If I have a <code>zipfile.ZipFile</code> object, how can I determine the file path of the zip file from the object? <pre class="prettyprint"><code>a = zipfile.ZipFile('C:\\path\\zipfile.zip') a.get_file_path() </code></pre> where <code>get_file_path()</code> should return <code>'C:\\path\\zipfile.zip'</code>.

You can access it using <code>filename</code> attribute: <pre class="prettyprint"><code>a.filename </code></pre>

How to get the filepath of a ZipFile object?

If I have a zipfile.ZipFile object, how can I determine the file path of the zip file from the object?

a = zipfile.ZipFile('C:\\path\\zipfile.zip')

a.get_file_path()

where get_file_path() should return 'C:\\path\\zipfile.zip'.

365

asked Oct 03 '22 10:10

You can access it using filename attribute:

a.filename

answered Oct 13 '22 10:10

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