How to get substring from column which contains records for filter and group by clause in AWS Redshift database.
I have table with records like:
Table_Id | Categories | Value
<ID> | ABC1; ABC1-1; XYZ | 10
<ID> | ABC1; ABC1-2; XYZ | 15
<ID> | XYZ | 5
.....
Now I want to filter records based on individual category like 'ABC1' or 'ABC1 and XYZ'
Expected output from query would like:
Table_Id | Categories | Value
<ID> | ABC1 | 25
<ID> | ABC1-1 | 10
<ID> | ABC1-2 | 15
<ID> | XYZ | 30
.....
So need to group results based on individual categories.
If you have at most 3 values in any "categories" cell you can unnest the cells, get the list of unique values and use that list in a join condition like this:
WITH
values as (
select distinct category
from (
select distinct split_part(categories,';',1) as category from your_table
union select distinct split_part(categories,';',2) from your_table
union select distinct split_part(categories,';',3) from your_table
)
where nullif(category,'') is not null
)
SELECT
t2.category
,sum(t1.value)
FROM your_table t1
JOIN values t2
ON split_part(categories,';',1)=t2.category
OR split_part(categories,';',2)=t2.category
OR split_part(categories,';',3)=t2.category
if you have more than 3 options just add another split_part level both in WITH part and the join condition
#JonScott, #AlexYes and other pals who struggle with similar kinda situations.
I found more better approach other than suggested by #AlexYes.
What I did, I flatter category column which result individual records.
Which I can further process.
Query:
select row_number() over(order by 1) as r1,
to_char(timestamptz 'epoch' + date_time * interval '1 second', 'yyyy-mm-dd') AS DAY,
split_part(categories, ';', numbers.n) as catg,
value
from <TABLE>
join numbers
on numbers.n <= regexp_count(category_string, ';') + 1 <OTHER_CONDITIONS>
Explanation:
Two functions are useful here: first, the split_part function, which takes a string, splits it on ';' delimiter, and returns the first, second, ... , nth value specified from the split string; second, regexp_count, which tells us how many times a particular pattern is found in our string.
To do this fully dynamically, you need to transpose or pivot values in "categories" column into separate rows.
Unfortunately, a "fully dynamic" solution (without knowing the different values beforehand) is NOT possible using redshift.
Your options are as follows:
Use the method suggested by AlexYes in another answer. This is
semi-dynamic and is probably your best option.
Outside of Redshift, run some ETL code to perform
the column -> multiple rows ETL.
Create a hardcoded type solution, and perform the pivot something like this:
select table_id,'ABC1' as category, case when concat(Categories,';') ilike '%ABC1;%' then value else 0 end as value from your_table
union all
select table_id,'ABC1-1' as category, case when concat(Categories,';')ilike '%ABC1-1;%' then value else 0 end as value from your_table
union all
etc
As the title suggests, I'd like to select the first row of each set of rows grouped with a GROUP BY.
Specifically, if I've got a purchases table that looks like this:
SELECT * FROM purchases;
My Output:
id
customer
total
1
Joe
5
2
Sally
3
3
Joe
2
4
Sally
1
I'd like to query for the id of the largest purchase (total) made by each customer. Something like this:
SELECT FIRST(id), customer, FIRST(total)
FROM purchases
GROUP BY customer
ORDER BY total DESC;
Expected Output:
FIRST(id)
customer
FIRST(total)
1
Joe
5
2
Sally
3
DISTINCT ON is typically simplest and fastest for this in PostgreSQL.
(For performance optimization for certain workloads see below.)
SELECT DISTINCT ON (customer)
id, customer, total
FROM purchases
ORDER BY customer, total DESC, id;
Or shorter (if not as clear) with ordinal numbers of output columns:
SELECT DISTINCT ON (2)
id, customer, total
FROM purchases
ORDER BY 2, 3 DESC, 1;
If total can be null, add NULLS LAST:
...
ORDER BY customer, total DESC NULLS LAST, id;
Works either way, but you'll want to match existing indexes
db<>fiddle here
Major points
DISTINCT ON is a PostgreSQL extension of the standard, where only DISTINCT on the whole SELECT list is defined.
List any number of expressions in the DISTINCT ON clause, the combined row value defines duplicates. The manual:
Obviously, two rows are considered distinct if they differ in at least
one column value. Null values are considered equal in this
comparison.
