ValueError：要解压的值太多（预期为 2）Django答案

【问题标题】：ValueError: too many values to unpack (expected 2) DjangoValueError：要解压的值太多（预期为 2）Django
【发布时间】：2021-09-02 07:17:02
【问题描述】：

不知道为什么，但我的代码出现以下错误：

ValueError: too many values to unpack (expected 2)

这是我的models.py：

class UserList(models.Model):
    list_name = models.CharField(max_length=255)
    user = models.ForeignKey(User, on_delete=models.CASCADE) 
    
    def __str__(self):
        return self.list_name

这是我的意见.py

def otherUserList(request):
    userName = request.GET.get('userName', None)
    print(userName)
    qs = UserList.objects.filter(user__username=userName)
    return qs

这是回溯：

Internal Server Error: /electra/otheruserlist/
Traceback (most recent call last):
  File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/core/handlers/exception.py", line 47, in inner
    response = get_response(request)
  File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/utils/deprecation.py", line 116, in __call__
    response = self.process_response(request, response)
  File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/middleware/clickjacking.py", line 26, in process_response
    if response.get('X-Frame-Options') is not None:
  File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/db/models/query.py", line 418, in get
    clone = self._chain() if self.query.combinator else self.filter(*args, **kwargs)
  File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/db/models/query.py", line 942, in filter
    return self._filter_or_exclude(False, *args, **kwargs)
  File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/db/models/query.py", line 962, in _filter_or_exclude
    clone._filter_or_exclude_inplace(negate, *args, **kwargs)
  File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/db/models/query.py", line 969, in _filter_or_exclude_inplace
    self._query.add_q(Q(*args, **kwargs))
  File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/db/models/sql/query.py", line 1358, in add_q
    clause, _ = self._add_q(q_object, self.used_aliases)
  File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/db/models/sql/query.py", line 1377, in _add_q
    child_clause, needed_inner = self.build_filter(
  File "/Library/Frameworks/Python.framework/Versions/3.8/lib/python3.8/site-packages/django/db/models/sql/query.py", line 1255, in build_filter
    arg, value = filter_expr
ValueError: too many values to unpack (expected 2)
[02/Sep/2021 05:38:43] "GET /electra/otheruserlist/?userName=alice HTTP/1.1" 500 94227

如果能提供任何帮助，我将不胜感激。

【问题讨论】：

你不能返回一个QuerySet，一个视图应该总是返回一个HTTP响应。

标签： django

【解决方案1】：

您的视图返回QuerySet，但这没有多大意义：视图应该返回HttpResponse。例如，您可以渲染模板、将其转换为 JSON 等。

例如，我们可以使用以下方式渲染模板：

from django.shortcuts import render

def otherUserList(request):
    userName = request.GET.get('userName', None)
    qs = UserList.objects.filter(user__username=userName)
    return render(request, 'some-template.html', {'lists': qs})

或者我们可以例如返回一个JsonResponse wit the list_name`s of user 具有给定用户名的用户：

from django.http import JsonField

def otherUserList(request):
    userName = request.GET.get('userName', None)
    qs = UserList.objects.filter(user__username=userName)
    return JsonResponse({'listnames': [list.name for list in qs]})

【讨论】：

【解决方案2】：

异常来自 Django 内部中间件，因为它试图将您返回的 qs 处理为响应。

您需要返回一个响应，而不仅仅是一个查询集，例如这个简单的示例返回用户 ID 列表。

from django.http import JsonResponse

def otherUserList(request):
    userName = request.GET.get('userName', None)
    print(userName)
    qs = UserList.objects.filter(user__username=userName)
    return JsonResponse({"ids": [user.id for user in qs]})

【讨论】：