I want to group by the foolowing table but I also need to group the column ID as depitected on the image.
SELECT SUM(ml),sku,name FROM consumos
GROUP BY sku,name
ORDER BY name
Any ideas?
Best regards
Looks like you want a JSON array. This can be done using jsonb_agg():
select name, sku, ml, jsonb_agg(id)
from the_table
group by name, sku, ml;
In standard SQL, this would look like:
select name, sku, ml,
'[' || listagg(id, ',') within group (order by id) || ']' as ids
from t
group by name, sku, ml
order by name, count(*);
However, not all databases support that standard operator || for string concatenation. And not all databases call their string aggregation operator listagg(). So you might need to tweak the query for your database.
EDIT:
In Postgres, this would be:
select name, sku, ml,
'[' || string_agg(id, ',' order by id) || ']' as ids
from t
group by name, sku, ml
order by name, count(*);
Related
I want to generate table and add all values per distinct id to one row using BigQuery
Example:
id label
000756f4-1af2-439b-b607-ce7384a6b8ee fast
000756f4-1af2-439b-b607-ce7384a6b8ee streaming
000756f4-1af2-439b-b607-ce7384a6b8ee other
0007bac4-1bed-4bf0-8b55-d21216723ef5 issue
000a03d2-f88c-4150-aa96-40b9fdaccb17 fast
000a03d2-f88c-4150-aa96-40b9fdaccb17 other
I would like to receive such table:
id label
000756f4-1af2-439b-b607-ce7384a6b8ee fast, streaming, other
0007bac4-1bed-4bf0-8b55-d21216723ef5 issue
000a03d2-f88c-4150-aa96-40b9fdaccb17 fast, other
Is it possible to achieve it with BigQuery?
You can just use string_agg():
select id, string_agg(label, ', ') as labels
from t
group by id;
Note that the ordering is arbitrary (and might even vary from one run to another). You might want to include an order by as well:
select id, string_agg(label, ', ' order by label) as labels
from t
group by id;
Update
Use string_agg:
select id, string_agg(label, ', ')
from mytable
group by id
Original answer
Use array_agg and array_to_string:
select id, array_to_string(array_agg(label), ', ')
from mytable
group by id
I would like to ask about a script in BigQuery. So, I tried to use a query below
SELECT id, STRING_AGG(DISTINCT status, ', ' ORDER BY timestamp) AS grouping
FROM table
GROUP BY id
But I couldn't run it since it gave me an error
An aggregate function that has both DISTINCT and ORDER BY arguments can only ORDER BY expressions that are arguments to the function
Could anyone help me to fix the error? Thank you in advance!
Do you want the distinct statuses ordered by timestamp?
If so, you can first order for each id the column status by timestamp, then aggregate.
WITH ordered as (
SELECT id, status
FROM table
ORDER BY id, row_number() over (partition by id ORDER BY timestamp)
)
SELECT id, STRING_AGG(DISTINCT status, ', ') AS grouping
FROM ordered
GROUP BY id
I have 3 columns in my table
Products (product_id, product_detail_name, product_detail_value)
For each product, I want all its details to be in a single cell, i.e in the below format:
product_id, concatenated_values
Use string aggregation function listagg():
select product_id, listagg(product_detail_value) product_details
from mytable
group by product_id
Or if you want to see the results as comma-separated 'name: value' pairs, then:
select
product_id,
listagg(product_detail_name || ': ' || product_detail_value) product_details
from mytable
group by product_id
listagg() also supports an order by clause (with the within group syntax), that is described in the documentation.
You can also use the second section of LISTAGG to define the delimiter.
Select product_id,
Listagg(product_detail_name || ':' || product_detail_value, ' ,') p_details
from mytable
Group By product_id;
So I am trying to use a LISTAGG function while using a WHERE statement.
What I want to do is search by the WHERE statement and the LISTAGG returns a list with in relations to other columns.
In other words, when I use the WHERE statement with LISTAGG, I only get the value that I'm searching for. The other values associated with the other columns don't show up.
My script is kind of like this:
WITH TEST AS
(
SELECT DISTINCT
LOCATION,
ID,
LISTAGG (TOMATOCOUNT, ', ')
WITHIN GROUP (ORDER BY TOMATOCOUNT)
OVER (PARTITION BY LOCATION, ID) TOMATOTYPES,
FROM TOMATOLAND
)
SELECT
*
FROM TEST
WHERE (:TOMATOTYPE = TOMATO
OR :TOMATOTYPE IS NULL)
Filter on the result of LISTAGG strikes me as a bit backwords. Generally, if you're going to aggregate several items into a list, you're doing it at the last minute, for display purposes. ideally you would filter prior to the aggregation.
The following will return one row for each location/ID that contains the specified tomatotype.
SELECT DISTINCT
location,
id,
LISTAGG (tomatotype, ', ')
WITHIN GROUP (ORDER BY tomatocount)
OVER (PARTITION BY location, id)
tomatotypes
FROM tomatoland
WHERE (location, id) IN (SELECT location, id
FROM tomatoland
WHERE :tomatotype = tomatotype)
OR :tomatotype IS NULL
I think you want something like this:
WITH TEST AS (
SELECT LOCATION, ID,
LISTAGG(TOMATOTYPE, ', ') WITHIN GROUP (ORDER BY TOMATOCOUNT) OVER (PARTITION BY LOCATION, ID) as TOMATOTYPES
FROM TOMATOLAND
)
SELECT *
FROM TEST
WHERE ', ' || :TOMATOTYPE || ', ' LIKE '%,' || TOMATOTYPES || ', %' OR
:TOMATOTYPE IS NULL
I need make a query that I get the result and put in one line separated per comma.
For example, I have this query:
SELECT
SIGLA
FROM
LANGUAGES
This query return the result below:
SIGLA
ESP
EN
BRA
I need to get this result in one single line that way:
SIGLA
ESP,EN,BRA
Can anyone help me?
Thank you!
SELECT LISTAGG(SIGLA, ', ') WITHIN GROUP (ORDER BY SIGLA) " As "S_List" FROM LANGUAGES
Should be the listagg sequence you are needing
try
SELECT LISTAGG( SIGLA, ',' ) within group (order by SIGLA) as NewSigla FROM LANGUAGES
If you want to get the values grouped together in the order that Oracle produces the rows then:
SELECT LISTAGG( SIGLA, ',' ) WITHIN GROUP ( ORDER BY ROWNUM ) AS SIGLA
FROM LANGUAGES;
If you want alphabetical ordering then replace ORDER BY ROWNUM with ORDER BY SIGLA (or, curiously, ORDER BY NULL).