sql - select recursive and order by -

i use below query select recursive top bottom
e.g
if tagid 1 rows 1 > 3,4, > 5 works fine,
want know how result order "name" @ each level(same parent id) rows 1 > 4,3 > 5 ?

i treid add order "name" after select * "tag" "tagid" = $1 not work .
, if add after select * tag_tree mess level become 1,4,5,3 not want .

create table if not exists "tag"( "tagid" serial not null, "parenttagid" integer, "name" varchar, primary key ("tagid") );  tagid | parenttagid | name | 1     |             |   | 2     |             | b   | 3     | 1           | b   | 4     | 1           |   | 5     | 3           |   |   var query = 'with recursive tag_tree (   (     select * "tag" "tagid" = $1   )   union   select child.* "tag" child     join tag_tree parent on parent."tagid" = child."parenttagid" ) select * tag_tree'; 

add order by coalesce():

with recursive tag_tree (   (     select * "tag" "tagid" = 1   )   union   select child.* "tag" child     join tag_tree parent on parent."tagid" = child."parenttagid" ) select * tag_tree order coalesce("parenttagid", 0), "name";   tagid | parenttagid | name  -------+-------------+------      1 |             |      4 |           1 |      3 |           1 | b      5 |           3 | (4 rows) 

for the documentation:

the coalesce function returns first of arguments not null. null returned if arguments null. used substitute default value null values when data retrieved display.

in case function changes null 0.