Django实现文件上传、下载

Django实现文件上传

1. 配置setting文件

1.1. 在setting里面配置文件上传路径,在setting末尾添加

#指定上传文件的存储相对路径(读取文件)
MEDIA_URL = '/media/'
#上传文件存储路径(存储路径)
MEDIA_ROOT = os.path.join(BASE_DIR,'media')

1.2. 在项目根目录新建media文件夹
1.3. 配置TEMPLATES,添加如下内容

'django.template.context_processors.media'

添加完如下所示

TEMPLATES = [
    {
        'BACKEND': 'django.template.backends.django.DjangoTemplates',
        'DIRS': [os.path.join(BASE_DIR, 'templates')]
        ,
        'APP_DIRS': True,
        'OPTIONS': {
            'context_processors': [
                'django.template.context_processors.debug',
                'django.template.context_processors.request',
                'django.contrib.auth.context_processors.auth',
                'django.contrib.messages.context_processors.messages',
                'django.template.context_processors.media'
            ],
        },
    },
]

2. 配置父路由urls

2.1. 将父路由映射到子路由的路径

urlpatterns = [
    path('admin/', admin.site.urls),
    path('student/', include('stu.urls'))
]

2.2. 配置通过数据加载加载图片

from django.views.static import serve
from .settings import MEDIA_ROOT, DEBUG

if DEBUG:
    urlpatterns += url(r'^media/(?P<path>.*)/$', serve, {'document_root': MEDIA_ROOT}),

3. 配置子路由urls

urlpatterns = [
    url(r'^$', views.index_view),
    url(r'^upload/', views.upload_view),
    url(r'^showall/', views.showall_view)
]

4. 构建模型数据库类models

class Student(models.Model):
    sno = models.AutoField(primary_key=True)
    sname = models.CharField(max_length=30)
    photo = models.ImageField(upload_to='imgs')

    def __str__(self):
        return u'Student:{}'.format(self.sname)

5.视图逻辑处理views

from django.http import HttpResponse
from django.shortcuts import render

# Create your views here.
from stu.models import Student


def index_view(request):
    if request.method == 'GET':
        return render(request, 'index.html')
    else:
        return HttpResponse('页面有误!')


def upload_view(request):
    uname=request.POST.get('uname','')
    photo=request.FILES.get('photo','')
    Student.objects.create(sname=uname,photo=photo)

    return HttpResponse('上传成功')


def showall_view(request):
    student=Student.objects.all()

    return render(request,'showall.html',{'student':student})

6.前端页面

6.1. index.html

<form action="/student/upload/" method="post" enctype="multipart/form-data">
        {% csrf_token %}
        <p>
            <label for="ua">姓名:</label>
            <input type="text" name="uname" id="ua">
        </p>
        <p>
            <label for="ph">头像:</label>
            <input type="file" name="photo" id="ph">
        </p>
        <p>
            &emsp;&emsp;&emsp;&emsp;<input type="submit" value="注册">
        </p>
    </form>

6.2. showall.html

<table width="500px" border="1" cellspacing="0">
    <tr>
        <th>学号</th>
        <th>姓名</th>
        <th>照片</th>
        <th>操作</th>
    </tr>
    {% for foo in student %}
    <tr>
        <td>{{ forloop.counter}}</td>
        <td>{{foo.sname}}</td>
        <td><img src="{{MEDIA_URL}}{{ foo.photo }}" style="width: 200px;"></img></td>
        <td>下载</td>
    </tr>
    {% endfor %}
</table>

Django文件下载

def download_view(request):
    photo = request.GET.get('photo', '')
    filename = photo[photo.rindex('/') + 1:]
    system_path = os.getcwd()
    print(photo)
    path = os.path.join(system_path, photo[1:].replace('/', '\\'))
    print(path)
    with open(path, 'rb') as fr:
        reponse = HttpResponse(fr.read())
        reponse['Content-Type'] = 'image/png'

        reponse['Content-Disposition'] = 'attachment;filename=' + filename

    return reponse
posted @ 2020-08-14 15:51  李东阳  阅读(988)  评论(0编辑  收藏  举报