Search code examples
rubyjekyllgithub-pagesjekyll-bootstrap

Jekyll private deployment?

I have created jekyll site. Regarding the deployment I don't want to host on github pages. To host private domain I came know from documentation to copy the all files from _site folder. That's all wicked.

Question:

  1. Each time I add new blog post, I am running cmd>jekyll build then I am copying newly created html to hosted domain. Is there any easy way to update without compiling each time ?

  2. The reason, Why I am asking is because it will updated by non technical person

Thanks for the help!!

Solution

  • If you don't want to use GitHub Pages, AFAIK there's no other way than to compile your site each time you make a change.

    But of course you can script/automate as much as possible.
    That's what I do with my own blog as well. I'm hosting it on my own webspace instead of GitHub Pages, so I need to do these steps for each update:

    1. Compile on local machine
    2. Upload via FTP

    I can do this with a single click (okay, a single double-click).

    Note: I'm on Windows, so the following solution is for Windows.
    But if you're using Linux/MacOS/whatever, of course you can use the tools given there to build something similar.

    I'm using a batch file (the Windows equivalent to a shell script) to compile my site and then call WinSCP, a free command-line FTP client.

    WinSCP allows me to store session configurations, so I saved the connection to my server there once.
    Because of this, I didn't want to commit WinSCP to my (public) repository, so my script expects WinSCP in the parent folder.

    The batch file looks like this:

    call jekyll build

echo If the build succeeded, press RETURN to upload!

pause

set uploadpath=%~dp0\_site
%~dp0\..\winscp.com /script=build-upload.txt /xmllog=build-upload.log

pause

    The first parameter in the WinSCP call (/script=build-upload.txt) specifies the script file which contains the actual WinSCP commands

    This is in the script file:

    option batch abort
option confirm off

open blog
synchronize remote -delete "%uploadpath%"

close
exit

    Some explanations:

    1. %~dp0 (in the batch file) is the folder where the current batch file is
    2. The set uploadpath=... line (in the batch file) saves the complete path to the generated site into an environment variable
    3. The open blog line (in the script file) opens a connection to the pre-saved session configuration (which I named blog)
    4. The synchronize remote ... line (in the script file) uses the synchronize command to sync from the local folder (saved in %uploadpath%, the environment variable from step 2) to the server.

    IMO this solution is suitable for non-technical persons as well.
    If the technical person in your case doesn't know how to use source control, you could even script committing & pushing, too.