Only if in “/” : alias ls='ls -I test'?
I have the following in my ~/.bashrc
to hide the listing of test
from the output of ls
:
alias ls='ls -I test'
But I want it to only hide test
if my current working directory is the root (/
) folder, not if I am in some other folder.
How may I achieve this?
bash ls alias
add a comment |
I have the following in my ~/.bashrc
to hide the listing of test
from the output of ls
:
alias ls='ls -I test'
But I want it to only hide test
if my current working directory is the root (/
) folder, not if I am in some other folder.
How may I achieve this?
bash ls alias
5
If the working directory is the root, or if the directory to be listed is the root? Whilst the former is what you asked, I'm not sure that's what you really want!
– Toby Speight
Mar 2 '17 at 11:06
1
Looks like someone is working on a user-level rootkit? :P
– Navin
Mar 3 '17 at 12:47
This is a bad idea. Better make a different alias and use that.
– reinierpost
Mar 3 '17 at 15:02
add a comment |
I have the following in my ~/.bashrc
to hide the listing of test
from the output of ls
:
alias ls='ls -I test'
But I want it to only hide test
if my current working directory is the root (/
) folder, not if I am in some other folder.
How may I achieve this?
bash ls alias
I have the following in my ~/.bashrc
to hide the listing of test
from the output of ls
:
alias ls='ls -I test'
But I want it to only hide test
if my current working directory is the root (/
) folder, not if I am in some other folder.
How may I achieve this?
bash ls alias
bash ls alias
edited Mar 2 '17 at 10:45
Tomasz
10.1k53068
10.1k53068
asked Mar 2 '17 at 9:00
user218642
5
If the working directory is the root, or if the directory to be listed is the root? Whilst the former is what you asked, I'm not sure that's what you really want!
– Toby Speight
Mar 2 '17 at 11:06
1
Looks like someone is working on a user-level rootkit? :P
– Navin
Mar 3 '17 at 12:47
This is a bad idea. Better make a different alias and use that.
– reinierpost
Mar 3 '17 at 15:02
add a comment |
5
If the working directory is the root, or if the directory to be listed is the root? Whilst the former is what you asked, I'm not sure that's what you really want!
– Toby Speight
Mar 2 '17 at 11:06
1
Looks like someone is working on a user-level rootkit? :P
– Navin
Mar 3 '17 at 12:47
This is a bad idea. Better make a different alias and use that.
– reinierpost
Mar 3 '17 at 15:02
5
5
If the working directory is the root, or if the directory to be listed is the root? Whilst the former is what you asked, I'm not sure that's what you really want!
– Toby Speight
Mar 2 '17 at 11:06
If the working directory is the root, or if the directory to be listed is the root? Whilst the former is what you asked, I'm not sure that's what you really want!
– Toby Speight
Mar 2 '17 at 11:06
1
1
Looks like someone is working on a user-level rootkit? :P
– Navin
Mar 3 '17 at 12:47
Looks like someone is working on a user-level rootkit? :P
– Navin
Mar 3 '17 at 12:47
This is a bad idea. Better make a different alias and use that.
– reinierpost
Mar 3 '17 at 15:02
This is a bad idea. Better make a different alias and use that.
– reinierpost
Mar 3 '17 at 15:02
add a comment |
3 Answers
3
active
oldest
votes
Something like this i thought it would work:
alias ls='[[ "$PWD" = "/" ]] && ls -I test ||ls'
$PWD
is the current working directory&&
has the action to perform if pwd is /
(condition check =true)||
has the action to perform if pwd is not /
(condition check=false)
But after carefull testing above solution IS NOT WORKING correctly.
On the other hand this will work ok as alternative to functions:
alias ls='{ [[ "$PWD" == "/" ]] && a="-I tmp2" ||a=""; };ls $a '
Or even better, similar to other answers but without the need of function:
alias lstest='{ [[ "$PWD" == "/" ]] && set -- "-I" "tmp2" || set --; }; ls "$@"'
Possible extra flags and/or full paths given in command line when invoking this alias are preserved and sent to ls.
