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Invoking pushd in a shell script
Different ways to execute a shell scriptchanging current working dir with a scriptTiming out in a shell scriptshell commands to check & create dirRule for invoking subshell in Bash?Interactive shell in background inside another shell scriptHow to run privileged bash script as non-root?Source shell script automatically in terminalHow can I search and execute a shell script from another?self killing shell scriptRun shell script at startup (Kali Linux)Importing environment variables from another script into the current shell
I have a simple script that I want to invoke 'pushed' followed by another command. But the 'pushd' command inside the script doesn't seem to stick past the script.
What's a way to make this execute in the shell terminal?
#!/bin/sh
pushd $1
time
What I really want to accomplish is invoke pushd followed by other-command with one command.
command-line shell
add a comment |
I have a simple script that I want to invoke 'pushed' followed by another command. But the 'pushd' command inside the script doesn't seem to stick past the script.
What's a way to make this execute in the shell terminal?
#!/bin/sh
pushd $1
time
What I really want to accomplish is invoke pushd followed by other-command with one command.
command-line shell
Related question: unix.stackexchange.com/q/2976/250
– phunehehe
Dec 29 '10 at 0:49
add a comment |
I have a simple script that I want to invoke 'pushed' followed by another command. But the 'pushd' command inside the script doesn't seem to stick past the script.
What's a way to make this execute in the shell terminal?
#!/bin/sh
pushd $1
time
What I really want to accomplish is invoke pushd followed by other-command with one command.
command-line shell
I have a simple script that I want to invoke 'pushed' followed by another command. But the 'pushd' command inside the script doesn't seem to stick past the script.
What's a way to make this execute in the shell terminal?
#!/bin/sh
pushd $1
time
What I really want to accomplish is invoke pushd followed by other-command with one command.
command-line shell
command-line shell
edited Dec 29 '10 at 0:40
asked Dec 28 '10 at 22:28
Sarah Soto
Related question: unix.stackexchange.com/q/2976/250
– phunehehe
Dec 29 '10 at 0:49
add a comment |
Related question: unix.stackexchange.com/q/2976/250
– phunehehe
Dec 29 '10 at 0:49
Related question: unix.stackexchange.com/q/2976/250
– phunehehe
Dec 29 '10 at 0:49
Related question: unix.stackexchange.com/q/2976/250
– phunehehe
Dec 29 '10 at 0:49
add a comment |
3 Answers
3
active
oldest
votes
A shell script normally executes in a separate instance of the shell program, /bin/sh in this case. Your pushd command affects that sub-shell's working directory only. If it were otherwise, any program you ran from the shell could mess with your shell's working directory.
To execute that script within the current shell, say this instead:
$ . my-command somedir
or, more verbosely:
$ source my-command somedir
To make it appear that your program works like any other, you can use an alias:
$ alias mycmd='source my-command'
$ mycmd /bin
$ pwd
/bin
answered Dec 28 '10 at 22:54
Warren YoungWarren Young
55.7k11143148
What I really want is execute 'pushd <params>' followed by another command that takes no params. I may not need a shell script for it - what's a good way to do this?
– Sarah Soto
Dec 29 '10 at 0:23
@sarah-soto All you need then ispushd <params>; other-command
– Shawn J. Goff
Dec 29 '10 at 0:26
I should have clarified: I want a shortcut so I am invoking both commands one after the after without implicitly calling both. I can't aliase the two commands to be alias p2='pushd; other-command' and then invoke 'p2 ~/temp' as the shell will return the error: '-bash: pushd: no other directory
– Sarah Soto
Dec 29 '10 at 0:38
You could use a shell function:function p2 () pushd "$@"; other-command;
– Steven D
Dec 29 '10 at 0:46
2
A shell script is a fine way to do this. It's inefficient, but this is not a case where optimization is worth bothering with. Defining a shell function in your~/.bash_profile(or whatever your shell uses, if not Bash) will also work. Your choice.
– Warren Young
Dec 29 '10 at 3:10
|
show 2 more comments
Scripts cannot alter their parent processe's environment. Because of this, any environment changes made in the script are lost.
