如何處理Django查詢集？

Question

嘿，我想處理一個查詢集內每個對象的一個​​屬性，之後我想返回JSON格式？如何處理？

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

例如，我想手動添加用戶的名稱字段「*」之後，再返回爲JSON格式？或保持查詢集類型？

Answer 1

你也可以遍歷一個查詢集，每個元素都是一個單獨的對象，所以是這樣的：在Django的外殼，它

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

發揮。

JSON格式？哦別人可以做到這一點！

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) 

關於JSON響應：Django Docs