Recursive method not working

Hi, I'm trying to display a hierarchical tree but there's something wrong with the method below.
The result var is 'lost' between the calls - I mean, each "#{result}" contains the right value at the end of the method (I checked that), but that value is lost when returning to the upper level.

I don't see why...

Thanks for your help !

def affiche_arbre(rubriques)
     for rubrique in rubriques
       result = '<br/>'
       0.step(rubrique.level) { result += '&nbsp;'}
       result += rubrique.libelle + ' ' + link_to('+', :action => 'new', :parent_id => rubrique) + ' '
       unless rubrique.parent_id == nil
         result += link_to('E', :action => 'edit', :id => rubrique) + ' ' + link_to('D', :action => 'delete', :id => rubrique )
       end
       if rubrique.children.size > 0
         result += affiche_arbre(rubrique.children)
       end
     end
     "#{result}"
end

Hi, I'm trying to display a hierarchical tree but there's something wrong with the method below.
The result var is 'lost' between the calls - I mean, each "#{result}" contains the right value at the end of the method (I checked that), but that value is lost when returning to the upper level.

I don't see why...

Thanks for your help !

def affiche_arbre(rubriques)

I'd try to insert

result=""

here.

    for rubrique in rubriques
      result = '<br/>'
      0.step(rubrique.level) { result += '&nbsp;'}
      result += rubrique.libelle + ' ' + link_to('+', :action => 'new', :parent_id => rubrique) + ' '
      unless rubrique.parent_id == nil
        result += link_to('E', :action => 'edit', :id => rubrique) + ' ' + link_to('D', :action => 'delete', :id => rubrique )
      end
      if rubrique.children.size > 0
        result += affiche_arbre(rubrique.children)
      end
    end
    "#{result}"

Why do you do that? You could simply return result.

end

A general note: using << instead of += is much more efficient. You can get even better by passing the result like this:

def meth(node, result="")
   ...
   result << "START"
   ...
   # recursion
   meth(another_node, result)
   ...
   result << "END"
   ...
   result
end

Kind regards

  robert

···

On 18.04.2007 18:01, Zouplaz wrote:

1> def affiche_arbre(rubriques)
2> for rubrique in rubriques
3> result = '<br/>'
4> 0.step(rubrique.level) { result += '&nbsp;'}
5> result += rubrique.libelle + ' ' + link_to('+', :action => 'new', :parent_id => rubrique) + ' '
6> unless rubrique.parent_id == nil
7> result += link_to('E', :action => 'edit', :id => rubrique) + ' ' + link_to('D', :action => 'delete', :id => rubrique )
8> end
9> if rubrique.children.size > 0
10> result += affiche_arbre(rubrique.children)
11> end
12> end
13> "#{result}"
14> end

Well, the only recursive call I can see is line 10. You append the return
values from these calls onto 'result'. But then in line 12 you go back
around the loop, and in line 3 you reinitialise result, thereby throwing
away everything you've calculated so far.

···

On Thu, Apr 19, 2007 at 01:05:05AM +0900, Zouplaz wrote:

Hi, I'm trying to display a hierarchical tree but there's something
wrong with the method below.
The result var is 'lost' between the calls - I mean, each "#{result}"
contains the right value at the end of the method (I checked that), but
that value is lost when returning to the upper level.

I don't see why...

Thanks for your help !

This code could also be:
============ code ============
links(rubrique)
  [ link_to('+', :action => 'new', :parent_id => rubrique) ] +
  ( rubrique.parent_id ?
    [ link_to('E', :action => 'edit', :id => rubrique),
      link_to('D', :action => 'delete', :id => rubrique) ] : )
end

def affiche_arbre(rubriques)
  rubriques.collect do |rubrique|
    '<br/>' + ('&nbsp;' * rubrique.level) +
    [ rubrique.libelle ].concat(links(rubrique)).join(' ') +
    ( rubrique.children.size.zero? ? '' :
affiche_arbre(rubrique.children) )
  end.join()
end
=========== /code ===============
Please excuse the Anglicized new function -- I know no French. Notice
that this code allows you to test correctness of link generation
separately from correctness of the generated HTML. By restricting
myself to expressions -- there are no variable assignments in this
code, let alone mutations -- I protect myself from the kind of bug
that marred the original.

···

On Apr 18, 9:01 am, Zouplaz <u...@domain.invalid> wrote:

def affiche_arbre(rubriques)
     for rubrique in rubriques
       result = '<br/>'
       0.step(rubrique.level) { result += '&nbsp;'}
       result += rubrique.libelle + ' ' + link_to('+', :action => 'new',
:parent_id => rubrique) + ' '
       unless rubrique.parent_id == nil
         result += link_to('E', :action => 'edit', :id => rubrique) + '
' + link_to('D', :action => 'delete', :id => rubrique )
       end
       if rubrique.children.size > 0
         result += affiche_arbre(rubrique.children)
       end
     end
     "#{result}"
end

> Hi, I'm trying to display a hierarchical tree but there's something
> wrong with the method below.
> The result var is 'lost' between the calls - I mean, each "#{result}"
> contains the right value at the end of the method (I checked that), but
> that value is lost when returning to the upper level.
>
> I don't see why...
>
> Thanks for your help !
>
> def affiche_arbre(rubriques)

I'd try to insert

result=""

here.

> for rubrique in rubriques
> result = '<br/>'
> 0.step(rubrique.level) { result += '&nbsp;'}
> result += rubrique.libelle + ' ' + link_to('+', :action => 'new',
> :parent_id => rubrique) + ' '
> unless rubrique.parent_id == nil
> result += link_to('E', :action => 'edit', :id => rubrique) + ' '
> + link_to('D', :action => 'delete', :id => rubrique )
> end
> if rubrique.children.size > 0
> result += affiche_arbre(rubrique.children)
> end
> end
> "#{result}"

Why do you do that? You could simply return result.

> end

A general note: using << instead of += is much more efficient. You can
get even better by passing the result like this:

def meth(node, result="")
   ...
   result << "START"
   ...
   # recursion
   meth(another_node, result)
   ...
   result << "END"
   ...
   result
end

Kind regards

        robert

More remarks:

you should change

> result = '<br/>'

to
result << '<br/>'
as well

now some enhacements:

> for rubrique in rubriques

rubriques.each do |rubrique|
-- it's a question of style/taste otherwise equal

> unless rubrique.parent_id == nil

unless rubrique.parent_id.nil?

> if rubrique.children.size > 0

unless rubrique.children.empty?
-- this one may be faster sometimes (if size computation is slow)
and it more explicitly shows what are you asking

> 0.step(rubrique.level) { result += '&nbsp;'}

(rubrique.level + 1).times { result << '&nbsp;'}
or
result << ('&nbsp;' * (rubrique.level +1))
-- the first one is more descriptive, IMO

Jano

···

On 4/18/07, Robert Klemme <shortcutter@googlemail.com> wrote:

On 18.04.2007 18:01, Zouplaz wrote:

You got it ! Thank you !

···

le 18/04/2007 21:18, Brian Candler nous a dit:

Well, the only recursive call I can see is line 10. You append the return
values from these calls onto 'result'. But then in line 12 you go back
around the loop, and in line 3 you reinitialise result, thereby throwing
away everything you've calculated so far.