Django 'model' 对象不可迭代

Question

我有一张表，上面显示了注册的员工。我想根据他们的数据库生成一个简单的 HTML 页面，其中包括他们的姓名、ID、名称等。
为此，我将一个 id 传递给 View ，以便它可以获取相应用户的详细信息并显示给我看。一切正常，直到出现错误 对象不可迭代 .这是我的代码report.html

{% if emp_item %}
 {% for some in emp_item %}
    <title> {{ some.employee_name }} Report</title>

    <h3>{{ some.employee_name }}</h3>

    <table style="width:30%" border="4">


        <td>{{some.id}}</td>
        <td>{{some.Annual_leave}} </td>
        <td>{{some.Sick_leave}} </td>
        <td>{{some.allowed}} </td>

        </table>

{% endfor %}

<h2>No User</h2>
{% else %}
{% endif %}

view.py

@staff_member_required  # for admin login required
def report(request, id):
    emp_item = Employee.objects.get(id=id)
    context = {'emp_item': emp_item}
    return render(request, 'projectfiles/report.html', context)

urls.py

    url(r'^(?i)Rejectleaves/$', views.rejected_leave_show,
        name='Reject_show'),  # user leaves

    url(r'^(?i)report/(?P<id>\d+)$', views.report,
        name='Report'),  # user Report

models.py

class Employee(models.Model):

    allowed = models.BooleanField(default=True)
    employee_name = models.OneToOneField(User, on_delete = models.CASCADE)
    employee_designation = models.CharField(max_length = 5)
    employee_department = models.CharField(max_length = 5)
    Annual_leave = models.PositiveSmallIntegerField(default=5)
    Sick_leave = models.PositiveSmallIntegerField(default=5)

我想根据他们所做的过程查看每个用户的数据。

Answer 1

更改 Employee.objects.get(id=id)至 Employee.objects.filter(id=id)“filter() 总是会给你一个 QuerySet” - 它是可迭代的get() - 返回单个对象，它不可迭代