To run the script in the same process, you can 'source' the script like this
. /path/to/script.sh args
answered Dec 28 '10 at 22:56
Shawn J. GoffShawn J. Goff
30k19112134
add a comment |
Instead of a shell script, function is more appropriate for this kind of operation, especially when using pushd.
Add this in .bashrc:
foo()
pushd $1
time
export foo
In shell:
$ foo mydir1
output:
~/mydir1 ~
real 0m0.000s
user 0m0.000s
sys 0m0.000s
answered 5 hours ago
otter.prootter.pro
11
New contributor
otter.pro 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 |
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3 Answers
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active
oldest
votes
3 Answers
3
active
oldest
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votes
A shell script normally executes in a separate instance of the shell program, /bin/sh in this case. Your pushd command affects that sub-shell's working directory only. If it were otherwise, any program you ran from the shell could mess with your shell's working directory.
To execute that script within the current shell, say this instead:
$ . my-command somedir
or, more verbosely:
$ source my-command somedir
To make it appear that your program works like any other, you can use an alias:
$ alias mycmd='source my-command'
$ mycmd /bin
$ pwd
/bin
answered Dec 28 '10 at 22:54
Warren YoungWarren Young
55.7k11143148
What I really want is execute 'pushd <params>' followed by another command that takes no params. I may not need a shell script for it - what's a good way to do this?
– Sarah Soto
Dec 29 '10 at 0:23
@sarah-soto All you need then ispushd <params>; other-command
– Shawn J. Goff
Dec 29 '10 at 0:26
I should have clarified: I want a shortcut so I am invoking both commands one after the after without implicitly calling both. I can't aliase the two commands to be alias p2='pushd; other-command' and then invoke 'p2 ~/temp' as the shell will return the error: '-bash: pushd: no other directory
– Sarah Soto
Dec 29 '10 at 0:38
You could use a shell function:function p2 () pushd "$@"; other-command;
– Steven D
Dec 29 '10 at 0:46
2
A shell script is a fine way to do this. It's inefficient, but this is not a case where optimization is worth bothering with. Defining a shell function in your~/.bash_profile(or whatever your shell uses, if not Bash) will also work. Your choice.
– Warren Young
Dec 29 '10 at 3:10
|
show 2 more comments
A shell script normally executes in a separate instance of the shell program, /bin/sh in this case. Your pushd command affects that sub-shell's working directory only. If it were otherwise, any program you ran from the shell could mess with your shell's working directory.
To execute that script within the current shell, say this instead:
$ . my-command somedir
or, more verbosely:
$ source my-command somedir
To make it appear that your program works like any other, you can use an alias:
$ alias mycmd='source my-command'
$ mycmd /bin
$ pwd
/bin
answered Dec 28 '10 at 22:54
Warren YoungWarren Young
55.7k11143148
What I really want is execute 'pushd <params>' followed by another command that takes no params. I may not need a shell script for it - what's a good way to do this?
– Sarah Soto
Dec 29 '10 at 0:23
@sarah-soto All you need then ispushd <params>; other-command
– Shawn J. Goff
Dec 29 '10 at 0:26
I should have clarified: I want a shortcut so I am invoking both commands one after the after without implicitly calling both. I can't aliase the two commands to be alias p2='pushd; other-command' and then invoke 'p2 ~/temp' as the shell will return the error: '-bash: pushd: no other directory
– Sarah Soto
Dec 29 '10 at 0:38
You could use a shell function:function p2 () pushd "$@"; other-command;
– Steven D
Dec 29 '10 at 0:46
2
A shell script is a fine way to do this. It's inefficient, but this is not a case where optimization is worth bothering with. Defining a shell function in your~/.bash_profile(or whatever your shell uses, if not Bash) will also work. Your choice.
– Warren Young
Dec 29 '10 at 3:10
|
show 2 more comments
A shell script normally executes in a separate instance of the shell program, /bin/sh in this case. Your pushd command affects that sub-shell's working directory only. If it were otherwise, any program you ran from the shell could mess with your shell's working directory.
To execute that script within the current shell, say this instead:
$ . my-command somedir
or, more verbosely:
$ source my-command somedir
To make it appear that your program works like any other, you can use an alias:
$ alias mycmd='source my-command'
$ mycmd /bin
$ pwd
/bin
answered Dec 28 '10 at 22:54
Warren YoungWarren Young
55.7k11143148
A shell script normally executes in a separate instance of the shell program, /bin/sh in this case. Your pushd command affects that sub-shell's working directory only. If it were otherwise, any program you ran from the shell could mess with your shell's working directory.