7
A problem with this is that usingls
from/
will always result in a listing of the current directory (/
), no matter what the command line says.
– Kusalananda
Mar 2 '17 at 9:30
@Kusalananda You are right , my solution was not correct. I would like you to have a look in my alternative. I tested it and seems to work fine in Debian Bash. Definetely my answer should not have been marked as accepted before , but with the troll presence things got mixed up. I don't know if there is a way to transfer the "solution" mark to your answer as it was in the very beginning (before troll attacks)
– George Vasiliou
Mar 2 '17 at 11:58
1
Ah!alias 'ls={ [[ "$PWD" == "/" ]] && set -- "-I" "tmp2"; }; ls "$@" '
, but that modifies the positional parameters instead...
– Kusalananda
Mar 2 '17 at 12:05
1
alias
es are nasty because they are some forms of macro expansion. Those particular alias can't stand for a realls
command. For instance,! ls
orfoo || ls
won't do what you think. This really calls for a function. Note that the first example would runls
twice if the first one exits with a non-zero exit status.
– Stéphane Chazelas
Mar 2 '17 at 23:05
3
"without the need of function" isn't really a plus. Both of the "corrected" functions unnecessarily overwrite global variables that might be in use.
– chepner
Mar 3 '17 at 3:29
|
show 4 more comments
ls () {
case "$PWD" in
/) command ls -I test "$@" ;;
*) command ls "$@" ;;
esac
}
The above shell function will test the current directory against /
and executes the GNU ls
command differently according to the outcome of the test.
"$@"
will be replaced by the command line options and operands on the original command line.
We need to use command ls
rather than just ls
in the function to bypass the shell function lookup for ls
(which would otherwise give us a nice infinite recursion).
Note that this will not use the ignore pattern if doing ls /
from somewhere else, and that it will use the ignore pattern if doing, e.g., ls /home/dvoo
from /
.
To make it easier to remember what this actually does (six months down the line when you wonder what it's doing), use the long options:
ls () {
case "$PWD" in
/) command ls --ignore='test' "$@" ;;
*) command ls "$@" ;;
esac
}
Alternative implementation of the above function that will only call ls
from one place (and which is shorter):
ls () {
[ "$PWD" = "/" ] && set -- --ignore='test' "$@"
command ls "$@"
}
add a comment |
Use a function that tests if you're in /
for ls
:
ls () {
if [[ "$PWD" == / ]]
then
command ls -I test "$@"
else
command ls "$@"
fi
}
This way, any arguments you pass to ls
will still be used.
Or:
ls () {
if [ "$PWD" == / ]
then
set -- -I test "$@"
fi
command ls "$@"
}
3
That was a nice way of sorting out only having one call tols
in the function!
– Kusalananda
Mar 2 '17 at 10:48
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
Something like this i thought it would work:
alias ls='[[ "$PWD" = "/" ]] && ls -I test ||ls'
$PWD
is the current working directory&&
has the action to perform if pwd is /
(condition check =true)||
has the action to perform if pwd is not /
(condition check=false)
But after carefull testing above solution IS NOT WORKING correctly.
On the other hand this will work ok as alternative to functions:
alias ls='{ [[ "$PWD" == "/" ]] && a="-I tmp2" ||a=""; };ls $a '
Or even better, similar to other answers but without the need of function:
alias lstest='{ [[ "$PWD" == "/" ]] && set -- "-I" "tmp2" || set --; }; ls "$@"'
Possible extra flags and/or full paths given in command line when invoking this alias are preserved and sent to ls.
7
A problem with this is that usingls
from/
will always result in a listing of the current directory (/
), no matter what the command line says.
– Kusalananda
Mar 2 '17 at 9:30
@Kusalananda You are right , my solution was not correct. I would like you to have a look in my alternative. I tested it and seems to work fine in Debian Bash. Definetely my answer should not have been marked as accepted before , but with the troll presence things got mixed up. I don't know if there is a way to transfer the "solution" mark to your answer as it was in the very beginning (before troll attacks)
– George Vasiliou
Mar 2 '17 at 11:58
1
Ah!alias 'ls={ [[ "$PWD" == "/" ]] && set -- "-I" "tmp2"; }; ls "$@" '
, but that modifies the positional parameters instead...
