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Asked: 2026-05-10T22:04:27+00:00

Edit: How to return/serve a file from a python controller (back end) over a

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Edit: How to return/serve a file from a python controller (back end) over a web server, with the file_name? as suggested by @JV

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  1. Fully supported in CherryPy using

    from cherrypy.lib.static import serve_file

    As documented in the CherryPy docs – FileDownload:

    import glob import os.path  import cherrypy from cherrypy.lib.static import serve_file   class Root:     def index(self, directory='.'):         html = '''<html><body><h2>Here are the files in the selected directory:</h2>         <a href='index?directory=%s'>Up</a><br />         ''' % os.path.dirname(os.path.abspath(directory))          for filename in glob.glob(directory + '/*'):             absPath = os.path.abspath(filename)             if os.path.isdir(absPath):                 html += '<a href='/index?directory=' + absPath + ''>' + os.path.basename(filename) + '</a> <br />'             else:                 html += '<a href='/download/?filepath=' + absPath + ''>' + os.path.basename(filename) + '</a> <br />'          html += '''</body></html>'''         return html     index.exposed = True  class Download:         def index(self, filepath):         return serve_file(filepath, 'application/x-download', 'attachment')         index.exposed = True  if __name__ == '__main__':     root = Root()     root.download = Download()     cherrypy.quickstart(root)
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