Use xargs to build and run a command list
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How to use xargs
to build and run a command list such as these:
#1
cmd1 <arg1> && cmd2 <arg1> && cmd3 <arg1>
#2
cmd1 <arg1> ; cmd2 <arg1>
xargs
add a comment |
How to use xargs
to build and run a command list such as these:
#1
cmd1 <arg1> && cmd2 <arg1> && cmd3 <arg1>
#2
cmd1 <arg1> ; cmd2 <arg1>
xargs
add a comment |
How to use xargs
to build and run a command list such as these:
#1
cmd1 <arg1> && cmd2 <arg1> && cmd3 <arg1>
#2
cmd1 <arg1> ; cmd2 <arg1>
xargs
How to use xargs
to build and run a command list such as these:
#1
cmd1 <arg1> && cmd2 <arg1> && cmd3 <arg1>
#2
cmd1 <arg1> ; cmd2 <arg1>
xargs
xargs
asked Mar 14 at 11:53
Tran TrietTran Triet
12610
12610
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
By starting a child shell for each line of input to xargs
:
xargs -I sh -c 'cmd1 "$1" && cmd2 "$1" && cmd3 "$1"' sh
xargs -I sh -c 'cmd1 "$1"; cmd2 "$1"' sh
This runs sh -c
which executes the given string as a shell script. The arguments to sh -c
, after the script itself, are given to $0
and $1
inside the script. The value of $0
should usually be the name of the shell, which is why we pass sh
as this argument (it will be used in error messages).
Alternatively,
xargs sh -c '
for arg do
cmd1 "$arg" && cmd2 "$arg" && cmd3 "$arg"
done' sh
xargs sh -c '
for arg do
cmd1 "$arg"
cmd2 "$arg"
done' sh
These variations will take as many arguments as possible and then apply the code to these in a loop inside the sh -c
scripts.
As always when using xargs
, care must be taken so that the arguments supplied to the given utility (sh -c
here) are delimited properly.
I'm actually using this method, but I wonder if there's any other ways that are more xargs native.
– Tran Triet
Mar 14 at 11:59
1
@TranTriet No, not really. This is a fairly common way of doing it. What are your issues with this way of doing it?
– Kusalananda♦
Mar 14 at 12:03
@TranTriet See updated answer.
– Kusalananda♦
Mar 14 at 12:08
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
By starting a child shell for each line of input to xargs
:
xargs -I sh -c 'cmd1 "$1" && cmd2 "$1" && cmd3 "$1"' sh
xargs -I sh -c 'cmd1 "$1"; cmd2 "$1"' sh
This runs sh -c
which executes the given string as a shell script. The arguments to sh -c
, after the script itself, are given to $0
and $1
inside the script. The value of $0
should usually be the name of the shell, which is why we pass sh
as this argument (it will be used in error messages).
Alternatively,
xargs sh -c '
for arg do
cmd1 "$arg" && cmd2 "$arg" && cmd3 "$arg"
done' sh
xargs sh -c '
for arg do
cmd1 "$arg"
cmd2 "$arg"
done' sh
These variations will take as many arguments as possible and then apply the code to these in a loop inside the sh -c
scripts.
As always when using xargs
, care must be taken so that the arguments supplied to the given utility (sh -c
here) are delimited properly.
I'm actually using this method, but I wonder if there's any other ways that are more xargs native.
– Tran Triet
Mar 14 at 11:59
1
@TranTriet No, not really. This is a fairly common way of doing it. What are your issues with this way of doing it?
– Kusalananda♦
Mar 14 at 12:03
@TranTriet See updated answer.
– Kusalananda♦
Mar 14 at 12:08
add a comment |
By starting a child shell for each line of input to xargs
:
xargs -I sh -c 'cmd1 "$1" && cmd2 "$1" && cmd3 "$1"' sh
xargs -I sh -c 'cmd1 "$1"; cmd2 "$1"' sh
This runs sh -c
which executes the given string as a shell script. The arguments to sh -c
, after the script itself, are given to $0
and $1
inside the script. The value of $0
should usually be the name of the shell, which is why we pass sh
as this argument (it will be used in error messages).
