很抱歉這個問題，但我是 Django 的初學者，我找不到任何像這種情況這樣的話題。這是代碼：views.pydef select_collections(request): listacolecao = Collection.objects.order_by('upload_date') listasubscription = Subscription.objects.filter(user=request.user) for obj in listacolecao: try: Subscription.objects.filter(user=request.user, collection=obj) except Subscription.DoesNotExist: print('not exist') else: print('Ok')它在終端打印這個結果：not existnot existOknot existOkOkOkOk我知道那不是列表，但我需要將其結果放入模板中。我怎樣才能做到這一點？
1 回答
三國紛爭
TA貢獻1804條經驗 獲得超7個贊
如果你只是想要你在那里擁有的相同但打印在你擁有的模板中，（以我的謙虛和初學者的觀點）你需要稍微修改你的視圖，創建一個你想要顯示它的模板并將 url 修改為在該模板中加載視圖，例如：
查看.py 變化：
def select_collections(request):
listacolecao = Collection.objects.order_by('upload_date')
listasubscription = Subscription.objects.filter(user=request.user)
a_list = [] #You would get something like: ['not exist', 'not exist', 'Ok', 'not exist', 'Ok','Ok','Ok'] which is what it was printed in your code
for obj in listacolecao:
try:
Subscription.objects.filter(user=request.user, collection=obj)
except Subscription.DoesNotExist:
#print('not exist') I would substitute it for .append, to add each value to the list as a new item
a_list.append('not exist')
else:
#print('Ok')
a_list.append('Ok')
#Now you pass that variable and sending it to your template, so you can use it there.
context = {
'a_list':a_list,
}
return render(request, 'your_template_name.html', context)
在你的 urls.py 中：
from .views import select_collections #Importing your recently created view
urlpatterns = [
path = ('the_url_where_you_want_it', select_collections, name="the_name_you_prefer" ),
]
現在在您的模板中：
#As you have already sent those variables here you can use Django's template tags
{% for each_obj in a_list %}
<h3> {{each_obj }} </h3> #If you change each_obj for a_list, you would get a QuerySet (fancy word for a list), with all the items in the "a_list" variable.
{% endfor %}
這應該允許您在模板中單獨查看列表中的每個項目。
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