To execute that script within the current shell, say this instead:
$ . my-command somedir
or, more verbosely:
$ source my-command somedir
To make it appear that your program works like any other, you can use an alias:
$ alias mycmd='source my-command'
$ mycmd /bin
$ pwd
/bin
answered Dec 28 '10 at 22:54
Warren YoungWarren Young
55.7k11143148
answered Dec 28 '10 at 22:54
Warren YoungWarren Young
55.7k11143148
answered Dec 28 '10 at 22:54
Warren YoungWarren Young
55.7k11143148
answered Dec 28 '10 at 22:54
Warren YoungWarren Young
55.7k11143148
55.7k11143148
What I really want is execute 'pushd <params>' followed by another command that takes no params. I may not need a shell script for it - what's a good way to do this?
– Sarah Soto
Dec 29 '10 at 0:23
@sarah-soto All you need then ispushd <params>; other-command
– Shawn J. Goff
Dec 29 '10 at 0:26
I should have clarified: I want a shortcut so I am invoking both commands one after the after without implicitly calling both. I can't aliase the two commands to be alias p2='pushd; other-command' and then invoke 'p2 ~/temp' as the shell will return the error: '-bash: pushd: no other directory
– Sarah Soto
Dec 29 '10 at 0:38
You could use a shell function:function p2 () pushd "$@"; other-command;
– Steven D
Dec 29 '10 at 0:46
2
A shell script is a fine way to do this. It's inefficient, but this is not a case where optimization is worth bothering with. Defining a shell function in your~/.bash_profile(or whatever your shell uses, if not Bash) will also work. Your choice.
– Warren Young
Dec 29 '10 at 3:10
|
show 2 more comments
What I really want is execute 'pushd <params>' followed by another command that takes no params. I may not need a shell script for it - what's a good way to do this?
– Sarah Soto
Dec 29 '10 at 0:23
@sarah-soto All you need then ispushd <params>; other-command
– Shawn J. Goff
Dec 29 '10 at 0:26
I should have clarified: I want a shortcut so I am invoking both commands one after the after without implicitly calling both. I can't aliase the two commands to be alias p2='pushd; other-command' and then invoke 'p2 ~/temp' as the shell will return the error: '-bash: pushd: no other directory
– Sarah Soto
Dec 29 '10 at 0:38
You could use a shell function:function p2 () pushd "$@"; other-command;
– Steven D
Dec 29 '10 at 0:46
2
A shell script is a fine way to do this. It's inefficient, but this is not a case where optimization is worth bothering with. Defining a shell function in your~/.bash_profile(or whatever your shell uses, if not Bash) will also work. Your choice.
– Warren Young
Dec 29 '10 at 3:10
What I really want is execute 'pushd <params>' followed by another command that takes no params. I may not need a shell script for it - what's a good way to do this?
– Sarah Soto
Dec 29 '10 at 0:23
What I really want is execute 'pushd <params>' followed by another command that takes no params. I may not need a shell script for it - what's a good way to do this?
– Sarah Soto
Dec 29 '10 at 0:23
@sarah-soto All you need then is
pushd <params>; other-command– Shawn J. Goff
Dec 29 '10 at 0:26
@sarah-soto All you need then is
pushd <params>; other-command– Shawn J. Goff
Dec 29 '10 at 0:26
I should have clarified: I want a shortcut so I am invoking both commands one after the after without implicitly calling both. I can't aliase the two commands to be alias p2='pushd; other-command' and then invoke 'p2 ~/temp' as the shell will return the error: '-bash: pushd: no other directory
– Sarah Soto
Dec 29 '10 at 0:38
I should have clarified: I want a shortcut so I am invoking both commands one after the after without implicitly calling both. I can't aliase the two commands to be alias p2='pushd; other-command' and then invoke 'p2 ~/temp' as the shell will return the error: '-bash: pushd: no other directory
– Sarah Soto
Dec 29 '10 at 0:38
You could use a shell function:
function p2 () pushd "$@"; other-command;– Steven D
Dec 29 '10 at 0:46
You could use a shell function:
function p2 () pushd "$@"; other-command;– Steven D
Dec 29 '10 at 0:46
2
2
A shell script is a fine way to do this. It's inefficient, but this is not a case where optimization is worth bothering with. Defining a shell function in your
~/.bash_profile (or whatever your shell uses, if not Bash) will also work. Your choice.– Warren Young
Dec 29 '10 at 3:10
A shell script is a fine way to do this. It's inefficient, but this is not a case where optimization is worth bothering with. Defining a shell function in your
~/.bash_profile (or whatever your shell uses, if not Bash) will also work. Your choice.– Warren Young
Dec 29 '10 at 3:10
|
show 2 more comments
Scripts cannot alter their parent processe's environment. Because of this, any environment changes made in the script are lost.