Alternatively,
xargs sh -c '
for arg do
cmd1 "$arg" && cmd2 "$arg" && cmd3 "$arg"
done' sh
xargs sh -c '
for arg do
cmd1 "$arg"
cmd2 "$arg"
done' sh
These variations will take as many arguments as possible and then apply the code to these in a loop inside the sh -c
scripts.
As always when using xargs
, care must be taken so that the arguments supplied to the given utility (sh -c
here) are delimited properly.
I'm actually using this method, but I wonder if there's any other ways that are more xargs native.
– Tran Triet
Mar 14 at 11:59
1
@TranTriet No, not really. This is a fairly common way of doing it. What are your issues with this way of doing it?
– Kusalananda♦
Mar 14 at 12:03
@TranTriet See updated answer.
– Kusalananda♦
Mar 14 at 12:08
add a comment |
By starting a child shell for each line of input to xargs
:
xargs -I sh -c 'cmd1 "$1" && cmd2 "$1" && cmd3 "$1"' sh
xargs -I sh -c 'cmd1 "$1"; cmd2 "$1"' sh
This runs sh -c
which executes the given string as a shell script. The arguments to sh -c
, after the script itself, are given to $0
and $1
inside the script. The value of $0
should usually be the name of the shell, which is why we pass sh
as this argument (it will be used in error messages).
Alternatively,
xargs sh -c '
for arg do
cmd1 "$arg" && cmd2 "$arg" && cmd3 "$arg"
done' sh
xargs sh -c '
for arg do
cmd1 "$arg"
cmd2 "$arg"
done' sh
These variations will take as many arguments as possible and then apply the code to these in a loop inside the sh -c
scripts.
As always when using xargs
, care must be taken so that the arguments supplied to the given utility (sh -c
here) are delimited properly.
By starting a child shell for each line of input to xargs
:
xargs -I sh -c 'cmd1 "$1" && cmd2 "$1" && cmd3 "$1"' sh
xargs -I sh -c 'cmd1 "$1"; cmd2 "$1"' sh
This runs sh -c
which executes the given string as a shell script. The arguments to sh -c
, after the script itself, are given to $0
and $1
inside the script. The value of $0
should usually be the name of the shell, which is why we pass sh
as this argument (it will be used in error messages).
Alternatively,
xargs sh -c '
for arg do
cmd1 "$arg" && cmd2 "$arg" && cmd3 "$arg"
done' sh
xargs sh -c '
for arg do
cmd1 "$arg"
cmd2 "$arg"
done' sh
These variations will take as many arguments as possible and then apply the code to these in a loop inside the sh -c
scripts.
As always when using xargs
, care must be taken so that the arguments supplied to the given utility (sh -c
here) are delimited properly.
edited Mar 14 at 12:08
answered Mar 14 at 11:57
Kusalananda♦Kusalananda
141k17263439
141k17263439
I'm actually using this method, but I wonder if there's any other ways that are more xargs native.
– Tran Triet
Mar 14 at 11:59
1
@TranTriet No, not really. This is a fairly common way of doing it. What are your issues with this way of doing it?
– Kusalananda♦
Mar 14 at 12:03
@TranTriet See updated answer.
– Kusalananda♦
Mar 14 at 12:08
add a comment |
I'm actually using this method, but I wonder if there's any other ways that are more xargs native.
– Tran Triet
Mar 14 at 11:59
1
@TranTriet No, not really. This is a fairly common way of doing it. What are your issues with this way of doing it?
– Kusalananda♦
Mar 14 at 12:03
@TranTriet See updated answer.
– Kusalananda♦
Mar 14 at 12:08
I'm actually using this method, but I wonder if there's any other ways that are more xargs native.
– Tran Triet
Mar 14 at 11:59
I'm actually using this method, but I wonder if there's any other ways that are more xargs native.
– Tran Triet
Mar 14 at 11:59
1
1
@TranTriet No, not really. This is a fairly common way of doing it. What are your issues with this way of doing it?
– Kusalananda♦
Mar 14 at 12:03
@TranTriet No, not really. This is a fairly common way of doing it. What are your issues with this way of doing it?
– Kusalananda♦
Mar 14 at 12:03
@TranTriet See updated answer.
– Kusalananda♦
Mar 14 at 12:08
@TranTriet See updated answer.
– Kusalananda♦
Mar 14 at 12:08
add a comment |
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