在django模板中播放<audio> </ audio>文件

Question

我一直在努力解决这个问题,以至于我接近抑郁症.

我有一个名为"Song"的模型,看起来像这样.

from django.db import models

class Song(models.Model):
    title = models.CharField(max_length=100)
    songfile = models.FileField()
    duration = models.FloatField()
    isPlaying = False

    def __str__(self):
        return self.title

当您从索引页面上传mp3文件时,它会创建此模型的实例,并使用此视图将文件存储在myapp/songdir /中:

def home(request):

    if request.method == 'POST':
        form = UploadForm(request.POST, request.FILES)
        if form.is_valid():
            song_title = request.POST.items()[1][1]
            song_address = 'upnplay/songdir/' + song_title + '.mp3'
            with open(song_address, 'wb+' ) as destination:
                for chunk in request.FILES['file'].chunks():
                    destination.write(chunk)
                audio = MP3(song_address)
                c = Song(title = song_title, songfile = song_address, duration = audio.info.length)
                c.save()
            return HttpResponseRedirect('')
    else:
        form = UploadForm()
    c = {'form': form}
    c.update(csrf(request))
    return render(request, 'index.html', {'form': form})

然后我有一个名为"choosesong"的模板,显示我从保存的模型实例中获得的歌曲列表:

{% extends 'index.html' %}


{% block content %}
<div class="row">
    {% for song in playlist %}
        <a href="playlist/{{song.title}}/"><h3>{{song.title}} -- {{song.duration}}</h3></a>
    {% endfor %}
</div>

{% endblock %}
{% block form %}{% endblock %}

当我点击其中一个链接时,我希望渲染一个新模板,其中一个元素播放我点击其名称的歌曲.我呈现的模板是这样的:

{% extends 'index.html' %}

{% block content %}

<div class='row'>
    {{link}}
    <audio controls>
        <source src="../../{{ link }}" type="audio/mpeg">
        Your browser does not support the audio element.
    </audio>
</div>

{% endblock %}

我用来提供它的视图如下:

def playAudioFile(request, songtitle):
    name = urllib.unquote(songtitle)
    song = get_object_or_404(Song, title=name )
    return render(request, 'playlist.html', {'link': song.songfile })

出于某种原因,我不能让它在音频元素中播放歌曲,我不知道还有什么可以尝试.

先谢谢你了.

Answer 1

您应该添加MEDIA_ROOT和MEDIA_URL配置.处理事情会很容易.这是您的问题的解决方案.

在settings.py中:

MEDIA_ROOT=os.path.join(BASE_DIR,"songdir")
MEDIA_URL='/media/'

同样在settings.py中,在TEMPLATES选项的context_processors中添加"django.template.context_processors.media".

在project/urls.py中:

from django.conf import settings
from django.conf.urls.static import static
urlpatterns += static(settings.MEDIA_URL, document_root=settings.MEDIA_ROOT)

然后你可以简单地使用:

{{link.url}} 

而不是在模板文件中硬编码.