我有一个表格,显示注册的员工。我想根据他们的数据库生成一个包括他们的姓名、ID、职位等信息的简单HTML页面。
为此,我向视图传递一个ID,以便它可以获取相应用户的详细信息并显示给我。一切正常,直到出现错误对象不可迭代。以下是我的代码。
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 %}
视图.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)
我想根据用户的进程查看每个用户的个人数据。