Questions tagged [window-functions]

A function that partitions a result set and computes something within that partition without rolling up the data in that partition.

A window function partitions a row set into blocks based on some grouping but does not roll up into that grouping. A local calculation can be done within that group that will reset when the system processes the next group.

The SQL syntax for a window function takes the form:

calculation()
   over (partition by foo,
         order by bar)

The order by clause is optional, but sorts rows within a partition. This can be useful for calculations that should reset across partitions. For example, one could use the windowing function to apply a row number for sorting within the partition or to calculate a running sum within that partition based a date ordering.

The partition by clause in the window function differs from the group by clause in SQL in that it does not actually aggregate the data in the partitions.

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What is the difference between `ORDER BY` and `PARTITION BY` arguments in the `OVER` clause?

These queries below both give me exactly the same results, which I assume is because of my dataset rather than how the arguments work. When using an OVER clause, what is the difference between ORDER BY and PARTITION BY. On a slightly different note,…
Zach Smith
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How can I use LEAD LAG-like functions on multiple columns simultaneously

Suppose I have the following table: start_data | end_date ---------------------- t1 | Null Null | t2 t3 | Null Null | t4 t5 | Null The output should be like this: start_data | end_date ---------------------- t1…
Fadi Bakoura
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SQL Query on partitioning does not run

When I run the following SQL command select payee_id, start_dt, row_number() over(partition by payee_id order by start_dt) as rn from xxx where rn = 1 I got an error : error Invalid identifier 'rn' Please advise.
Ahmed
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Filtering SQL Search Query

How can I filter results in a SQL query? Let’s say I have Products A-Z, and each product has a monthly observation for X Months, but for most products the value in the observation stays the same for that particular product, but I need to find the…
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Display Difference of Date (days) Between Last and Last-1 Issue Of Each KEY1

Table Definition CREATE TABLE TEST ( KEY1 VARCHAR NULL Dt DATE NULL, ISSUE INTEGER NULL, ); Example Data INSERT INTO TEST VALUES (100, '9/01/2010', 5), (100, '14/01/2010', 10), (100, '13/01/2010', 9), …
Arunava
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