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how to copy multiple files to a directory and simultaneously enter in that directory?

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}

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

share|improve this question

edited 16 hours ago

Davide Luise

asked 16 hours ago

Davide LuiseDavide Luise

183

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  Please format the code using the code icon { } above the edit question window.
  
  – user68186
  16 hours ago
  

  
  
  
  

add a comment | 

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

share|improve this question

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.

• 
  
  
  

  
  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

share|improve this question

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.

share|improve this question

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.

share|improve this question

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

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

1 Answer
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0

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]

share|improve this answer

answered 15 hours ago

EliasElias

2458

add a comment | 

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0

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]

share|improve this answer

answered 15 hours ago

EliasElias

2458

add a comment | 

0

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]

share|improve this answer

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]

share|improve this answer

answered 15 hours ago

EliasElias

2458

cp "$@" "${!#}"

cp "$@"

./yourscript.sh [arguments]

. yourscript.sh [arguments]

share|improve this answer

answered 15 hours ago

EliasElias

2458

answered 15 hours ago

EliasElias

2458

answered 15 hours ago

EliasElias

2458