I want to write a script that will change to different directories depending on my input. something like this:
test.sh:
#!/bin/bash
ssh machine001 '(chdir ~/dev$1; pwd)'
But as I run ./test.sh 2 it still goes to ~/dev. It seems that my argument gets ignored. Am I doing anything very stupid here?
Bash ignores any variable syntax inside the single-quoted(') strings. You need double quotes(") in order to make a substitution:
#!/bin/bash
ssh machine001 "(chdir ~/dev$1; pwd)"
The parameter is enclosed in single quotes, so it isn't expanded on the local side. Use double-quotes instead.
#!/bin/bash
ssh machine001 "chdir ~/dev$1; pwd"
There's no need for the (...), since you are only running the pair of commands then exiting.
Related
Currently trying to move all of my aliases from .bash_profile to .zshrc. However, found a problem with one of the longer aliases I use for substituting root to ubuntu when passing a command to access AWS instances.
AWS (){
cd /Users/user/aws_keys
cmd=$(echo $# | sed "s/root/ubuntu/g")
$cmd[#]
}
The error I get is AWS:5: command not found ssh -i keypair.pem ubuntu#ec1.compute.amazonaws.com
I would really appreciate any suggestions!
The basic problem is that the cmd=$(echo ... line is mashing all the arguments together into a space-delimited string, and you're depending on word-splitting to split it up into a command and its arguments. But word-splitting is usually more of a problem than anything else, so zsh doesn't do it by default. This means that rather than trying to run the command named ssh with arguments -i, keypair.pem, etc, it's treating the entire string as the command name.
The simple solution is to avoid mashing the arguments together, so you don't need word-splitting to separate them out again. You can use a modifier to the parameter expansion to replace "root" with "ubuntu". BTW, I also strongly recommend checking for error when using cd, and not proceeding if it gets an error.
So something like this:
AWS (){
cd /Users/user/aws_keys || return $?
"${#//root/ubuntu}"
}
This syntax will work in bash as well as zsh (the double-quotes prevent unexpected word-splitting in bash, and aren't really needed in zsh).
BTW, I'm also a bit nervous about just blindly replacing "root" with "ubuntu" in the arguments; what if it occurs somewhere other than the username, like as part of a filename or hostname?
In BASH, I use "pushd . " command to save the current directory on the stack.
After issuing this command in couple of different directories, I have multiple directories saved on the stack which I am able to see by issuing command "dirs".
For example, the output of "dirs" command in my current bash session is given below -
0 ~/eclipse/src
1 ~/eclipse
2 ~/parboil/src
Now, to switch to 0th directory, I issue a command "cd ~0".
I want to create a bash alias command or a function for this command.
Something like "xya 0", which will switch to 0th directory on stack.
I wrote following function to achieve this -
xya(){
cd ~$1
}
Where "$1" in above function, is the first argument passed to the function "xya".
But, I am getting the following error -
-bash: cd: ~1: No such file or directory
Can you please tell what is going wrong here ?
Generally, bash parsing happens in the following order:
brace expansion
tilde expansion
parameter, variable, arithmetic expansion; command substitution (same phase, left-to-right)
word splitting
pathname expansion
Thus, by the time your parameter is expanded, tilde expansion is already finished and will not take place again, without doing something explicit like use of eval.
If you know the risks and are willing to accept them, use eval to force parsing to restart at the beginning after the expansion of $1 is complete. The below tries to mitigate the damage should something that isn't eval-safe is passed as an argument:
xya() {
local cmd
printf -v cmd 'cd ~%q' "$1"
eval "$cmd"
}
...or, less cautiously (which is to say that the below trusts your arguments to be eval-safe):
xya() {
eval "cd ~$1"
}
You can let dirs print the absolute path for you:
xya(){
cd "$(dirs -${1-0} -l)"
}
I am confused by how to use variables in Bash. Please see the following example. I am not able to figure out why Bash isn't able to recognize the variable within (). Can anybody please help me understand what is going on.
$echo $SHELL
/bin/bash
$export TestC=/Users
$echo $TestC
/Users
$export TestD=$TestC/ABCD
$echo $TestD
/Users/ABCD
$export TestD=$(TestC)/ABCD
-bash: TestC: command not found
Thanks for your help
When referencing a bash variable you either use $ then the name, as in $TestC or you can put braces around the name like ${TestC}.
$(...) is a subshell syntax called command substitution that will execute the command inside the parens then "return" the stdout of that command.
Read all about parameter/variable expansion here, which also shows a lot of the extra things you can do with the parameter expansion when using braces.
I have an rsync command that works as expected when I type it directly into a terminal. The command includes several --include='blah' and --exclude='foo' type arguments. However, if I save that command to a one-line file called "myfile" and I try `cat myfile` (or, equivalently $(cat myfile)), the rsync command behaves differently.
I'm sure it is the exact same command in both cases.
Is this behavior expected/explainable?
I've found the answer to this question. The point is that the cat command takes the contents of the file and treats it like a string. Any string operators (like the escape operator, ) are executed. Then, the final string output is what is passed to a command via the backticks.
As a solution, I've just made "myfile" a shell script that I can execute rather than trying to use cat.
i have unix shell script which is need to be run like below
test_sh XYZ=KLMN
the content of the script is
#!/bin/ksh
echo $XYZ
for using the value of XYZ i have do set -k before i run the script.
is there a way where i can do this without doint set -k before running the script. or is there something that i can do in the script where i can use value of the parameter given while running the script in the below way
test_sh XYZ=KLMN
i am using ksh.
Any help is appreciated.
How about running this?
XYZ=KLMN ./test_sh //running from directory where test_sh is
If your script needs no other arguments, a quick and dirty way do to it is to put
eval "$#"
at the start of your script. This will evaluate the command line arguments as shell commands. If those commands are to assign a shell/environment variable, then that's what it will do.
It's quick-and-dirty since anything could be put on the command line, causing problems from a syntax error to a bad security hole (if the script is trusted).
I'm not sure if "$#" means the same in ksh as it does in bash - using just $* (without quotes) would work too, but is even dirtier.
It looks like you are trying to use the environment variable "INSTANCE" in your script.
For that, the environment variable must be set in advance of executing your script. Using the "set" command sets exportable environment variables. Incidentally, my version of ksh dates from 1993 and the "-k" option was obsolete back then.
To set an environment variable so that it is exported into spawned shells, simply use the "export" command like so:
export INSTANCE='whatever you want to put here'
If you want to use a positional parameter for your script -- that is have the "KLMN" value accessed within your script, and assuming it is the first parameter, then you do the following in your script:
#!/bin/ksh
echo $1
You can also assign the positional parameter to a local variable for later use in your script like so:
#!/bin/ksh
param_one=$1
echo $param_one
You can call this with:
test_sh KLMN
Note that the spacing in the assignment is important -- do not use spaces.
I am tring this option
#!/bin/ksh
echo $1
awk '{FS="=";print $2}' $1
and on the command line
test_sh INSTANCE=LSN_MUM
but awk is failing.is there any problem over here?
Probably #!/bin/ksh -k will work (untested).