创建我自己的登录表单django(creating my own login form django)
一直在寻找2天现在,我只需要创建我自己的登录表单..我需要在forms.py中创建表单,然后连接到views.py中的视图,并从urls.py中的URL调用它,并使用所有我的自定义用户模型作为后端
这里是我的models.py ,其中包含用户模型:
class UserModelManager(BaseUserManager): def create_user(self, email, password, pseudo): user = self.model() user.name = name user.email = self.normalize_email(email=email) user.set_password(password) user.save() return user def create_superuser(self, email, password): ''' Used for: python manage.py createsuperuser ''' user = self.model() user.name = 'admin-yeah' user.email = self.normalize_email(email=email) user.set_password(password) user.is_staff = True user.is_superuser = True user.save() return user class UserModel(AbstractBaseUser, PermissionsMixin): ## Personnal fields. email = models.EmailField(max_length=254, unique=True) name = models.CharField(max_length=16) ## [...] ## Django manage fields. date_joined = models.DateTimeField(auto_now_add=True) is_active = models.BooleanField(default=True) is_staff = models.BooleanField(default=False) USERNAME_FIELD = 'email' REQUIRED_FIELD = ['email', 'name'] objects = UserModelManager() def __str__(self): return self.email def get_short_name(self): return self.name[:2].upper() def get_full_name(self): return self.name
Been looking for 2 days now, I simply need to create my own login form .. I need to create the form in forms.py then connect to a view in views.py and call it from an URL in urls.py and all use my custom user model as backend
here are my models.py that contains user model :
class UserModelManager(BaseUserManager): def create_user(self, email, password, pseudo): user = self.model() user.name = name user.email = self.normalize_email(email=email) user.set_password(password) user.save() return user def create_superuser(self, email, password): ''' Used for: python manage.py createsuperuser ''' user = self.model() user.name = 'admin-yeah' user.email = self.normalize_email(email=email) user.set_password(password) user.is_staff = True user.is_superuser = True user.save() return user class UserModel(AbstractBaseUser, PermissionsMixin): ## Personnal fields. email = models.EmailField(max_length=254, unique=True) name = models.CharField(max_length=16) ## [...] ## Django manage fields. date_joined = models.DateTimeField(auto_now_add=True) is_active = models.BooleanField(default=True) is_staff = models.BooleanField(default=False) USERNAME_FIELD = 'email' REQUIRED_FIELD = ['email', 'name'] objects = UserModelManager() def __str__(self): return self.email def get_short_name(self): return self.name[:2].upper() def get_full_name(self): return self.name
原文:https://stackoverflow.com/questions/50898193
最满意答案
我没有你的
$options
变量,所以我创建了自己的问题表来模拟你的情况。要选择一个
radio button
我们只需要使用这个jQuery$("input[type=radio]:checked")
然后我们必须使用.each()
函数检查这些选中的单选按钮的每个人。以下是您应该做的完整示例:
$("#submit").on("click",function(){ $("input[type=radio]:checked").each(function(){ console.log($(this).val()); }); });
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <div class="radio" id="form1"> <label>Question 1 <input type="radio" class="optionChecked" name="name1" id="'id1" value="value1" data-id="option1" data-title="question1"></label><br> <label>Question 2 <input type="radio" class="optionChecked" name="name2" id="'id2" value="value2" data-id="option2" data-title="question2"></label><br> <label>Question 3 <input type="radio" class="optionChecked" name="name3" id="'id3" value="value3" data-id="option3" data-title="question3"></label><br> <label>Question 4 <input type="radio" class="optionChecked" name="name4" id="'id4" value="value4" data-id="option4" data-title="question4"></label><br> <label>Question 5 <input type="radio" class="optionChecked" name="name5" id="'id5" value="value5" data-id="option5" data-title="question5"></label><br> <button type="button" id="submit"> Submit</button> </div>
I don't have your
$options
variable, so I created my own questions form to simulate your situation.To select a
radio button
we have only to use this jQuery$("input[type=radio]:checked")
then we must check everyone of these checked radio buttons using the function.each()
.Here is a full example of what you should do:
$("#submit").on("click",function(){ $("input[type=radio]:checked").each(function(){ console.log($(this).val()); }); });
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script> <div class="radio" id="form1"> <label>Question 1 <input type="radio" class="optionChecked" name="name1" id="'id1" value="value1" data-id="option1" data-title="question1"></label><br> <label>Question 2 <input type="radio" class="optionChecked" name="name2" id="'id2" value="value2" data-id="option2" data-title="question2"></label><br> <label>Question 3 <input type="radio" class="optionChecked" name="name3" id="'id3" value="value3" data-id="option3" data-title="question3"></label><br> <label>Question 4 <input type="radio" class="optionChecked" name="name4" id="'id4" value="value4" data-id="option4" data-title="question4"></label><br> <label>Question 5 <input type="radio" class="optionChecked" name="name5" id="'id5" value="value5" data-id="option5" data-title="question5"></label><br> <button type="button" id="submit"> Submit</button> </div>
相关问答
更多-
JavaScript无线电输入(JavaScript radio input)[2022-06-26]
document.querySelector('.rad')只能引用第一个单选按钮(一个是19 ,而另一个则不是)。 你可能想要事件代表团 。 替换document.querySelector('.rad').addEventListener("change", nextQuestion); 通过document.getElementById("question").addEventListener("change", nextQuestion); 。 这是因为change事件冒泡到question元素。 ... -
在Deif的解决方案上构建,这将在单击div时切换检查状态 小提琴Some content