Bold emphasis mine.
DISTINCT ON can be combined with ORDER BY. Leading expressions in ORDER BY must be in the set of expressions in DISTINCT ON, but you can rearrange order among those freely. Example.
You can add additional expressions to ORDER BY to pick a particular row from each group of peers. Or, as the manual puts it:
The DISTINCT ON expression(s) must match the leftmost ORDER BY
expression(s). The ORDER BY clause will normally contain additional
expression(s) that determine the desired precedence of rows within
each DISTINCT ON group.
I added id as last item to break ties:
"Pick the row with the smallest id from each group sharing the highest total."
To order results in a way that disagrees with the sort order determining the first per group, you can nest above query in an outer query with another ORDER BY. Example.
If total can be null, you most probably want the row with the greatest non-null value. Add NULLS LAST like demonstrated. See:
Sort by column ASC, but NULL values first?
The SELECT list is not constrained by expressions in DISTINCT ON or ORDER BY in any way:
You don't have to include any of the expressions in DISTINCT ON or ORDER BY.
You can include any other expression in the SELECT list. This is instrumental for replacing complex subqueries and aggregate / window functions.
I tested with Postgres versions 8.3 – 15. But the feature has been there at least since version 7.1, so basically always.
Index
The perfect index for the above query would be a multi-column index spanning all three columns in matching sequence and with matching sort order:
CREATE INDEX purchases_3c_idx ON purchases (customer, total DESC, id);
May be too specialized. But use it if read performance for the particular query is crucial. If you have DESC NULLS LAST in the query, use the same in the index so that sort order matches and the index is perfectly applicable.
Effectiveness / Performance optimization
Weigh cost and benefit before creating tailored indexes for each query. The potential of above index largely depends on data distribution.
The index is used because it delivers pre-sorted data. In Postgres 9.2 or later the query can also benefit from an index only scan if the index is smaller than the underlying table. The index has to be scanned in its entirety, though. Example.
For few rows per customer (high cardinality in column customer), this is very efficient. Even more so if you need sorted output anyway. The benefit shrinks with a growing number of rows per customer.
Ideally, you have enough work_mem to process the involved sort step in RAM and not spill to disk. But generally setting work_mem too high can have adverse effects. Consider SET LOCAL for exceptionally big queries. Find how much you need with EXPLAIN ANALYZE. Mention of "Disk:" in the sort step indicates the need for more:
Configuration parameter work_mem in PostgreSQL on Linux
Optimize simple query using ORDER BY date and text
For many rows per customer (low cardinality in column customer), an "index skip scan" or "loose index scan" would be (much) more efficient. But that's not implemented up to Postgres 15. Serious work to implement it one way or another has been ongoing for years now, but so far unsuccessful. See here and here.
For now, there are faster query techniques to substitute for this. In particular if you have a separate table holding unique customers, which is the typical use case. But also if you don't:
SELECT DISTINCT is slower than expected on my table in PostgreSQL
Optimize GROUP BY query to retrieve latest row per user
Optimize groupwise maximum query
Query last N related rows per row
Benchmarks
See separate answer.
On databases that support CTE and windowing functions:
WITH summary AS (
SELECT p.id,
p.customer,
p.total,
ROW_NUMBER() OVER(PARTITION BY p.customer
ORDER BY p.total DESC) AS rank
FROM PURCHASES p)
SELECT *
FROM summary
WHERE rank = 1
Supported by any database:
But you need to add logic to break ties:
SELECT MIN(x.id), -- change to MAX if you want the highest
x.customer,
x.total
FROM PURCHASES x
JOIN (SELECT p.customer,
MAX(total) AS max_total
FROM PURCHASES p
GROUP BY p.customer) y ON y.customer = x.customer
AND y.max_total = x.total
GROUP BY x.customer, x.total
Benchmarks
I tested the most interesting candidates:
Initially with Postgres 9.4 and 9.5.
Added accented tests for Postgres 13 later.
