我正在尝试在django中获取某个对象的id但是我一直收到以下错误 异常值:QuerySet;对象没有属性ID。 我在views.py中的功能
@csrf_exempt
def check_question_answered(request):
userID = request.POST['userID']
markerID = request.POST['markerID']
title=request.POST['question']
m = Marker.objects.get(id=markerID)
u = App_User.objects.get(id=userID)
print userID
print markerID
print title
# userID='1'
# markerID='1'
# title='Hello'
at = AttachedInfo.objects.filter(attachedMarker=m.id, title=title)
print 'user'
print u.id
print 'marker'
print m.id
print 'att'
print at
#print at.id
if(Answer.objects.filter(marker=m.id, user=u.id, attachedInfo=at.id)):
print 'pass'
return HttpResponse('already answered')
else:
print 'not'
return HttpResponse('not answered yet')
错误发生在此部分的if条件中(attachedInfo = at.id)。我检查过,因为当我从条件中移除它时,一切都工作正常。
这是models.py
class AttachedInfo(models.Model):
title = models.CharField(max_length=200)
helpText = models.CharField(max_length=200, null=True, blank=True)
type = models.CharField(max_length=200)
attachedMarker = models.ForeignKey(Marker)
answer1 = models.CharField(max_length=200, null=True, blank=True)
answer2 = models.CharField(max_length=200, null=True, blank=True)
answer3 = models.CharField(max_length=200, null=True, blank=True)
answer4 = models.CharField(max_length=200, null=True, blank=True)
correctAnswer = models.CharField(max_length=50, null=True, blank=True)
optionalMessage = models.CharField(max_length=200, null=True, blank=True)
def __unicode__(self):
return self.title
class Answer(models.Model):
user = models.ForeignKey(App_User)
app = models.ForeignKey(App, null=True, blank=True)
marker = models.ForeignKey(Marker)
attachedInfo = models.ForeignKey(AttachedInfo)
textAnswer = models.CharField(max_length=200, null=True, blank=True)
mcqAnswer = models.CharField(max_length=200, null=True, blank=True)
answered = models.BooleanField(default=False)
def __unicode__(self):
return self.attachedInfo.title
为什么我收到此错误的任何帮助?!
答案 0 :(得分:31)
这行代码
at = AttachedInfo.objects.filter(attachedMarker=m.id, title=title)
返回queryset
并且您正在尝试访问它的一个字段(不存在)。
您可能需要的是
at = AttachedInfo.objects.get(attachedMarker=m.id, title=title)
答案 1 :(得分:16)
您收到错误的原因是at
是QuerySet
即:列表。您可以执行at[0].id
之类的操作或使用get
代替filter
来获取at
对象。
希望它有所帮助!
答案 2 :(得分:2)
在大多数情况下,您不希望不处理这样的现有对象。而不是
ad[0].id
使用
get_object_or_404(AttachedInfo, attachedMarker=m.id, title=title)
这是推荐的Django shortcut。
答案 3 :(得分:0)
我将近2天收到此错误,此错误的主要问题仅取决于两个文件,即
models.py&views.py
我收到此错误消息是因为我想从电子邮件ID创建会话,但显示它们不是属性电子邮件,因此未获取任何str对象。
解决方案:-
models.py
class Register(models.Model):
userid = models.AutoField(primary_key=True)
name = models.CharField(max_length=100)
email = models.EmailField(max_length=200)
password = models.CharField(max_length=100)
def __str__(self):
return "%s %s" %(self.name, self.email)
为以下项目创建一个字符串,以根据项目从中获取数据。
views.py
if request.method == "POST":
emailx1 = request.POST['emailx']
passwordx1 = request.POST['passwordx']
if (Register.objects.filter(email=emailx1, password=passwordx1)).exists():
a = Register.objects.filter(email=emailx1).first()
request.session['session_name'] = a.email
request.session['session_id'] = a.userid
return render(request, "index.html", {"a": a})
将.first()方法与Model.objects方法一起使用。这已经解决了我的问题,希望也能解决您的问题。