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Just got setup with Awesome, after initial what-on-earth-is-this-thing confusion, I find myself liking it ;-)

One issue that I'm struggling with is a feature that many likely prefer in a multi monitor environment: switching between workspaces in one screen does not switch workspaces in other screens.

In my screen real estate limited laptop + external monitor (15"/23") setup with Compiz, I grouped applications by workspace, spanning both screens; then when switching to another workspace both screens were updated with applications relevant to the target workspace.

I can understand maintaining independent screens with 3+ monitors, but in my setup I find myself switching workspaces much more in the primary monitor than the laptop, which has very much become a secondary and somewhat limited option (as I have to ctr + mod + j over to the laptop, then mod + left/right arrow to wherever the application is that corresponds with application(s) on the primary monitor).

Is there any way to toggle workspaces simultaneously across multiple screens in Awesome? Feeling like my workflow is somewhat disjointed compared to previous Compiz setup.

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Thanks to @Cris9288, this works:

awful.key({ modkey, "Control"   }, "Left", 
    for i = 1, screen.count() do
  end ),

awful.key({ modkey, "Control"   }, "Right", 
    for i = 1, screen.count() do
  end ),
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There's a follow-up question here. Perhaps you can help? – Karan Jun 10 '13 at 17:44
Sorry, using i3 instead of Awesome. Also, realized I actually want independent screen switching ;-) Good luck – virtualeyes Jun 10 '13 at 19:38
Oh it wasn't for me, I was just posting on behalf of a new user who couldn't comment here. I've recommended to him that he ask on the same forum you did. Perhaps he'll find an answer there too, unless of course someone posts a solution for his question here. – Karan Jun 10 '13 at 20:34

As an addition to @virtualeyes, for mod+Ctrl+[number] add within a For i=1, screen.count() do loop:

    awful.key({ modkey, "Control" }, "#" .. i + 9,
              function ()
                    for j = 1, screen.count() do
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