To run the script in the same process, you can 'source' the script like this
. /path/to/script.sh args
answered Dec 28 '10 at 22:56
Shawn J. GoffShawn J. Goff
30k19112134
add a comment |
Scripts cannot alter their parent processe's environment. Because of this, any environment changes made in the script are lost.
To run the script in the same process, you can 'source' the script like this
. /path/to/script.sh args
answered Dec 28 '10 at 22:56
Shawn J. GoffShawn J. Goff
30k19112134
add a comment |
Scripts cannot alter their parent processe's environment. Because of this, any environment changes made in the script are lost.
To run the script in the same process, you can 'source' the script like this
. /path/to/script.sh args
answered Dec 28 '10 at 22:56
Shawn J. GoffShawn J. Goff
30k19112134
Scripts cannot alter their parent processe's environment. Because of this, any environment changes made in the script are lost.
To run the script in the same process, you can 'source' the script like this
. /path/to/script.sh args
answered Dec 28 '10 at 22:56
Shawn J. GoffShawn J. Goff
30k19112134
edited Dec 29 '10 at 0:24
answered Dec 28 '10 at 22:56
Shawn J. GoffShawn J. Goff
30k19112134
answered Dec 28 '10 at 22:56
Shawn J. GoffShawn J. Goff
30k19112134
answered Dec 28 '10 at 22:56
Shawn J. GoffShawn J. Goff
30k19112134
30k19112134
add a comment |
add a comment |
Instead of a shell script, function is more appropriate for this kind of operation, especially when using pushd.
Add this in .bashrc:
foo()
pushd $1
time
export foo
In shell:
$ foo mydir1
output:
~/mydir1 ~
real 0m0.000s
user 0m0.000s
sys 0m0.000s
answered 5 hours ago
otter.prootter.pro
11
New contributor
otter.pro 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 |
Instead of a shell script, function is more appropriate for this kind of operation, especially when using pushd.
Add this in .bashrc:
foo()
pushd $1
time
export foo
In shell:
$ foo mydir1
output:
~/mydir1 ~
real 0m0.000s
user 0m0.000s
sys 0m0.000s
answered 5 hours ago
otter.prootter.pro
11
New contributor
otter.pro 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 |
Instead of a shell script, function is more appropriate for this kind of operation, especially when using pushd.
Add this in .bashrc:
foo()
pushd $1
time
export foo
In shell:
$ foo mydir1
output:
~/mydir1 ~
real 0m0.000s
user 0m0.000s
sys 0m0.000s
answered 5 hours ago
otter.prootter.pro
11
New contributor
otter.pro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Instead of a shell script, function is more appropriate for this kind of operation, especially when using pushd.
Add this in .bashrc:
foo()
pushd $1
time
export foo
In shell:
$ foo mydir1
output:
~/mydir1 ~
real 0m0.000s
user 0m0.000s
sys 0m0.000s
answered 5 hours ago
otter.prootter.pro
11
New contributor
otter.pro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 5 hours ago
otter.prootter.pro
11
New contributor
otter.pro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 5 hours ago
otter.prootter.pro
11
answered 5 hours ago
otter.prootter.pro
11
11
New contributor
otter.pro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
otter.pro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
otter.pro 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 |
add a comment |
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Related question: unix.stackexchange.com/q/2976/250
– phunehehe
Dec 29 '10 at 0:49