– Kusalananda
Mar 2 '17 at 12:05
1
alias
es are nasty because they are some forms of macro expansion. Those particular alias can't stand for a realls
command. For instance,! ls
orfoo || ls
won't do what you think. This really calls for a function. Note that the first example would runls
twice if the first one exits with a non-zero exit status.
– Stéphane Chazelas
Mar 2 '17 at 23:05
3
"without the need of function" isn't really a plus. Both of the "corrected" functions unnecessarily overwrite global variables that might be in use.
– chepner
Mar 3 '17 at 3:29
|
show 4 more comments
Something like this i thought it would work:
alias ls='[[ "$PWD" = "/" ]] && ls -I test ||ls'
$PWD
is the current working directory&&
has the action to perform if pwd is /
(condition check =true)||
has the action to perform if pwd is not /
(condition check=false)
But after carefull testing above solution IS NOT WORKING correctly.
On the other hand this will work ok as alternative to functions:
alias ls='{ [[ "$PWD" == "/" ]] && a="-I tmp2" ||a=""; };ls $a '
Or even better, similar to other answers but without the need of function:
alias lstest='{ [[ "$PWD" == "/" ]] && set -- "-I" "tmp2" || set --; }; ls "$@"'
Possible extra flags and/or full paths given in command line when invoking this alias are preserved and sent to ls.
7
A problem with this is that usingls
from/
will always result in a listing of the current directory (/
), no matter what the command line says.
– Kusalananda
Mar 2 '17 at 9:30
@Kusalananda You are right , my solution was not correct. I would like you to have a look in my alternative. I tested it and seems to work fine in Debian Bash. Definetely my answer should not have been marked as accepted before , but with the troll presence things got mixed up. I don't know if there is a way to transfer the "solution" mark to your answer as it was in the very beginning (before troll attacks)
– George Vasiliou
Mar 2 '17 at 11:58
1
Ah!alias 'ls={ [[ "$PWD" == "/" ]] && set -- "-I" "tmp2"; }; ls "$@" '
, but that modifies the positional parameters instead...
– Kusalananda
Mar 2 '17 at 12:05
1
alias
es are nasty because they are some forms of macro expansion. Those particular alias can't stand for a realls
command. For instance,! ls
orfoo || ls
won't do what you think. This really calls for a function. Note that the first example would runls
twice if the first one exits with a non-zero exit status.
– Stéphane Chazelas
Mar 2 '17 at 23:05
3
"without the need of function" isn't really a plus. Both of the "corrected" functions unnecessarily overwrite global variables that might be in use.
– chepner
Mar 3 '17 at 3:29
|
show 4 more comments
Something like this i thought it would work:
alias ls='[[ "$PWD" = "/" ]] && ls -I test ||ls'
$PWD
is the current working directory&&
has the action to perform if pwd is /
(condition check =true)||
has the action to perform if pwd is not /
(condition check=false)
But after carefull testing above solution IS NOT WORKING correctly.
On the other hand this will work ok as alternative to functions:
alias ls='{ [[ "$PWD" == "/" ]] && a="-I tmp2" ||a=""; };ls $a '
Or even better, similar to other answers but without the need of function:
alias lstest='{ [[ "$PWD" == "/" ]] && set -- "-I" "tmp2" || set --; }; ls "$@"'
Possible extra flags and/or full paths given in command line when invoking this alias are preserved and sent to ls.
Something like this i thought it would work:
alias ls='[[ "$PWD" = "/" ]] && ls -I test ||ls'
$PWD
is the current working directory&&
has the action to perform if pwd is /
(condition check =true)||
has the action to perform if pwd is not /
(condition check=false)
But after carefull testing above solution IS NOT WORKING correctly.
