在我的Django 1.11模板中播放mp3文件

Question

所以我在播放我的Django应用程序中的mp3文件时遇到了问题。

settings.py - 增加MEDIA_ROOT和MEDIA_URLS

MEDIA_ROOT = os.path.join(BASE_DIR, 'uploads') 
MEDIA_URL = '/uploads/' 

models.py - 我有一个简单的音频类设置

class Audio(models.Model): 
    author = models.ForeignKey('auth.User') 
    title = models.CharField(max_length=200) 
    audio_file = models.FileField(upload_to='audio/', blank=True) 
    description = models.TextField() 
    featured_image = models.URLField(max_length=500) 
    created_date = models.DateTimeField(default=timezone.now) 
    published_date = models.DateTimeField(blank=True, null=True) 

views.py - 这是我的看法

class AudioListView(ListView): 
context_object_name = 'audio' 
template_name = 'media/audio/audio_list.html' 
model = Audio 

audio_list.html - 这是我的模板

{% block content %} 
{% for aud in audio %} 
<p>Audio Title: {{ aud.title }}</p> 
<img src="{{ aud.featured_image}}" alt=""> 
<hr> 
<audio controls> 
<source src="{{ aud.audio_file }}" type="audio/mp3"> 
</audio> 
<hr> 
<p>Audio Description: {{ aud.description }}</p> 
{% endfor %} 
{% endblock %} 

我可以进去，并添加音频文件，并将其上传到以下目录

# myapp -> uploads -> audio -> audio.mp3 

通过以下设置HTML5的音频播放器甚至不显示了。当我检查页面的源时，我得到以下内容：

<audio controls> 
<source src="audio/audio.mp3" type="audio/mp3"> 
</audio> 

我梳理了Django文档。我搜遍了谷歌。我不认为这会很难拉出一个文件路径并播放它。

任何意见将不胜感激！

编辑：加入我的网址文件

# mainapp/urls.py 
from django.conf.urls import include, url 
from django.contrib import admin 
from django.contrib.flatpages import views 
from home.views import HomeTemplateView 


urlpatterns = [ 
url(r'^admin/', admin.site.urls), 
url(r'', include('home.urls')), 
url(r'^blog/', include('blog.urls')), 
url(r'^media/', include('media.urls')), 
url(r'^about/', views.flatpage, {'url': '/about/'}, name='about'), 
url(r'^contact/', views.flatpage, {'url': '/contact/'}, name='contact'), 
url(r'^donations/', views.flatpage, {'url': '/donations/'}, 
name='donations'),] 

# media/urls.py 
from django.conf import settings 
from django.conf.urls.static import static 
from django.conf.urls import include, url 
from media.views import AudioListView, VideoListView 

urlpatterns = [ 
url(r'^video/$', VideoListView.as_view(), name='video'), 
url(r'^audio/$', AudioListView.as_view(), name='audio'), 
] 

Answer 1

第一：你需要让所有音频文件

class AudioListView(ListView): 
    template_name = 'media/audio/audio_list.html' 
    model = Audio 

    def get_context_data(self, **kwargs): 
     context = super(AudioListView, self).get_context_data(**kwargs) 
     context['audio'] = Audio.objects.all() 
     return context 

然后，你需要的网址：

<source src="{{ aud.audio_file.url }}" type="audio/mp3"> 

更改目录的权限以便在页面上查看玩家。

chmod a+r -R uploads