNT(*) makes a full table scan to return the row count, it can take very long time for a large table. You can use the sysindexes system table instead in this case. There is a ROWS column that contains the total row count for each table in your database. You can use the following select statement:
SELECT rows FROM sysindexes WHERE id = OBJECT_ID('table_name') AND indid < 2
This will drastically reduce the time your query takes.
You can use this for get first record where has clause
SELECT TOP(1) * , ROW_NUMBER() OVER(ORDER BY UserId) AS rownum
FROM Users
WHERE UserName = 'Joe'
ORDER BY rownum ASC
ROW_NUMBER() returns a unique number for each row starting with 1. You can easily use this by simply writing:
ROW_NUMBER() OVER (ORDER BY 'Column_Name' DESC) as ROW_NUMBER
May not be related to the question here. But I found it could be useful when using ROW_NUMBER -
SELECT *,
ROW_NUMBER() OVER (ORDER BY (SELECT 100)) AS Any_ID
FROM #Any_Table
select
Ml.Hid,
ml.blockid,
row_number() over (partition by ml.blockid order by Ml.Hid desc) as rownumber,
H.HNAME
from MIT_LeadBechmarkHamletwise ML
join [MT.HAMLE] h on ML.Hid=h.HID
SELECT num, UserName FROM
(SELECT UserName, ROW_NUMBER() OVER(ORDER BY UserId) AS num
From Users) AS numbered
WHERE UserName='Joe'
You can use Row_Number for limit query result.
Example:
SELECT * FROM (
select row_number() OVER (order by createtime desc) AS ROWINDEX,*
from TABLENAME ) TB
WHERE TB.ROWINDEX between 0 and 10
--
With above query, I will get PAGE 1 of results from TABLENAME.
If you absolutely want to use ROW_NUMBER for this (instead of count(*)) you can always use:
SELECT TOP 1 ROW_NUMBER() OVER (ORDER BY Id)
FROM USERS
ORDER BY ROW_NUMBER() OVER (ORDER BY Id) DESC
Need to create virtual table by using WITH table AS, which is mention in given Query.
By using this virtual table, you can perform CRUD operation w.r.t row_number.
QUERY:
WITH table AS
-
(SELECT row_number() OVER(ORDER BY UserId) rn, * FROM Users)
-
SELECT * FROM table WHERE UserName='Joe'
-
You can use INSERT, UPDATE or DELETE in last sentence by in spite of SELECT.
SQL Row_Number() function is to sort and assign an order number to data rows in related record set. So it is used to number rows, for example to identify the top 10 rows which have the highest order amount or identify the order of each customer which is the highest amount, etc.
If you want to sort the dataset and number each row by seperating them into categories we use Row_Number() with Partition By clause. For example, sorting orders of each customer within itself where the dataset contains all orders, etc.
SELECT
SalesOrderNumber,
CustomerId,
SubTotal,
ROW_NUMBER() OVER (PARTITION BY CustomerId ORDER BY SubTotal DESC) rn
FROM Sales.SalesOrderHeader
But as I understand you want to calculate the number of rows of grouped by a column. To visualize the requirement, if you want to see the count of all orders of the related customer as a seperate column besides order info, you can use COUNT() aggregation function with Partition By clause
For example,
SELECT
SalesOrderNumber,
CustomerId,
COUNT(*) OVER (PARTITION BY CustomerId) CustomerOrderCount
FROM Sales.SalesOrderHeader
This query:
SELECT ROW_NUMBER() OVER(ORDER BY UserId) From Users WHERE UserName='Joe'
will return all rows where the UserName is 'Joe' UNLESS you have no UserName='Joe'
They will be listed in order of UserID and the row_number field will start with 1 and increment however many rows contain UserName='Joe'
If it does not work for you then your WHERE command has an issue OR there is no UserID in the table. Check spelling for both fields UserID and UserName.