Tim and Nis,
Okay thanks for the help. My view is definitily better now than
before. Here is my new view
def searchresult(request):
if request.method == 'POST':
myset = set()
NOT_PICKED = "---------------"
y = Choice.objects.all()
if ('price' in request.POST and request.POST['price'] <>
NOT_PICKED):
y = y.filter(price__price_cat=request['price'])
if ('size' in request.POST and request.POST['size'] <>
NOT_PICKED):
y = y.filter(size__size_cat=request['size'])
for q in y:
styles = Choice.objects.get(id=q.id).style_set.all()
if ('color' in request.POST) and (request.POST['color']
<>
NOT_PICKED):
styles =
styles.filter(color_cat=request['color'])
for style in styles:
myset.add(style)
if myset == set([]):
return render_to_response('searchresult_none.html', {'s': "No
product available"})
else:
return render_to_response('searchresult.html', {'s': myset})
////////////////////
Does that look any better? I have two issues with this.
1) Is there anyway to get it so that I don't have to use the following
for loop:
for style in styles:
myset.add(style)
2) Is still adds duplicate products. If a product contains two
choices that have a price of 149 and 199. Then when a user searches
by price only (100-199) then the product is displayed twice in the
result set.
////////
Thank you SO much for your help!!! This is my first time developing
in Python and I'm learning quite a bit.
On Aug 9, 8:37 am, Tim Chase <[EMAIL PROTECTED]> wrote:
> > NO_COLOR = "---------------"
> > styles = Choice.objects.get(id=h.id).style_set.all()
> > if ('color' in request.POST) and (request.POST['color'] <> NO_COLOR):
> > styles = styles.filter(color_cat=request['color'])
> > for u in styles:
> > dict[u] = u
>
> > ////
>
> > I did notice that whenever I do a search for just the color 'brown'.
> > Then the result set will bring back the same style however many
> > different sizes that are in the style. So if an area rug sells two
> > choices (2'x3' 39.00, 4'x6' 149.00). Then that style will show up
> > twice in the result set. Is there anyway better to filter out the
> > styles that have already been added to the dictionary. My previous
> > code worked..but not sure it's the best way to do it. Here it is:
>
> > num = 0
> > for a in dict:
> > if a == j: # 'j' being the name of the style and 'a' is the name of
> > the style that is already in the dictionary
> > num = 1
> > if num == 0:
> > dict[j] = j
>
> I second the suggestion by Nis to make meaningful variable names,
> as well as the suggestion to use sets rather than abusing a
> dictionary (and tromping on the namespace with "dict"...just got
> bitten by this yesterday, a "zip"-code variable shadowed the
> built-in zip() command causing some confusing errors)
>
> It looks like you could just do something like
>
> results = Choice.objects.get(id=h.id).style_set.all()
> # filter results
> results = set(results)
>
> I'm not sure on the performance of set creation, so you might
> compare the results with
>
> results = set(list(results))
>
> or
>
> results = set(tuple(results))
>
> -tim
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