python中的文件上传

Django实现


form表单实现文件的上传

/form_upload/urls.py

from django.conf.urls import url
from django.contrib import admin
from app01 import views

urlpatterns = [
    url(r'^admin/', admin.site.urls),
    url(r'^upload/', views.upload),
]

/app01/views.py

from django.shortcuts import render
import os
from django.conf import settings
import datetime
# Create your views here.


def upload(request):
    # 1. 第一次GET请求来, 应该给用户返回一个页面
    if request.method == 'POST':
        # print(request.FILES)
        file_obj = request.FILES.get('file_name')  # 文件对象
        # print(type(file_obj))  # <class 'django.core.files.uploadedfile.InMemoryUploadedFile'>
        # print(type(file_obj.name))  # <class 'str'>

        file_name = file_obj.name
        if os.path.exists(os.path.join(settings.BASE_DIR, file_name)):
            # 如果存在重名的文件
            name, suffix = file_name.split('.')
            datatime_obj = datetime.datetime.now()
            year, month, day, hour, minute,second = datatime_obj.year, datatime_obj.month, datatime_obj.day,datatime_obj.hour, datatime_obj.minute, datatime_obj.second
            name += '_%s-%s-%s %s:%s:%s'%(year, month, day, hour, minute, second)
            file_name = '%s.%s'%(name, suffix)


        # 从上传文件对象里 一点一点读取数据, 写到本地
        # with open(file_obj.name, 'wb') as f:
        #     for line in file_obj:
        #         f.write(line)

        # 官方推荐写法, file_obj.chunks() 可以指定一次读多少数据
        with open(file_name, 'wb') as f:
            for chunk in file_obj.chunks(2200):
                f.write(chunk)

    return render(request, 'upload.html')

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posted @ 2019-10-24 17:39  cjw1219  阅读(411)  评论(0编辑  收藏  举报