[英]Django request.user Not Resolving SimpleLazyObject
我的Django應用程序中有一個使用用戶ID查找對象的方法。 通過AJAX調用來調用此方法。 登錄到有效的用戶帳戶后,無論我如何嘗試, request.user
評估為django.utils.functional.SimpleLazyObject
對象,並且無法檢索所需的數據(當絕對存在時, SkillEntry匹配查詢不存在) 。 我在Django中嘗試了該解決方案:按用戶過濾草稿會導致錯誤 ,但無濟於事。
如何讓用戶引用用戶對象的實際實例?
查看代碼:
@login_required
def skill_set(request, name):
skill = Skill.objects.get(slug=name) # Found.
level = 0
user = request.user
if request.method == 'POST':
if user.is_authenticated():
entry = SkillEntry.objects.get(user=user.pk, skill=skill) # Not found.
entry.level = request.POST['level']
entry.save()
return HttpResponse(status=200)
else:
return HttpResponseForbidden()
JavaScript客戶端代碼:
function setSkill(skill, value) {
var req = new XMLHttpRequest();
req.open("POST", "/skill/" + skill + "/set/", true);
csrftoken = getCookie('csrftoken');
req.setRequestHeader("X-CSRFToken", csrftoken);
req.send('level=' + value);
var elem = document.getElementById('level');
elem.innerHTML = "My skill level is " + value + ".";
}
可能需要在請求中設置一些內容來維護會話信息嗎? 我可以發誓,在更早版本的Django上,我已經成功完成了類似的工作。 我正在使用1.4.3。
編輯:
這是我的models.py中的模型定義:
class Skill(models.Model):
name = models.CharField(max_length=100)
slug = models.CharField(max_length=100, blank=True)
keywords = models.CharField(max_length=120, blank=True, help_text='List of additional keywords this skill should show up for in search')
description = models.TextField(null=True, blank=True)
parent = models.ForeignKey('Skill', null=True, blank=True, help_text='Parent skill. Leave blank if this is a root category.')
def save(self, *args, **kwargs):
if not self.id:
self.slug = slugify(self.name)
super(Skill, self).save(*args, **kwargs)
def __unicode__(self):
return self.name
class Meta:
ordering = ('name',)
class SkillEntry(models.Model):
skill = models.ForeignKey('Skill')
level = models.IntegerField()
user = models.ForeignKey(User)
last_updated = models.DateField(auto_now=True)
class Meta:
verbose_name_plural = 'Skill Entries'
ordering = ('skill__name',)
def __unicode__(self):
return self.user.username + ' knows ' + self.skill.name + ' at level ' + str(self.level)
正如羅漢所說,您不需要使用user.pk
更改此查詢
entry = SkillEntry.objects.get(user=user.pk, skill=skill)
至
entry = SkillEntry.objects.get(user=user, skill=skill)
仍然,如果它不能正常工作,請分享您的模型。
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