That's it I'll try it thanks a lot!
On 29 nov, 00:48, mdipierro <[email protected]> wrote: > I assume you are trying to build a hierarchial tree of articles and > you want them all from the parent. > > I would do this something like this: > > db.define_table('article',Field('f_parent','reference > article'),Field('title')) > > def index(): > rows = dict((r.id,r) for r in db(db.article).select()) > for id,row in rows.items(): row.children=[] > for id,row in rows.items(): > if row.f_parent==0: root=row > else: rows[row.f_parent].append(row) > def tree(row): > return DIV(H1(row.title),UL(*[LI(tree(child)) for child in > row.children]))) > return tree(root) > > On Nov 28, 5:00 pm, JmiXIII <[email protected]> wrote: > > > Hello, > > > I'm facing I guess a well known programming problem. It is not really > > related to web2by, but since I'm using web2py I hope somebody here can > > help. > > > Here is my controller : > > > def global_view(): > > parents = db(~(db.t_article.id==db.t_list.f_article)).select() > > tree={} > > for parent in parents: > > tree[parent.t_article.id]={} > > sublevel=db(parent.t_article.id==db.t_list.f_parent).select() > > for son in sublevel: > > tree[parent.t_article.id][son.f_article]=son.f_article > > tree[parent.t_article.id][son.f_article]={} > > sublevel2=db(son.f_article==db.t_list.f_parent).select() > > for son2 in sublevel2: > > tree[parent.t_article.id][son.f_article] > > [son2.f_article]=son2.f_article > > tree[parent.t_article.id][son.f_article][son2.f_article]={} > > return dict(tree=tree) > > > So in fact I'd like to go on for son3, son4, based on the same model > > as sublevel2... until sublevelx=None. > > I guess I should use some kind of recursive or loop function. > > I've read different things using right and left variable for > > hierarchical tree, but it seems to me that it is possible to build a > > loop in my controller without modifying my model. > > > Hope I do not disturb this groupes with this kind of question. > >
