I'm looking for a method of adding a custom shortcut in Fedora (17). /g/ was no help, and seemed to think I don't know how to use google, but the only things I have managed to find is 1) A method of doing it for commands that are already extant, 2) A method of doing it on Ubuntu, or 3) http://pnmyers.blogspot.co.uk/2012/05/fedora-17-no-more-launch-terminal.html
Unfortunately, I have absolutely no idea what the code in 3 means, or what to do with it. All I'm trying to do is have something like ctrl + t to open terminal, ctrl + f to open firefox, etc.
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Anonymous2012-06-04 14:36
Your window or context manager should provide support, maybe graphical, for shortcuts.
I don't actually know; it's a fresh install of Fedora, so whatever comes with it by default. I've only ever used Ubuntu before, and even then I treated it like Windows.
Not sure what you mean by 'on gnome'? I tried that command with alt f2, and it said it command not found, and I got the same thing in terminal?
Following a lead on >>2 I have installed openbox and found a file in /etc/xdg/openbox/rc.xml with code that looks like " <keybind key="A-F4">
<action name="Close"/>
</keybind>
<keybind key="A-Escape">
<action name="Lower"/>
<action name="FocusToBottom"/>
<action name="Unfocus"/>
</keybind>
<keybind key="A-space">
<action name="ShowMenu"><menu>client-menu</menu></action>
</keybind>"
I'm assuming I could edit this file to do what I'm trying to get at? If so, what would I do to, for example, open firefox with ctrl + f?
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Anonymous2012-06-04 14:56
>>6
if you are not on gnome you must be in either kde or xfce, google how to set keyboard shortcuts on those window managers
openbox's keybindings won't be listened to if your session is using a different wm.
Is there a way of knowing for sure whether it's in kde or xfce? Something like a command in terminal that will give me a straightforward answer? Though I only downloaded the live CD today, and I am positive it was the Gnome one. I'm not sure how I could have fucked that up.
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Anonymous2012-06-04 15:03
>>8
if you open a terminal and type 'gno' and the 'tab' key, for auto complete, you should see a list of programs installed in the interesting directories that begin with 'gno', same with 'xf', 'kde' and so on, since window managers usually install little binaries that facilitate your use through a different name, like a default way of calling the program set as default in that wm.
I get plenty of things listed for the gnome one, but zero for the others. Considering the 'gnome-keybinding-properties' did nothing, does this mean I fucked up the installation?
I'll give it a read. Before I do, one last question and then I'll leave you alone; I've tried the same thing with terminal, but it doesn't seem to come up. Is there a reason for this? Nevermind if it'd be too long-winded to explain; you've helped enough.
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Anonymous2012-06-04 15:59
>>23
what do you mean, if you write firefox on a terminal nothing happens?
I mean, adding the shortcut "Firefox" with the command "firefox" opens up FF just like I want it to, but adding the shortcut "Terminal" with the command "terminal returns nothing". I was using the shortcut ctrl + t if it matters.
There is a transfinite sequence of cardinal numbers:
0, 1, 2, 3, ..., n; ℵ0, ℵ0, ℵ1, ... ℵα, ...
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Anonymous2013-08-31 23:31
(κ ≤ μ) → (κν ≤ μν).
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Anonymous2013-09-01 0:16
It is therefore assumed by physicists that no measurable quantity could have an infinite value, for instance by taking an infinite value in an extended real number system, or by requiring the counting of an infinite number of events.
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Anonymous2013-09-01 1:01
Each set in this hierarchy is assigned (by transfinite recursion) an ordinal number α, known as its rank. The rank of a pure set X is defined to be the least upper bound of all successors of ranks of members of X.
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Anonymous2013-09-01 1:47
Many discussions of "classes" in the 19th century and earlier are really referring to sets, or perhaps to a more ambiguous concept.
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Anonymous2013-09-01 2:32
But some subsets of the real numbers do not have least elements. For example, the open interval (0,1) does not have a least element: if x is in (0,1), then so is x/2, and x/2 is always strictly smaller than x. So this attempt also fails.
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Anonymous2013-09-01 3:17
If S is a set of sentences of first-order logic and B is a consistent subset of S, then B is included in a set that is maximal among consistent subsets of S. The special case where S is the set of all first-order sentences in a given signature is weaker, equivalent to the Boolean prime ideal theorem; see the section "Weaker forms" below.