如何在用户inpu上重新加载表单无效的页面

Question

我正在使用django为新用户创建一个注册表单。如果用户输入无效，我希望页面重新加载无效表单和自定义错误消息。因此，我基本上是想找到一种方法，在else语句上返回无效表单和页面重载

<div id="signupboxelementscontainer">
     <form action="" method="post" name="signupform">
       {% csrf_token%}
     <div id="namerow" class="row">
       <input name="firstname"id="firstname" class="formelements" placeholder="First name" type="text">
       <input name="lastname" id="lastname" class="formelements" placeholder="Last name" type="text">
     </div>
     <input name="email" id="email" class="formelements" placeholder="Email address" type="email">
     <br>
    <input name="storename" id="storename" class="formelements" placeholder="Your store name" type="text">
     <input  name="password" id="password" class="formelements" placeholder="Password" type="password">
     <br>
     <input name="passwordagain" id="passwordagain" class="formelements" placeholder="Retype password" type="password">
     <br>
     <div class="row">
     </div>
     <button id="submitbutton" type="submit"> Submit</button>
     </div>
     <p>Already have an account? <a href="{% url 'login'%}">Sign in</a> </p>
     <p id="error">{{ error_message }}</p>
   </div>

def signup(request):
if request.method == "POST":
    signupform = signupform(request.POST)
    first_name = request.POST.get("firstname")
    last_name = request.POST.get("lastname")
    email = request.POST.get("email")
    storename = request.POST.get("storename")
    password = request.POST.get("password")
    passwordagain = request.POST.get("passwordagain")
    if password == passwordagain:
        user = User.objects.create_user(first_name=first_name,last_name=last_name,email = email, password = password, username=storename)
    else:
        error_message = "Your passwords do not match"
        return render(request,'main/signup.html',  {"error_message": error_message,"signupform": signupform,})
return render(request, "main/signup.html",)

Answer 1

尝试使用Django的表单库。参见https://docs.djangoproject.com/en/2.1/topics/forms/上的示例

这将做你想做的，并使你的代码更简单。其他的事情会复杂得多