On the other hand this will work ok as alternative to functions:
alias ls='{ [[ "$PWD" == "/" ]] && a="-I tmp2" ||a=""; };ls $a '
Or even better, similar to other answers but without the need of function:
alias lstest='{ [[ "$PWD" == "/" ]] && set -- "-I" "tmp2" || set --; }; ls "$@"'
Possible extra flags and/or full paths given in command line when invoking this alias are preserved and sent to ls.
edited Mar 2 '17 at 12:27
answered Mar 2 '17 at 9:09
George VasiliouGeorge Vasiliou
5,70531029
5,70531029
7
A problem with this is that usingls
from/
will always result in a listing of the current directory (/
), no matter what the command line says.
– Kusalananda
Mar 2 '17 at 9:30
@Kusalananda You are right , my solution was not correct. I would like you to have a look in my alternative. I tested it and seems to work fine in Debian Bash. Definetely my answer should not have been marked as accepted before , but with the troll presence things got mixed up. I don't know if there is a way to transfer the "solution" mark to your answer as it was in the very beginning (before troll attacks)
– George Vasiliou
Mar 2 '17 at 11:58
1
Ah!alias 'ls={ [[ "$PWD" == "/" ]] && set -- "-I" "tmp2"; }; ls "$@" '
, but that modifies the positional parameters instead...
– Kusalananda
Mar 2 '17 at 12:05
1
alias
es are nasty because they are some forms of macro expansion. Those particular alias can't stand for a realls
command. For instance,! ls
orfoo || ls
won't do what you think. This really calls for a function. Note that the first example would runls
twice if the first one exits with a non-zero exit status.
– Stéphane Chazelas
Mar 2 '17 at 23:05
3
"without the need of function" isn't really a plus. Both of the "corrected" functions unnecessarily overwrite global variables that might be in use.
– chepner
Mar 3 '17 at 3:29
|
show 4 more comments
7
A problem with this is that usingls
from/
will always result in a listing of the current directory (/
), no matter what the command line says.
– Kusalananda
Mar 2 '17 at 9:30
@Kusalananda You are right , my solution was not correct. I would like you to have a look in my alternative. I tested it and seems to work fine in Debian Bash. Definetely my answer should not have been marked as accepted before , but with the troll presence things got mixed up. I don't know if there is a way to transfer the "solution" mark to your answer as it was in the very beginning (before troll attacks)
– George Vasiliou
Mar 2 '17 at 11:58
1
Ah!alias 'ls={ [[ "$PWD" == "/" ]] && set -- "-I" "tmp2"; }; ls "$@" '
, but that modifies the positional parameters instead...
– Kusalananda
Mar 2 '17 at 12:05
1
alias
es are nasty because they are some forms of macro expansion. Those particular alias can't stand for a realls
command. For instance,! ls
orfoo || ls
won't do what you think. This really calls for a function. Note that the first example would runls
twice if the first one exits with a non-zero exit status.
– Stéphane Chazelas
Mar 2 '17 at 23:05
3
"without the need of function" isn't really a plus. Both of the "corrected" functions unnecessarily overwrite global variables that might be in use.
– chepner
Mar 3 '17 at 3:29
7
7
A problem with this is that using
ls
from /
will always result in a listing of the current directory (/
), no matter what the command line says.– Kusalananda
Mar 2 '17 at 9:30
A problem with this is that using
ls
from /
will always result in a listing of the current directory (/
), no matter what the command line says.– Kusalananda
Mar 2 '17 at 9:30
@Kusalananda You are right , my solution was not correct. I would like you to have a look in my alternative. I tested it and seems to work fine in Debian Bash. Definetely my answer should not have been marked as accepted before , but with the troll presence things got mixed up. I don't know if there is a way to transfer the "solution" mark to your answer as it was in the very beginning (before troll attacks)
– George Vasiliou
Mar 2 '17 at 11:58
@Kusalananda You are right , my solution was not correct. I would like you to have a look in my alternative. I tested it and seems to work fine in Debian Bash. Definetely my answer should not have been marked as accepted before , but with the troll presence things got mixed up. I don't know if there is a way to transfer the "solution" mark to your answer as it was in the very beginning (before troll attacks)
– George Vasiliou
Mar 2 '17 at 11:58
1
1
Ah!
alias 'ls={ [[ "$PWD" == "/" ]] && set -- "-I" "tmp2"; }; ls "$@" '
, but that modifies the positional parameters instead...– Kusalananda
Mar 2 '17 at 12:05
Ah!
