1

我的 Django 应用程序中有一个使用用户 ID 查找对象的方法。此方法通过 AJAX 调用调用。当登录到有效的用户帐户时，无论我尝试什么，都会request.user评估为一个django.utils.functional.SimpleLazyObject对象并且无法检索到我想要的数据（SkillEntry 匹配查询绝对存在时不存在）。我在 Django 中尝试了解决方案：按用户过滤草稿导致错误无济于事。

如何让用户引用用户对象的实际实例？

查看代码：

@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 客户端代码：

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 + ".";
}

也许我需要在请求中设置一些东西来维护会话信息？我可以发誓我以前在更早版本的 Django 上成功地做过类似的事情。我正在使用 1.4.3。

编辑：

以下是我的 models.py 中的模型定义：

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)
4

1 回答 1

0

正如rohan所说，你不需要使用user.pk

更改此查询

entry = SkillEntry.objects.get(user=user.pk, skill=skill)

到

    entry = SkillEntry.objects.get(user=user, skill=skill)

如果它不起作用，请分享您的 models.py

于 2013-03-16T06:11:12.350 回答