如何编写代码，以便在使用 django 的模型中只选择一个单选按钮。？

Question

#models.py
from django.db import models
from django.contrib.auth.models import User

CHOICES = (('1','Earned Leave'),('2','Casual Leave'),('3','Sick Leave'),('4','Paid Leave'))

class Leave(models.Model):

    employee_ID = models.CharField(max_length = 20)

    name = models.CharField(max_length = 50)

    user = models.ForeignKey(User, on_delete = models.CASCADE, null =True)

    department = models.CharField(max_length = 50)

    designation = models.CharField(max_length = 50)

    type_of_leave = models.CharField(max_length = 15, choices = CHOICES)

    from_date = models.DateField(help_text = 'mm/dd/yy')

    to_date = models.DateField(help_text = 'mm/dd/yy')

    reporting_manager = models.CharField(max_length = 50, default = None, help_text = '0001_manager, 0002_manager')

    reason = models.CharField(max_length= 180)

    accepted = models.BooleanField(('accept'), default= False)

    rejected = models.BooleanField(('reject'), default = False)

    reason_reject = models.CharField(('reason for rejection'),max_length=50) // this didn't help me.

    def __str__(self):
        return self.name

这是一个休假申请表，其中只有两个字段 (accepted, rejected) 应该被选中，如果拒绝字段没有被选中，那么 reason_reject 不应显示在/admin面板。

#forms.py
from django import forms
from lrequests import models

class LeaveRequestForm(forms.ModelForm): 
    
    class Meta:
        fields = ("name", "employee_ID", "department", "designation", "type_of_leave", "from_date", "to_date", "reporting_manager", "reason")
        model = models.Leave

用户填写表格并提交。所以，现在管理员必须接受或拒绝它。只有在管理员选择 rejected 字段后，reason_reject 字段才会出现在管理员面前。这一切应该只发生在管理员端。

#admin.py
from django.contrib import admin
from . import models

@admin.register(models.Leave)
class LeaveAdmin(admin.ModelAdmin):
    list_display = ["name"]
    #"employee_ID", "department", "designation", "type_of_leave", "from_date", "to_date", "reporting_manager", "reason", "accepted", "rejected", "reason_reject"
    
    list_filter = ['department','type_of_leave']

    def get_queryset(self, request):
        qs = super().get_queryset(request)
        if request.user.is_superuser:
            return qs
        return (qs.filter(reporting_manager=request.user.username) or qs.filter(employee_ID=request.user.username))
    


 # def get_readonly_fields(self, request, obj=None):
# # the logged in user can be accessed through the request object 
#     if obj and request.user.is_staff:
#         readonly_fields = [f.name for f in self.opts.fields]
#         readonly_fields.remove('accepted')
#         readonly_fields.remove('rejected')
#     return readonly_fields
#the above chunk of get_readonly_fields was written so that the admin could only either 'accept' or 'reject' the form.

    def get_fields(self, request, obj=None):
         fields = [f.name for f in self.opts.fields]
         if obj and obj.rejected:
            fields.append('reason_reject')
         return fields

通过运行上面的代码我得到 KeyError at /admin/lrequests/leave/15/change/ 和

"Key 'id' not found in 'LeaveForm'. Choices are: accepted, department, designation, employee_ID, from_date, name, reason, reason_reject, rejected, reporting_manager, to_date, type_of_leave, user."

Answer 1

整个前提都是有缺陷的。你不能在模型 definition 中执行这种逻辑，因为那时没有实例，所以没有值可以比较。即使可以，它仍然没有意义，因为模型定义定义了模型具有的数据库列；你不能让不同的实例在数据库中有不同的列。模型定义适用于整个模型。您需要在 class 定义中包含 reason_reject。

您可能想要根据模型中的值更改表单。您可以通过覆盖 get_fields 方法在管理员中执行此操作。

class LeaveAdmin(admin.ModelAdmin):
    ...
    def get_fields(self, request, obj=None):
        fields = [....list of fields for the form...]
        if obj and obj.rejected:
            fields.append('reason_reject')
        return fields

请注意，您可能需要考虑是否真的需要单独的 accepted/rejected 字段；最好为申请状态设置一个字段：接受或拒绝。您可以将它们表示为单独的单选按钮或下拉菜单。

STATUS_CHOICES = (
    (1, 'Accepted'),
    (0, 'Rejected'),
)
status = models.IntegerField(choices=STATUS_CHOICES, blank=True, null=True)

Answer 2

首先，将其定义为下拉菜单。那就是像

这样的选择

STATUS_CHOICES = (('0', 'Rejected'),('1', 'Accepted'),)
class Leave(models.Model):
     ...
     status = models.CharField(max_length = 15, choices = STATUS_CHOICES)

然后在admin.py

class LeaveAdmin(admin.ModelAdmin):
     ...
     radio_fields = {"status": admin.HORIZONTAL}

如何编写代码，以便在使用 django 的模型中只选择一个单选按钮。？

How to write code such that only one of the radio button should be selected in models with django.?

python

django

model

admin

view