如果想要ajax提交数据到django后台，对页面进行跳转，需要在ajax 中进行跳转，在后台进行跳转不生效
示例代码
views部分
rom django.shortcuts import render, redirect, HttpResponse
# Create your views here.
from day60app.form_cls import AForm
import json
from django.forms.utils import ErrorDict
def ajax(request):
if request.method == "GET":
obj = AForm()
return render(request, "ajax.html", {"obj": obj})
elif request.method == "POST":
obj = AForm(request.POST)
ret = {"status": None, "message": None}
print("31----", obj)
if obj.is_valid():
ret["status"] = "成功"
print("35", ret)
return HttpResponse(json.dumps(ret))
else:
# err = obj.errors
ret["message"] = obj.errors
return HttpResponse(json.dumps(ret))
from 部分
from django.forms import Form
from django.forms import widgets
from django.forms import fields
from django.forms.models import ModelChoiceField
from day60app import models
class AForm(Form):
user_id = fields.IntegerField(
widget=widgets.Select(choices=[(1, "赖英"), (2, "春云"), (3, "星星")])
)
age = fields.IntegerField()
前端页面部分
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Title</title>
</head>
<body>
<form id="fm" action="/ajax/" method="post" novalidate>
{{ obj.as_p }}
<input type="button" value="ajax提交" id="but">
</form>
<script src="/static/jquery-3.1.1.js"></script>
<script>
$(function () {
$("#but").click(function () {
$.ajax({
url:"/ajax/",
type:"POST",
data:$("#fm").serialize(),
dataType:"JSON",
success:function(arg){
console.log("26----",arg);
if (arg["status"] == "成功"){
window.location.href = "http://www.jd.com"
}
{# else:#}
}
})
})
})
</script>
</body>
</html>
时间： 2024-08-06 23:48:53