Basic test setup
Main table: purchases:
CREATE TABLE purchases (
id serial -- PK constraint added below
, customer_id int -- REFERENCES customer
, total int -- could be amount of money in Cent
, some_column text -- to make the row bigger, more realistic
);
Dummy data (with some dead tuples), PK, index:
INSERT INTO purchases (customer_id, total, some_column) -- 200k rows
SELECT (random() * 10000)::int AS customer_id -- 10k distinct customers
, (random() * random() * 100000)::int AS total
, 'note: ' || repeat('x', (random()^2 * random() * random() * 500)::int)
FROM generate_series(1,200000) g;
ALTER TABLE purchases ADD CONSTRAINT purchases_id_pkey PRIMARY KEY (id);
DELETE FROM purchases WHERE random() > 0.9; -- some dead rows
INSERT INTO purchases (customer_id, total, some_column)
SELECT (random() * 10000)::int AS customer_id -- 10k customers
, (random() * random() * 100000)::int AS total
, 'note: ' || repeat('x', (random()^2 * random() * random() * 500)::int)
FROM generate_series(1,20000) g; -- add 20k to make it ~ 200k
CREATE INDEX purchases_3c_idx ON purchases (customer_id, total DESC, id);
VACUUM ANALYZE purchases;
customer table - used for optimized query:
CREATE TABLE customer AS
SELECT customer_id, 'customer_' || customer_id AS customer
FROM purchases
GROUP BY 1
ORDER BY 1;
ALTER TABLE customer ADD CONSTRAINT customer_customer_id_pkey PRIMARY KEY (customer_id);
VACUUM ANALYZE customer;
In my second test for 9.5 I used the same setup, but with 100000 distinct customer_id to get few rows per customer_id.
Object sizes for table purchases
Basic setup: 200k rows in purchases, 10k distinct customer_id, avg. 20 rows per customer.
For Postgres 9.5 I added a 2nd test with 86446 distinct customers - avg. 2.3 rows per customer.
Generated with a query taken from here:
Measure the size of a PostgreSQL table row
Gathered for Postgres 9.5:
what | bytes/ct | bytes_pretty | bytes_per_row
-----------------------------------+----------+--------------+---------------
core_relation_size | 20496384 | 20 MB | 102
visibility_map | 0 | 0 bytes | 0
free_space_map | 24576 | 24 kB | 0
table_size_incl_toast | 20529152 | 20 MB | 102
indexes_size | 10977280 | 10 MB | 54
total_size_incl_toast_and_indexes | 31506432 | 30 MB | 157
live_rows_in_text_representation | 13729802 | 13 MB | 68
------------------------------ | | |
row_count | 200045 | |
live_tuples | 200045 | |
dead_tuples | 19955 | |
Queries
1. row_number() in CTE, (see other answer)
WITH cte AS (
SELECT id, customer_id, total
, row_number() OVER (PARTITION BY customer_id ORDER BY total DESC) AS rn
FROM purchases
)
SELECT id, customer_id, total
FROM cte
WHERE rn = 1;
2. row_number() in subquery (my optimization)
SELECT id, customer_id, total
FROM (
SELECT id, customer_id, total
, row_number() OVER (PARTITION BY customer_id ORDER BY total DESC) AS rn
FROM purchases
) sub
WHERE rn = 1;
3. DISTINCT ON (see other answer)
SELECT DISTINCT ON (customer_id)
id, customer_id, total
FROM purchases
ORDER BY customer_id, total DESC, id;
4. rCTE with LATERAL subquery (see here)
WITH RECURSIVE cte AS (
( -- parentheses required
SELECT id, customer_id, total
FROM purchases
ORDER BY customer_id, total DESC
LIMIT 1
)
UNION ALL
SELECT u.*
FROM cte c
, LATERAL (
SELECT id, customer_id, total
FROM purchases
WHERE customer_id > c.customer_id -- lateral reference
ORDER BY customer_id, total DESC
LIMIT 1
) u
)
SELECT id, customer_id, total
FROM cte
ORDER BY customer_id;
5. customer table with LATERAL (see here)
SELECT l.*
FROM customer c
, LATERAL (
SELECT id, customer_id, total
FROM purchases
WHERE customer_id = c.customer_id -- lateral reference
ORDER BY total DESC
LIMIT 1
) l;
6. array_agg() with ORDER BY (see other answer)
SELECT (array_agg(id ORDER BY total DESC))[1] AS id
, customer_id
, max(total) AS total
FROM purchases
GROUP BY customer_id;
Results
Execution time for above queries with EXPLAIN (ANALYZE, TIMING OFF, COSTS OFF, best of 5 runs to compare with warm cache.
