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hello,
i have the problem, that when i want to remove many packets i must type the whole name.
for example: kdemod-kdegraphics-kpdf
can i write something like kdemo*kpdf or something other?
lg
ps
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Why don't you use bash or zsh tab completion to make that much easier?
pacman roulette : pacman -S $(pacman -Slq | LANG=C sort -R | head -n $((RANDOM % 10)))
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this doesn't work. nothing happen when i use tab. -or what should i do?
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I suppose you use bash then? In this case, try this :
http://wiki.archlinux.org/index.php/Bas … Completion
pacman roulette : pacman -S $(pacman -Slq | LANG=C sort -R | head -n $((RANDOM % 10)))
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ah.. this was it. thanks... so easy if you know it ![]()
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hej,
i cleaup my system at the moment, and bash-compleshion isn't enough...
for example when i go to my starmenu and see kppp, which i don't need any more. but i can't write pacman -R *kppp*. i must write first pacman -Ss kppp and than i must write pacman -R kdemod[...]kppp
is there an other alternative???
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You can try this
pacman -Rs $(pacman -Qsq kppp)This just combines the two commands you need into one. But it only works when the search is precise enough, and only returns the packages you want to remove.
pacman roulette : pacman -S $(pacman -Slq | LANG=C sort -R | head -n $((RANDOM % 10)))
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hm, this looks not easy to use, but its much better.. thanks.
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You could always add this to your user .bashrc :
pacR() {
sudo pacman -Rs $(pacman -Qsq $1)
}or the same without sudo to your root .bashrc, if you run pacman as root.
Then you can just do
pacR kppppacman roulette : pacman -S $(pacman -Slq | LANG=C sort -R | head -n $((RANDOM % 10)))
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hehe, this looks nice... i didn't thought that such is so easy with the bash... very cool ![]()
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