alias 'ls={ [[ "$PWD" == "/" ]] && set -- "-I" "tmp2"; }; ls "$@" '
, but that modifies the positional parameters instead...– Kusalananda
Mar 2 '17 at 12:05
1
1
alias
es are nasty because they are some forms of macro expansion. Those particular alias can't stand for a real ls
command. For instance, ! ls
or foo || ls
won't do what you think. This really calls for a function. Note that the first example would run ls
twice if the first one exits with a non-zero exit status.– Stéphane Chazelas
Mar 2 '17 at 23:05
alias
es are nasty because they are some forms of macro expansion. Those particular alias can't stand for a real ls
command. For instance, ! ls
or foo || ls
won't do what you think. This really calls for a function. Note that the first example would run ls
twice if the first one exits with a non-zero exit status.– Stéphane Chazelas
Mar 2 '17 at 23:05
3
3
"without the need of function" isn't really a plus. Both of the "corrected" functions unnecessarily overwrite global variables that might be in use.
– chepner
Mar 3 '17 at 3:29
"without the need of function" isn't really a plus. Both of the "corrected" functions unnecessarily overwrite global variables that might be in use.
– chepner
Mar 3 '17 at 3:29
|
show 4 more comments
ls () {
case "$PWD" in
/) command ls -I test "$@" ;;
*) command ls "$@" ;;
esac
}
The above shell function will test the current directory against /
and executes the GNU ls
command differently according to the outcome of the test.
"$@"
will be replaced by the command line options and operands on the original command line.
We need to use command ls
rather than just ls
in the function to bypass the shell function lookup for ls
(which would otherwise give us a nice infinite recursion).
Note that this will not use the ignore pattern if doing ls /
from somewhere else, and that it will use the ignore pattern if doing, e.g., ls /home/dvoo
from /
.
To make it easier to remember what this actually does (six months down the line when you wonder what it's doing), use the long options:
ls () {
case "$PWD" in
/) command ls --ignore='test' "$@" ;;
*) command ls "$@" ;;
esac
}
Alternative implementation of the above function that will only call ls
from one place (and which is shorter):
ls () {
[ "$PWD" = "/" ] && set -- --ignore='test' "$@"
command ls "$@"
}
add a comment |
ls () {
case "$PWD" in
/) command ls -I test "$@" ;;
*) command ls "$@" ;;
esac
}
The above shell function will test the current directory against /
and executes the GNU ls
command differently according to the outcome of the test.
"$@"
will be replaced by the command line options and operands on the original command line.
We need to use command ls
rather than just ls
in the function to bypass the shell function lookup for ls
(which would otherwise give us a nice infinite recursion).
Note that this will not use the ignore pattern if doing ls /
from somewhere else, and that it will use the ignore pattern if doing, e.g., ls /home/dvoo
from /
.
To make it easier to remember what this actually does (six months down the line when you wonder what it's doing), use the long options:
ls () {
case "$PWD" in
/) command ls --ignore='test' "$@" ;;
*) command ls "$@" ;;
esac
}
Alternative implementation of the above function that will only call ls
from one place (and which is shorter):
ls () {
[ "$PWD" = "/" ] && set -- --ignore='test' "$@"
command ls "$@"
}
add a comment |
ls () {
case "$PWD" in
/) command ls -I test "$@" ;;
*) command ls "$@" ;;
esac
}
The above shell function will test the current directory against /
and executes the GNU ls
command differently according to the outcome of the test.
"$@"
will be replaced by the command line options and operands on the original command line.
We need to use command ls
rather than just ls
in the function to bypass the shell function lookup for ls
(which would otherwise give us a nice infinite recursion).
Note that this will not use the ignore pattern if doing ls /
from somewhere else, and that it will use the ignore pattern if doing, e.g., ls /home/dvoo
from /
.