All queries used an Index Only Scan on purchases2_3c_idx (among other steps). Some only to benefit from the smaller size of the index, others more effectively.
A. Postgres 9.4 with 200k rows and ~ 20 per customer_id
1. 273.274 ms
2. 194.572 ms
3. 111.067 ms
4. 92.922 ms -- !
5. 37.679 ms -- winner
6. 189.495 ms
B. Same as A. with Postgres 9.5
1. 288.006 ms
2. 223.032 ms
3. 107.074 ms
4. 78.032 ms -- !
5. 33.944 ms -- winner
6. 211.540 ms
C. Same as B., but with ~ 2.3 rows per customer_id
1. 381.573 ms
2. 311.976 ms
3. 124.074 ms -- winner
4. 710.631 ms
5. 311.976 ms
6. 421.679 ms
Retest with Postgres 13 on 2021-08-11
Simplified test setup: no deleted rows, because VACUUM ANALYZE cleans the table completely for the simple case.
Important changes for Postgres:
General performance improvements.
CTEs can be inlined since Postgres 12, so query 1. and 2. now perform mostly identical (same query plan).
D. Like B. ~ 20 rows per customer_id
1. 103 ms
2. 103 ms
3. 23 ms -- winner
4. 71 ms
5. 22 ms -- winner
6. 81 ms
db<>fiddle here
E. Like C. ~ 2.3 rows per customer_id
1. 127 ms
2. 126 ms
3. 36 ms -- winner
4. 620 ms
5. 145 ms
6. 203 ms
db<>fiddle here
Accented tests with Postgres 13
1M rows, 10.000 vs. 100 vs. 1.6 rows per customer.
F. with ~ 10.000 rows per customer
1. 526 ms
2. 527 ms
3. 127 ms
4. 2 ms -- winner !
5. 1 ms -- winner !
6. 356 ms
db<>fiddle here
G. with ~ 100 rows per customer
1. 535 ms
2. 529 ms
3. 132 ms
4. 108 ms -- !
5. 71 ms -- winner
6. 376 ms
db<>fiddle here
H. with ~ 1.6 rows per customer
1. 691 ms
2. 684 ms
3. 234 ms -- winner
4. 4669 ms
5. 1089 ms
6. 1264 ms
db<>fiddle here
Conclusions
DISTINCT ON uses the index effectively and typically performs best for few rows per group. And it performs decently even with many rows per group.
For many rows per group, emulating an index skip scan with an rCTE performs best - second only to the query technique with a separate lookup table (if that's available).
The row_number() technique demonstrated in the currently accepted answer never wins any performance test. Not then, not now. It never comes even close to DISTINCT ON, not even when the data distribution is unfavorable for the latter. The only good thing about row_number(): it does not scale terribly, just mediocre.
More benchmarks
Benchmark by "ogr" with 10M rows and 60k unique "customers" on Postgres 11.5. Results are in line with what we have seen so far:
Proper way to access latest row for each individual identifier?
Original (outdated) benchmark from 2011
I ran three tests with PostgreSQL 9.1 on a real life table of 65579 rows and single-column btree indexes on each of the three columns involved and took the best execution time of 5 runs.
Comparing #OMGPonies' first query (A) to the above DISTINCT ON solution (B):
Select the whole table, results in 5958 rows in this case.
A: 567.218 ms
B: 386.673 ms
Use condition WHERE customer BETWEEN x AND y resulting in 1000 rows.
A: 249.136 ms
B: 55.111 ms
Select a single customer with WHERE customer = x.
A: 0.143 ms
B: 0.072 ms
Same test repeated with the index described in the other answer:
CREATE INDEX purchases_3c_idx ON purchases (customer, total DESC, id);
1A: 277.953 ms
1B: 193.547 ms
2A: 249.796 ms -- special index not used
2B: 28.679 ms
3A: 0.120 ms
3B: 0.048 ms
This is common greatest-n-per-group problem, which already has well tested and highly optimized solutions. Personally I prefer the left join solution by Bill Karwin (the original post with lots of other solutions).
