一、基于form表单上传文件
1、html里是有一个input type="file" 和 ‘submit’的标签
2、vies.py
def fileupload(request): ???if request.method == ‘POST‘: ???????print(request.POST) ???????print(request.FILES) ???????# from django.core.files.uploadedfile import InMemoryUploadedFile ???????print(type(request.FILES.get(‘myfile‘))) ???????myfile = request.FILES.get(‘myfile‘) ???????name = myfile.name ???????print(name) ???????with open(name,‘wb‘) as f: ???????????# for line in myfile.chunks(): ???????????# ????f.write(line) ???????????for line in myfile: ???????????????f.write(line) ???????return HttpResponse(‘文件上传成功‘)
二、基于JSON上传文件
$("#ajax_button").click(function () { ???????var formdata=new FormData() ???????formdata.append(‘name‘,$("#id_name2").val()) ???????formdata.append(‘myfile‘,$("#myfile")[0].files[0]) ???????$.ajax({ ???????????url:‘‘, ???????????type:‘post‘, ???????????processData:false, //告诉jQuery不要去处理发送的数据 ???????????contentType:false,// 告诉jQuery不要去设置Content-Type请求头 ???????????data:formdata, ???????????success:function (data) { ???????????????console.log(data) ???????????} ???????}) ???})def fileupload(request): ???if request.method == ‘POST‘: ???????print(request.POST) ???????print(request.FILES) ???????# from django.core.files.uploadedfile import InMemoryUploadedFile ???????print(type(request.FILES.get(‘myfile‘))) ???????myfile = request.FILES.get(‘myfile‘) ???????name = myfile.name ???????print(name) ???????with open(name,‘wb‘) as f: ???????????# for line in myfile.chunks(): ???????????# ????f.write(line) ???????????for line in myfile: ???????????????f.write(line) ???????return HttpResponse(‘文件上传成功‘)
三、综上所述
基于form表单上传文件和基于ajax上传文件后端view界面一样的不需要改动，前台需呀改动
django 基于form表单上传文件和基于ajax上传文件
原文地址：https://www.cnblogs.com/di2wu/p/10068511.html