Programmatically fetching the current Zoom version number | Community
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Newcomer
November 5, 2023
Question

Programmatically fetching the current Zoom version number

  • November 5, 2023
  • 4 replies
  • 30 views

I used a simple script to find the current Zoom version number in the page "Release Notes for Linux" (https://support.zoom.com/hc/en/article?id=zm_kb&sysparm_article=KB0068973). Recently, the page has changed to be JavaScript heavy and the latest version can't be extracted from it programmatically (or at least, I can't do it). Is there some place where I can get the version number?
I use it for later automation of installation of the client on several different machines.

    4 replies

    Ray_Harwood
    Community Super Champion | Customer
    Community Super Champion | Customer
    November 6, 2023

    This is a question that might be more appropriate for the Developers Forum at https://developers.zoom.us/

    Newcomer
    January 24, 2024

    Does https://zoom.us/client/latest/zoom_amd64.deb not work? I've been using that URL in a bash script for years.

    Newcomer
    March 21, 2024

    I am currently in this same situation. I hope zoom will add some way of doing this. API? Static release page?

    Were you able to resolve this?

    I may have to use python. 

    Ray_Harwood
    Community Super Champion | Customer
    Community Super Champion | Customer
    March 21, 2024

    Welcome to the Zoom Community, @ZM6

     

    I can’t predict what Zoom will/won’t do. 

    You might ask in the Developers Forum at https://developers.zoom.us/

    Newcomer
    April 7, 2024

    Same problem, but was parsing a different page. I fixed it this way. The the "latest" URL redirects to the actual version download, which has the version number in the URL.

     

    So with curl

    curl -sI https://zoom.us/client/latest/zoom_amd64.deb | awk -F/ '/location/{ print $5 }'

     

    Or python

    import requests response = requests.head('https://zoom.us/client/latest/zoom_amd64.deb') version = response.headers['location'].split('/')[-2]