GROUP BY but get all values from other column

Question

I''ll explain what I need to do on example. First of all, we have a simple table like this one, named table:

id | name
===+=====
1  | foo
1  | bar
1  | foobar
2  | foo
2  | bar
2  | foobar

Now the query:

SELECT t.* FROM table t GROUP BY t.id

Will get us result similar to this one:

id | name
===+=====
1  | foo
2  | foo

But is it possible, to collect all values of name to have result like this?

id | name
===+=================
1  | foo, bar, foobar
2  | foo, bar, foobar

Answer 1

Using MySQL you can use GROUP_CONCAT(expr)

This function returns a string result with the concatenated non-NULL values from a group. It returns NULL if there are no non-NULL values. The full syntax is as follows:

GROUP_CONCAT([DISTINCT] expr [,expr ...]
             [ORDER BY {unsigned_integer | col_name | expr}
                 [ASC | DESC] [,col_name ...]]
             [SEPARATOR str_val])

Something like

SELECT ID, GROUP_CONCAT(name) GroupedName
FROM Table1
GROUP BY ID

SQL Fiddle DEMO

Answer 2

For SQL Server (before 2017) use FOR XML clause and STUFF() function for that:

SELECT distinct id, name = 
    STUFF((SELECT ' , ' + name
           FROM Table1 b 
           WHERE b.id = a.id 
          FOR XML PATH('')), 1, 2, '')
FROM Table1 a
GROUP BY id;

UPDATE

With SQL Server 2017, you can simply use STRING_AGG() function to achieve that:

SELECT ID, STRING_AGG (name, ', ') AS Name
FROM Table1
GROUP BY ID

See this SQLFiddle

Answer 3

For Postgres use string_agg()

select id, 
       string_agg(name, ',' order by name) as name_list
from the_table
group by id
order by id;

Answer 4

Below query works for you if you are using db2:

SELECT DISTINCT id, LISTAGG(names, ',')
FROM tb
GROUP BY id;