Question

嘿，我想在一个查询集中处理每个对象的一个属性，之后我想返回JSON格式？该怎么办？

results = Sample.objects.filter(user=user)

例如，我想在用户名字段之后手动添加'*'，然后以JSON格式返回？或保持查询集类型？

Answer 1

您可以遍历查询集，每个元素都是一个对象，如下所示：

starnames = [ n.username+"*" for n in results]

在Django shell玩它。

JSON格式？哦别人可以做到！

Answer 2

class ProcessQuerySet(object):
"""
A control that allow to add extra attributes for each object inside queryset.
"""
def process_queryset(self, queryset):
    """ queryset is a QuerySet or iterable object. """
    return map(self.extra, queryset) # Using map instead list you can save memory.

def extra(self, obj):
    """ Hook method to add extra attributes to each object inside queryset. """
    current_user = self.request.user # You can use `self` to access current view object
    obj.username += '*'
    return obj

用法：

class YourView(ProcessQuerySet, AnyDjangoGenericView):
def get_queryset(self):
    queryset = SomeModel.objects.all()
    return self.process_queryset(queryset)

关于JSON响应：Django Docs

如何处理django查询集？

2 个答案: