[英]Fetch parent child relation with level from the same table
嗨,我是 django 和 python 的新手。
# function for fecthing the child
def get_childern(request, username):
Q = list(Profile.objects.filter(sponsor_id__exact=username))
populate(request, Q, username)
# Iterate the Queryset
def populate(request, quesryset, username):
if quesryset:
messages.success(request, username+' has '+str(len(quesryset))+' child')
messages.info(request, quesryset)
for user in quesryset:
get_childern(request, user.user_id)
else:
messages.warning(request, 'User '+username+' has no child')
return False
# calling all children
get_childern(request, username)
我想添加級別,如何進一步分解它。
請幫助我,投入大量時間和精力。 謝謝大家:)
如果有人想提供一些反饋或想法以獲得更好的解決方案,我找到了解決方案,請在此處發表評論。 給大家點贊:)
# for return a child query-set
def get_childern(request, username):
Q = Profile.objects.filter(sponsor_id__exact=username)
return Q
level = int(1) # set according to need
tree_set = {}
# iterate the queryset and call itself
# recursion - set the level to control
def iterate_object(request, quesryset):
global tree_set
outer_count = int(0)
inner_count = int(0)
level_tree = []
for users in quesryset:
outer_count = outer_count + 1
# messages.info(request, users.user_id)
Qs = get_childern(request, users.user_id)
if Qs:
for child in Qs:
inner_count = inner_count + 1
level_tree.append(child)
# count = count + len(Qs)
# messages.warning(request, Qs)
else:
# messages.warning(request, 'No child')
pass
else:
# messages.error(request, 'Total child at level 2 => ' + str(count))
# messages.info(request, outer_count)
# messages.success(request, inner_count)
global level
level = level + 1
# set the numeric value to control the level
if level <= 10:
tree_set.update({level: level_tree})
iterate_object(request, level_tree)
return tree_set
else:
# messages.warning(request, 'else part')
pass
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