如何在Django中获取文件扩展名？ [英] How to get the extension of a file in Django?

问题描述

我正在Django中构建一个Web应用程序。
我有一个将文件发送到views.py的表单。

I'm building a web app in Django. I have a form that sends a file to views.py.

视图：

@login_required(login_url=login_url)
def addCancion(request):
    if request.method == 'POST':
        form2 = UploadSong(request.POST, request.FILES)
        if form2.is_valid():
            if(handle_uploaded_song(request.FILES['file'])):
                path = '%s' % (request.FILES['file'])
                ruta =  "http://domain.com/static/canciones/%s" % path
                usuario = Usuario.objects.get(pk=request.session['persona'])
                song = Cancion(autor=usuario, cancion=ruta)
                song.save()
                return HttpResponse(ruta)
            else:
                return HttpResponse("-3")
        else:
            return HttpResponse("-2")
    else:
        return HttpResponse("-1")   

我正在尝试仅上传MP3文件，但是我不知道如何制作此过滤器。
我尝试了一个名为 ContentTypeRestrictedFileField（FileField）：的类，但无法正常工作。

I'm trying to upload only the MP3 files, but I don't know how to make this filter. I tried a class named "ContentTypeRestrictedFileField(FileField):" and doesn't work.

如何在views.py中获取文件类型？

How can I get the file type in views.py?

谢谢！

推荐答案

您还可以使用clean（）方法从窗体，用于验证它。因此，您可以拒绝不是mp3的文件。像这样的东西：

You could also use the clean() method from the form, which is used to validate it. Thus, you can reject files that are not mp3. Something like this:

class UploadSong(forms.Form):
    [...]

    def clean(self):
        cleaned_data = super(UploadSong, self).clean()
        file = cleaned_data.get('file')

        if file:
            filename = file.name
            print filename
            if filename.endswith('.mp3'):
                print 'File is a mp3'
            else:
                print 'File is NOT a mp3'
                raise forms.ValidationError("File is not a mp3. Please upload only mp3 files")

        return file

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