To make it easier to remember what this actually does (six months down the line when you wonder what it's doing), use the long options:
ls () {
case "$PWD" in
/) command ls --ignore='test' "$@" ;;
*) command ls "$@" ;;
esac
}
Alternative implementation of the above function that will only call ls
from one place (and which is shorter):
ls () {
[ "$PWD" = "/" ] && set -- --ignore='test' "$@"
command ls "$@"
}
ls () {
case "$PWD" in
/) command ls -I test "$@" ;;
*) command ls "$@" ;;
esac
}
The above shell function will test the current directory against /
and executes the GNU ls
command differently according to the outcome of the test.
"$@"
will be replaced by the command line options and operands on the original command line.
We need to use command ls
rather than just ls
in the function to bypass the shell function lookup for ls
(which would otherwise give us a nice infinite recursion).
Note that this will not use the ignore pattern if doing ls /
from somewhere else, and that it will use the ignore pattern if doing, e.g., ls /home/dvoo
from /
.
To make it easier to remember what this actually does (six months down the line when you wonder what it's doing), use the long options:
ls () {
case "$PWD" in
/) command ls --ignore='test' "$@" ;;
*) command ls "$@" ;;
esac
}
Alternative implementation of the above function that will only call ls
from one place (and which is shorter):
ls () {
[ "$PWD" = "/" ] && set -- --ignore='test' "$@"
command ls "$@"
}
edited 4 hours ago
answered Mar 2 '17 at 9:13
KusalanandaKusalananda
135k17255422
135k17255422
add a comment |
add a comment |
Use a function that tests if you're in /
for ls
:
ls () {
if [[ "$PWD" == / ]]
then
command ls -I test "$@"
else
command ls "$@"
fi
}
This way, any arguments you pass to ls
will still be used.
Or:
ls () {
if [ "$PWD" == / ]
then
set -- -I test "$@"
fi
command ls "$@"
}
3
That was a nice way of sorting out only having one call tols
in the function!
– Kusalananda
Mar 2 '17 at 10:48
add a comment |
Use a function that tests if you're in /
for ls
:
ls () {
if [[ "$PWD" == / ]]
then
command ls -I test "$@"
else
command ls "$@"
fi
}
This way, any arguments you pass to ls
will still be used.
Or:
ls () {
if [ "$PWD" == / ]
then
set -- -I test "$@"
fi
command ls "$@"
}
3
That was a nice way of sorting out only having one call tols
in the function!
– Kusalananda
Mar 2 '17 at 10:48
add a comment |
Use a function that tests if you're in /
for ls
:
ls () {
if [[ "$PWD" == / ]]
then
command ls -I test "$@"
else
command ls "$@"
fi
}
This way, any arguments you pass to ls
will still be used.
Or:
ls () {
if [ "$PWD" == / ]
then
set -- -I test "$@"
fi
command ls "$@"
}
Use a function that tests if you're in /
for ls
:
ls () {
if [[ "$PWD" == / ]]
then
command ls -I test "$@"
else
command ls "$@"
fi
}
This way, any arguments you pass to ls
will still be used.
Or:
ls () {
if [ "$PWD" == / ]
then
set -- -I test "$@"
fi
command ls "$@"
}
edited Mar 2 '17 at 10:49
Kusalananda
135k17255422
135k17255422
answered Mar 2 '17 at 9:10
murumuru
1
1
3
That was a nice way of sorting out only having one call tols
in the function!
– Kusalananda
Mar 2 '17 at 10:48
add a comment |
3
That was a nice way of sorting out only having one call tols
in the function!
– Kusalananda
Mar 2 '17 at 10:48
3
3
That was a nice way of sorting out only having one call to
ls
in the function!– Kusalananda
Mar 2 '17 at 10:48
That was a nice way of sorting out only having one call to
ls
in the function!– Kusalananda
Mar 2 '17 at 10:48
add a comment |
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5
If the working directory is the root, or if the directory to be listed is the root? Whilst the former is what you asked, I'm not sure that's what you really want!
– Toby Speight
Mar 2 '17 at 11:06
1
Looks like someone is working on a user-level rootkit? :P
– Navin
Mar 3 '17 at 12:47
This is a bad idea. Better make a different alias and use that.
– reinierpost
Mar 3 '17 at 15:02