django实现文件上传
一、实现文件上传
1定义模型
#上传作品表:作品名称，用户名
class Upload(models.Model):
#注意这里的用户名是FileField类型
name = models.FileField(upload_to = './upload/')
username = models.CharField(max_length = 20)
def __str__(self):
return self.name
2定义模版
<form method="post" enctype="multipart/form-data" >
{{uf.as_p}}
<input type="submit" value="上传"/>
</form>
3定义视图
def user(request):
print "user"
if request.method == "POST":
uf = UploadFileForm(request.POST,request.FILES)
if uf.is_valid():
#获取表单信息
#username = uf.cleaned_data['username']
headImg = uf.cleaned_data['headImg']
print str(headImg)
#写入数据库
upload = Upload()
upload.name = headImg
upload.username = request.user.username
upload.save()
return HttpResponse('upload ok!')
else:
return HttpResponse("表单验证失败")
else:
uf = UploadFileForm()
return render_to_response('user.html',{'uf':uf , "user":request.user})
4配置url
url(r'^user/$' , 'account.views.user'),
这样就实现了文件上传功能
二、展示上传的文件
(例如图片等)
根据django官方文档的说明:https://docs.djangoproject.com/en/1.7/topics/files/#using-files-in-models
我们定义的模型中一个字段name是文件字段，但是其实存储的是上传的图片或文件的路径
name = models.FileField(upload_to = './upload/')
我们需要配置upload上传路径
1在settings.py中添加
MEDIA_ROOT = os.path.join(BASE_DIR, 'upload') MEDIA_URL = '/upload/'
2在url中配置路径：
达成的效果是: 让 MEDIA_URL 等同于 MEDIAA_ROOT
from django.conf import settings
from django.conf.urls.static import static
urlpatterns = patterns('',
# Examples:
url(r'^admin/', include(admin.site.urls)),
) + static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)
注意:在 patterns(....)后面加上 static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)
3在模板中设置图片或文件路径
<body>
{% if upload %}
<!--在你的字段中获取url属性进行展示-->
<img src="{{upload.name.url}}" />
{% endif %}
</body>
关键在于: 比如你从后台获取查询的数据叫做upload , upload.name是FileField的字段，只需要获取其url属性进行展示即可