我的程序的目标是在输入数据时将名字和姓氏放入我的数据库,并由用户点击Submit
按钮,这样当我手动检入Terminal
数据库中的内容时,我可以看到用户输入和提交的内容。我需要了解如何将我的Submit
按钮连接到我的views.py
文件以便它可以通过,排序类似于onClick()
,但这次要转到.py
文件。 (如果我对这段思路错了,请纠正我。)
我将如何实现这一目标?
这是我的views.py
文件:
from django.http import HttpResponse
from django.shortcuts import render
from .models import Person
def index(request):
if request.method == 'POST':
first_name = request.POST.get('firstName')
last_name = request.POST.get('lastName')
if first_name and last_name:
user = Person.objects.create(firstName=first_name, lastName=last_name)
user.save()
return render(request, 'music/index.html')
def detail(request, user_id): # Testing out page 2
return HttpResponse("<h2>Page # (testing this out) " + str(user_id) + "</h2>")
这是我的index.html
文件:
<!DOCTYPE html>
<html lang="en">
<head>
<title>The Page</title>
<meta charset="utf-8">
<meta name="viewport" content="width=device-width, initial-scale=1">
<link rel="stylesheet" href="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/css/bootstrap.min.css">
<link rel="stylesheet" href="index.css">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
<script src="https://maxcdn.bootstrapcdn.com/bootstrap/3.3.7/js/bootstrap.min.js"></script>
</head>
<body>
<div class="container">
<form action="#">
<div class="form-group">
<label for="firstName">First Name:</label>
<input type="email" class="form-control" id="firstName" placeholder="Enter first name" name="firstName">
</div>
<div class="form-group">
<label for="">Last Name:</label>
<input type="email" class="form-control" id="lastName" placeholder="Enter last name" name="lastName">
</div>
</form>
<div class="checkbox">
<label><input type="checkbox" name="remember">Remember me</label></div></br>
<button type="submit" class="btn btn-default">Submit</button>
</div>
</div>
</body>
</html>
答案 0 :(得分:1)
您的表单应声明为:
<form method="POST">
,您的<button type="submit">
应位于<form></form>
。
由于同一视图处理GET和POST方法,您应该删除action="#"
属性,这样操作将指向同一视图。