<pre class="prettyprint lang-none prettyprint-override"><code>file:///home/ashu/Music/Collections/randomPicks/ipod%20on%20sep%2009/Coldplay-Sparks.mp3 </code></pre> <p>How can I convert a string like the above to get the normal file path which I can pass to <code>open()</code> function?</p>

<p>This is called unquote. Avaiable from urllib.</p> <pre class="prettyprint"><code>import urllib urllib.unquote('%20') </code></pre>

How to remove %20 from the file path?

file:///home/ashu/Music/Collections/randomPicks/ipod%20on%20sep%2009/Coldplay-Sparks.mp3

How can I convert a string like the above to get the normal file path which I can pass to open() function?

310

asked Nov 27 '22 22:11

Have a look at url2pathname:

import urllib2

path = urllib2.url2pathname("file:///home/ashu/Music/Collections/randomPicks/ipod%20on%20sep%2009/Coldplay-Sparks.mp3")

162

answered Dec 06 '22 15:12

This is called unquote. Avaiable from urllib.

import urllib
urllib.unquote('%20')

answered Dec 06 '22 15:12

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