Note that bunch of solutions to this common problem can surprisingly be found in the MySQL manual -- even though your problem is in Postgres, not MySQL, the solutions given should work with most SQL variants. See Examples of Common Queries :: The Rows Holding the Group-wise Maximum of a Certain Column.
In Postgres you can use array_agg like this:
SELECT customer,
(array_agg(id ORDER BY total DESC))[1],
max(total)
FROM purchases
GROUP BY customer
This will give you the id of each customer's largest purchase.
Some things to note:
array_agg is an aggregate function, so it works with GROUP BY.
array_agg lets you specify an ordering scoped to just itself, so it doesn't constrain the structure of the whole query. There is also syntax for how you sort NULLs, if you need to do something different from the default.
Once we build the array, we take the first element. (Postgres arrays are 1-indexed, not 0-indexed).
You could use array_agg in a similar way for your third output column, but max(total) is simpler.
Unlike DISTINCT ON, using array_agg lets you keep your GROUP BY, in case you want that for other reasons.
The Query:
SELECT purchases.*
FROM purchases
LEFT JOIN purchases as p
ON
p.customer = purchases.customer
AND
purchases.total < p.total
WHERE p.total IS NULL
HOW DOES THAT WORK! (I've been there)
We want to make sure that we only have the highest total for each purchase.
Some Theoretical Stuff (skip this part if you only want to understand the query)
Let Total be a function T(customer,id) where it returns a value given the name and id
To prove that the given total (T(customer,id)) is the highest we have to prove that
We want to prove either
∀x T(customer,id) > T(customer,x) (this total is higher than all other
total for that customer)
OR
¬∃x T(customer, id) < T(customer, x) (there exists no higher total for
that customer)
The first approach will need us to get all the records for that name which I do not really like.
The second one will need a smart way to say there can be no record higher than this one.
Back to SQL
If we left joins the table on the name and total being less than the joined table:
LEFT JOIN purchases as p
ON
p.customer = purchases.customer
AND
purchases.total < p.total
we make sure that all records that have another record with the higher total for the same user to be joined:
+--------------+---------------------+-----------------+------+------------+---------+
| purchases.id | purchases.customer | purchases.total | p.id | p.customer | p.total |
+--------------+---------------------+-----------------+------+------------+---------+
| 1 | Tom | 200 | 2 | Tom | 300 |
| 2 | Tom | 300 | | | |
| 3 | Bob | 400 | 4 | Bob | 500 |
| 4 | Bob | 500 | | | |
| 5 | Alice | 600 | 6 | Alice | 700 |
| 6 | Alice | 700 | | | |
+--------------+---------------------+-----------------+------+------------+---------+
That will help us filter for the highest total for each purchase with no grouping needed:
WHERE p.total IS NULL
+--------------+----------------+-----------------+------+--------+---------+
| purchases.id | purchases.name | purchases.total | p.id | p.name | p.total |
+--------------+----------------+-----------------+------+--------+---------+
| 2 | Tom | 300 | | | |
| 4 | Bob | 500 | | | |
| 6 | Alice | 700 | | | |
+--------------+----------------+-----------------+------+--------+---------+
And that's the answer we need.
The solution is not very efficient as pointed by Erwin, because of presence of SubQs
select * from purchases p1 where total in
(select max(total) from purchases where p1.customer=customer) order by total desc;
I use this way (postgresql only): https://wiki.postgresql.org/wiki/First/last_%28aggregate%29
-- Create a function that always returns the first non-NULL item
CREATE OR REPLACE FUNCTION public.first_agg ( anyelement, anyelement )
RETURNS anyelement LANGUAGE sql IMMUTABLE STRICT AS $$
SELECT $1;
$$;
-- And then wrap an aggregate around it
CREATE AGGREGATE public.first (
sfunc = public.first_agg,
basetype = anyelement,
stype = anyelement
);
-- Create a function that always returns the last non-NULL item
CREATE OR REPLACE FUNCTION public.last_agg ( anyelement, anyelement )
RETURNS anyelement LANGUAGE sql IMMUTABLE STRICT AS $$
SELECT $2;
$$;
-- And then wrap an aggregate around it
CREATE AGGREGATE public.last (
sfunc = public.last_agg,
basetype = anyelement,
stype = anyelement
);
Then your example should work almost as is:
SELECT FIRST(id), customer, FIRST(total)
FROM purchases
GROUP BY customer
ORDER BY FIRST(total) DESC;
CAVEAT: It ignore's NULL rows
Edit 1 - Use the postgres extension instead
Now I use this way: http://pgxn.org/dist/first_last_agg/
To install on ubuntu 14.04:
apt-get install postgresql-server-dev-9.3 git build-essential -y
git clone git://github.com/wulczer/first_last_agg.git
cd first_last_app
make && sudo make install
psql -c 'create extension first_last_agg'
It's a postgres extension that gives you first and last functions; apparently faster than the above way.
