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how to copy multiple files to a directory and simultaneously enter in that directory?
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.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ margin-bottom:0;
}
I am trying to write a bash function in my .bashrc which copy multiple files in a directory and enter it, the problem is that it copies the files but do not enter the directory and says:"omitted directory"
cpcd() if [ -d "${!#}" ]
then
cp "$@" "${!#}" && cd "${!#}" && ls
else
print "last argument is not a directory or does not exist"
fi
copy
asked 16 hours ago
Davide LuiseDavide Luise
183
New contributor
Davide Luise is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I am trying to write a bash function in my .bashrc which copy multiple files in a directory and enter it, the problem is that it copies the files but do not enter the directory and says:"omitted directory"
cpcd() if [ -d "${!#}" ]
then
cp "$@" "${!#}" && cd "${!#}" && ls
else
print "last argument is not a directory or does not exist"
fi
copy
asked 16 hours ago
Davide LuiseDavide Luise
183
New contributor
Davide Luise is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
Please format the code using the code icon { } above the edit question window.
– user68186
16 hours ago
add a comment |
I am trying to write a bash function in my .bashrc which copy multiple files in a directory and enter it, the problem is that it copies the files but do not enter the directory and says:"omitted directory"
cpcd() if [ -d "${!#}" ]
then
cp "$@" "${!#}" && cd "${!#}" && ls
else
print "last argument is not a directory or does not exist"
fi
copy
asked 16 hours ago
Davide LuiseDavide Luise
183
New contributor
Davide Luise is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am trying to write a bash function in my .bashrc which copy multiple files in a directory and enter it, the problem is that it copies the files but do not enter the directory and says:"omitted directory"
cpcd() if [ -d "${!#}" ]
then
cp "$@" "${!#}" && cd "${!#}" && ls
else
print "last argument is not a directory or does not exist"
fi
copy
copy
asked 16 hours ago
Davide LuiseDavide Luise
183
New contributor
Davide Luise is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 16 hours ago
Davide LuiseDavide Luise
183
New contributor
Davide Luise is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 16 hours ago
Davide Luise
asked 16 hours ago
Davide LuiseDavide Luise
183
New contributor
Davide Luise is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 16 hours ago
Davide LuiseDavide Luise
183
asked 16 hours ago
Davide LuiseDavide Luise
183
183
New contributor
Davide Luise is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Davide Luise is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Davide Luise is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
Please format the code using the code icon { } above the edit question window.
– user68186
16 hours ago
add a comment |
2
Please format the code using the code icon { } above the edit question window.
– user68186
16 hours ago
2
2
Please format the code using the code icon { } above the edit question window.
– user68186
16 hours ago
Please format the code using the code icon { } above the edit question window.
– user68186
16 hours ago
add a comment |
1 Answer
1
active
oldest
votes
When you do this:
cp "$@" "${!#}"
that has the effect that the last argument (the directory name) is repeated.
So, I think what you want is probably this instead (just remove the "${!#}"):
cp "$@"
The other issue is that you say it does not enter the directory. A script that you just execute normally will not change your current working directory; if you want that to happen you need to source the script instead. So, instead of running it like this:
./yourscript.sh [arguments]
you would then source it, like this:
. yourscript.sh [arguments]
answered 15 hours ago
EliasElias
2458
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
When you do this:
cp "$@" "${!#}"
that has the effect that the last argument (the directory name) is repeated.
So, I think what you want is probably this instead (just remove the "${!#}"):
cp "$@"
The other issue is that you say it does not enter the directory. A script that you just execute normally will not change your current working directory; if you want that to happen you need to source the script instead. So, instead of running it like this:
./yourscript.sh [arguments]
you would then source it, like this:
. yourscript.sh [arguments]
answered 15 hours ago
EliasElias
2458
add a comment |
When you do this:
cp "$@" "${!#}"
that has the effect that the last argument (the directory name) is repeated.
So, I think what you want is probably this instead (just remove the "${!#}"):
cp "$@"
The other issue is that you say it does not enter the directory. A script that you just execute normally will not change your current working directory; if you want that to happen you need to source the script instead. So, instead of running it like this:
./yourscript.sh [arguments]
you would then source it, like this:
. yourscript.sh [arguments]
answered 15 hours ago
EliasElias
2458
add a comment |
When you do this:
cp "$@" "${!#}"
that has the effect that the last argument (the directory name) is repeated.
So, I think what you want is probably this instead (just remove the "${!#}"):
cp "$@"
The other issue is that you say it does not enter the directory. A script that you just execute normally will not change your current working directory; if you want that to happen you need to source the script instead. So, instead of running it like this:
./yourscript.sh [arguments]
you would then source it, like this:
. yourscript.sh [arguments]
answered 15 hours ago
EliasElias
2458
When you do this:
cp "$@" "${!#}"
that has the effect that the last argument (the directory name) is repeated.
So, I think what you want is probably this instead (just remove the "${!#}"):
cp "$@"
The other issue is that you say it does not enter the directory. A script that you just execute normally will not change your current working directory; if you want that to happen you need to source the script instead. So, instead of running it like this:
./yourscript.sh [arguments]
you would then source it, like this:
. yourscript.sh [arguments]
answered 15 hours ago
EliasElias
2458
answered 15 hours ago
EliasElias
2458
answered 15 hours ago
EliasElias
2458
answered 15 hours ago
EliasElias
2458
2458
add a comment |
add a comment |
Davide Luise is a new contributor. Be nice, and check out our Code of Conduct.
Davide Luise is a new contributor. Be nice, and check out our Code of Conduct.
Davide Luise is a new contributor. Be nice, and check out our Code of Conduct.
Davide Luise is a new contributor. Be nice, and check out our Code of Conduct.
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2
Please format the code using the code icon { } above the edit question window.
– user68186
16 hours ago