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Asked: 2026-06-09T00:37:50+00:00

I am writing following program : ***import os filepath=r’C:\TestData\openfolder.html’ abc=open(filepath,’w’) abc.writelines(‘<html><head></head><body>’) abc.writelines(‘<a href=os.startfile(filepath)>First Link</a>\n’)

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I am writing following program :

***import os
filepath=r'C:\TestData\openfolder.html'
abc=open(filepath,'w')
abc.writelines('<html><head></head><body>')

abc.writelines('<a href="os.startfile(filepath)">First Link</a>\n')

abc.writelines('</body></html>')***

What I want to do is if I click First Link on a browser, I should be able to open the folder having path as “Filepath”. os.startfile works perfect for opening a folder but I don’t know how to implement this inside some link.
Thanks.

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1 Answer

  1. Editorial Team

    Try to use URI with file: scheme like file:///C:/TestData/openfolder.html in your html:

    <a href="file:///C:/TestData/openfolder.html">Link to test data</a>

    Here is article on using file URIs in Windows.

    UPD (extraction from comments): Each browser has its own way to handle such urls. At least Internet Explorer 8 under Windows 7 opens links in Windows Explorer as was required by jags.

    Finally, for dynamic pages the web server is required. If one is needed take a look at discussion on creating simple web services using python.

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