Edit 2 - Ordering and filtering
If you use aggregate functions (like these), you can order the results, without the need to have the data already ordered:
http://www.postgresql.org/docs/current/static/sql-expressions.html#SYNTAX-AGGREGATES
So the equivalent example, with ordering would be something like:
SELECT first(id order by id), customer, first(total order by id)
FROM purchases
GROUP BY customer
ORDER BY first(total);
Of course you can order and filter as you deem fit within the aggregate; it's very powerful syntax.
Use ARRAY_AGG function for PostgreSQL, U-SQL, IBM DB2, and Google BigQuery SQL:
SELECT customer, (ARRAY_AGG(id ORDER BY total DESC))[1], MAX(total)
FROM purchases
GROUP BY customer
In SQL Server you can do this:
SELECT *
FROM (
SELECT ROW_NUMBER()
OVER(PARTITION BY customer
ORDER BY total DESC) AS StRank, *
FROM Purchases) n
WHERE StRank = 1
Explaination:Here Group by is done on the basis of customer and then order it by total then each such group is given serial number as StRank and we are taking out first 1 customer whose StRank is 1
Very fast solution
SELECT a.*
FROM
purchases a
JOIN (
SELECT customer, min( id ) as id
FROM purchases
GROUP BY customer
) b USING ( id );
and really very fast if table is indexed by id:
create index purchases_id on purchases (id);
Snowflake/Teradata supports QUALIFY clause which works like HAVING for windowed functions:
SELECT id, customer, total
FROM PURCHASES
QUALIFY ROW_NUMBER() OVER(PARTITION BY p.customer ORDER BY p.total DESC) = 1
In PostgreSQL, another possibility is to use the first_value window function in combination with SELECT DISTINCT:
select distinct customer_id,
first_value(row(id, total)) over(partition by customer_id order by total desc, id)
from purchases;
I created a composite (id, total), so both values are returned by the same aggregate. You can of course always apply first_value() twice.
This way it work for me:
SELECT article, dealer, price
FROM shop s1
WHERE price=(SELECT MAX(s2.price)
FROM shop s2
WHERE s1.article = s2.article
GROUP BY s2.article)
ORDER BY article;
Select highest price on each article
This is how we can achieve this by using windows function:
create table purchases (id int4, customer varchar(10), total integer);
insert into purchases values (1, 'Joe', 5);
insert into purchases values (2, 'Sally', 3);
insert into purchases values (3, 'Joe', 2);
insert into purchases values (4, 'Sally', 1);
select ID, CUSTOMER, TOTAL from (
select ID, CUSTOMER, TOTAL,
row_number () over (partition by CUSTOMER order by TOTAL desc) RN
from purchases) A where RN = 1;
The accepted OMG Ponies' "Supported by any database" solution has good speed from my test.
Here I provide a same-approach, but more complete and clean any-database solution. Ties are considered (assume desire to get only one row for each customer, even multiple records for max total per customer), and other purchase fields (e.g. purchase_payment_id) will be selected for the real matching rows in the purchase table.
Supported by any database:
select * from purchase
join (
select min(id) as id from purchase
join (
select customer, max(total) as total from purchase
group by customer
) t1 using (customer, total)
group by customer
) t2 using (id)
order by customer
This query is reasonably fast especially when there is a composite index like (customer, total) on the purchase table.
Remark:
t1, t2 are subquery alias which could be removed depending on database.
Caveat: the using (...) clause is currently not supported in MS-SQL and Oracle db as of this edit on Jan 2017. You have to expand it yourself to e.g. on t2.id = purchase.id etc. The USING syntax works in SQLite, MySQL and PostgreSQL.
If you want to select any (by your some specific condition) row from the set of aggregated rows.
