如何通过django(python)中的http响应播放音频文件 [英] How to play a audio file through http response in django(python)
本文介绍了如何通过django(python)中的http响应播放音频文件的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
现在我硬编码代码中的文件名。
url:http:// localhost / playfile / audiofile_name
def playAudioFile(request):
try:
fname =C:\\test\\audio\\t.mp3
wrapper = FileWrapper (file(fname))
print content_type
response = HttpResponse(wrapper,content_type =audio / mpeg)
打印响应
响应['Content-Length'] = os .path.getsize(fname)
返回响应
除了:
返回HttpResponse()
提前感谢
解决方案
我找到答案.....
def playAudioFile(request):
fname =C:\\test\\audio\\audio.mp3
f = open(fname,rb)
response = HttpResponse()
响应。 write(f.read())
response ['Content-Type'] ='audio / mp3'
response ['Content-Length'] = os.path.getsize(fname)
返回响应
I want to make request to url and django view should read the file and send the http response back to play the same file in browser.I got the following code but it does't play anything please anyone help me.. Right now i hard coded the file name in the code.
url: http://localhost/playfile/audiofile_name
def playAudioFile(request):
try:
fname="C:\\test\\audio\\t.mp3"
wrapper = FileWrapper(file(fname))
print content_type
response = HttpResponse(wrapper, content_type="audio/mpeg")
print response
response['Content-Length'] =os.path.getsize(fname )
return response
except:
return HttpResponse()
Thank in advance..
解决方案
I found the answer.....
def playAudioFile(request):
fname="C:\\test\\audio\\audio.mp3"
f = open(fname,"rb")
response = HttpResponse()
response.write(f.read())
response['Content-Type'] ='audio/mp3'
response['Content-Length'] =os.path.getsize(fname )
return response
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