how to use concat, like and subconsultas with where? - google-chrome

Question

SELECT asset_tag.asset_id, LEFT(asset_tag,SUBSTRING(asset_tag)-1 AS 'ETIQ'
from (SELECT DISTINCT S2.asset_id + ',' AS etiquetas
(SELECT S1.tag_id
FROM asset_tag AS S1
WHERE S1.tag_id
ORDER BY S1.tag_id
FOR XML PATH (''),TYPE
).VALUE('TEXT(1)`[1]','ninteger(MAX')[aset_tag] FROM asset_tag AS S2 ) asset_tag;
I have to group by asset and the asset 1 have in one column 1,2,3,4,5 or the tag that it have
how to use heidisql functions, on dbforge? I know but here not I use heidisql version 12. and is my first time working with this
The objective is that the source table that has two columns, group by column 1 and that a new column indicate separated by commas what column 1 has in column 2 (of origin).
columna 1 - 1 1 2 2 3 3 4 4
columna 2 - a b c a d a f g
and in a new column or table 1 - a b / 2 - b c
I see this answer on this page: https://stackoverflow.com/a/545672/20100117 But i don´t know what mean "st1" or [text()] the alias?
SELECT Main.SubjectID,
LEFT(Main.Students,Len(Main.Students)-1) As "Students" FROM
(
SELECT DISTINCT ST2.SubjectID,
(
SELECT ST1.StudentName + ',' AS [text()]
FROM dbo.Students ST1
WHERE ST1.SubjectID = ST2.SubjectID
ORDER BY ST1.SubjectID
FOR XML PATH (''), TYPE
).value('text()[1]','nvarchar(max)') [Students]
FROM dbo.Students ST2
) [Main]

Answer 1

If you use HeidiSQL, you can use the completion proposal to help finding the right syntax for the various functions. Just type some first characters and press Ctrl+Space:
Here's a basic example of how SUBSTRING() works:
SELECT SUBSTRING(name, 2, 3) FROM mytable;

Answer 2

How to concatenate rows of the same column in PostgreSQL?
Given that you want to concatenate rows of the same column, and not different columns (which is what you do when using CONCAT_WS()), what you would really be looking for is to use the ARRAY_AGG aggregation function within the ARRAY_TO_STRING function.
Documentation: https://www.postgresql.org/docs/13/functions-array.html
solution:
SELECT
a.asset_id, ARRAY_TO_STRING(ARRAY_AGG(a.tag_id), ',') AS etiqueta
FROM public.asset_tag AS a
GROUP BY a.asset_id;
Result:
on asset_id 1 | 1,3,5 tag_id
on asset_id 6 | 1,2 tag_id
If you insert this:
CREATE TABLE asset_tag ( asset_id INT,tag_id INT);
INSERT INTO asset_tag VALUES (1,1);
INSERT INTO asset_tag VALUES (1,3);
INSERT INTO asset_tag VALUES (1,5);
INSERT INTO asset_tag VALUES (6,1);
INSERT INTO asset_tag VALUES (6,2);
thanks to the person who gave me this answer .