If you want to use another (sum/avg) aggregation function in addition to max/min. Thus you can not use clue with DISTINCT ON
You can use next subquery:
SELECT
(
SELECT **id** FROM t2
WHERE id = ANY ( ARRAY_AGG( tf.id ) ) AND amount = MAX( tf.amount )
) id,
name,
MAX(amount) ma,
SUM( ratio )
FROM t2 tf
GROUP BY name
You can replace amount = MAX( tf.amount ) with any condition you want with one restriction: This subquery must not return more than one row
But if you wanna to do such things you probably looking for window functions
For SQl Server the most efficient way is:
with
ids as ( --condition for split table into groups
select i from (values (9),(12),(17),(18),(19),(20),(22),(21),(23),(10)) as v(i)
)
,src as (
select * from yourTable where <condition> --use this as filter for other conditions
)
,joined as (
select tops.* from ids
cross apply --it`s like for each rows
(
select top(1) *
from src
where CommodityId = ids.i
) as tops
)
select * from joined
and don't forget to create clustered index for used columns
This can be achieved easily by MAX FUNCTION on total and GROUP BY id and customer.
SELECT id, customer, MAX(total) FROM purchases GROUP BY id, customer
ORDER BY total DESC;
My approach via window function dbfiddle:
Assign row_number at each group: row_number() over (partition by agreement_id, order_id ) as nrow
Take only first row at group: filter (where nrow = 1)
with intermediate as (select
*,
row_number() over ( partition by agreement_id, order_id ) as nrow,
(sum( suma ) over ( partition by agreement_id, order_id ))::numeric( 10, 2) as order_suma,
from <your table>)
select
*,
sum( order_suma ) filter (where nrow = 1) over (partition by agreement_id)
from intermediate
This question already has answers here:
How can I combine multiple rows into a comma-delimited list in Oracle? [duplicate]
(11 answers)
Closed 8 years ago.
I have data that looks like
CUSTOMER, CUSTOMER_ID, PRODUCT
ABC INC 1 XYX
ABC INC 1 ZZZ
DEF CO 2 XYX
DEF CO 2 ZZZ
DEF CO 2 WWW
GHI LLC 3 ZYX
I'd like to write a query that'd make the data look like this:
CUSTOMER, CUSTOMER_ID, PRODUCTS
ABC INC 1 XYX, ZZZ
DEF CO 2 XYX, ZZZ, WWW
GHI LLC 3 ZYX
Using Oracle 10g if helps. I saw something that would work using MYSQL, but I need a plain SQL or ORACLE equivalent. I've also seen examples of stored procs that could be made, however, I cannot use a stored proc with the product i'm using.
Here's how'd it work in MySQL if I were using it
SELECT CUSTOMER,
CUSTOMER_ID,
GROUP_CONCAT( PRODUCT )
FROM MAGIC_TABLE
GROUP BY CUSTOMER, CUSTOMER_ID
Thank you.
I think LISTAGG is the best aggregate group by function to use in this situation:
SELECT CUSTOMER, CUSTOMER_ID,
LISTAGG(PRODUCT, ', ') WITHIN GROUP (ORDER BY PRODUCT)
FROM SOME_TABLE
GROUP BY CUSTOMER, CUSTOMER_ID
ORDER BY 1, 2
This link refers to a number of examples of different ways to do this on Oracle. See if there's something there that you have permissions on your database to do.
The oracle user function 'wm_concat' works the same way as LISTAGG except you cannot specify a delimiter ',' by default or a sort order. It is however compatible with 10g.
Thanks Nigel,
My SQL is not as elegant as could be, but I needed a solution that required SQL only, not PLSQL or TSQL, so it ended up looking like this:
SELECT CUSTOMER, CUSTOMER_ID, COUNT(PRODUCT) PROD_COUNT,
RTRIM(
XMLAGG( XMLELEMENT (C, PRODUCT || ',') ORDER BY PRODUCT
).EXTRACT ('//text()'), ','
) AS PRODUCTS FROM (
SELECT DISTINCT CUSTOMER, CUSTOMER_ID, PRODUCT
FROM MAGIC_TABLE
) GROUP BY CUSTOMER, CUSTOMER_ID ORDER BY 1 , 2
Still not exactly sure what the XML functions do exactly, but I'll dig in when the need arrises.