I'm running a shell script using bash --init-file script.sh that runs some commands, then leaves an interactive session open. How can I pass arguments to this init file from the process that runs the initial bash command? bash --init-file 'script.sh arg' doesn't work.
Interestingly, if the script contains echo "$# $*", passing an argument as I did above causes it to print nothing, while not passing an argument prints '0'.
Create a file with the content:
#!/bin/bash
script.sh arg
Pass that file to bash: bash --init-file thatfile
I'd like the arg to come from the command that runs bash with the
Create a file from the command line and pass it:
arg="$1"
cat >thatfile <<EOF
$(declare -p arg)
script.sh \"\$arg\"
EOF
bash --init-file thatfile
You might be interested in researching what is a process substitution in bash.
Related
Normally, we use
sh script.sh 1>t.log 2>t.err
to redirect log.
How can I use variant to log:
string="1>t.log 2>t.err"
sh script.sh $string
You need to use 'eval' shell builtin for this purpose. As per man page of bash command:
eval [arg ...]
The args are read and concatenated together into a single command. This command is then read and exe‐
cuted by the shell, and its exit status is returned as the value of eval. If there are no args, or
only null arguments, eval returns 0.
Run your command like below:
eval sh script.sh $string
However, do you really need to run script.sh through sh command? If you instead put sh interpreter line (using #!/bin/sh as the first line in your shell script) in your script itself and give it execute permission, that would let you access return code of ls command. Below is an example of using sh and not using sh. Notice the difference in exit codes.
Note: I had only one file try.sh in my current directory. So ls command was bound to exit with return code 2.
$ ls try1.sh try1.sh.backup 1>out.txt 2>err.txt
$ echo $?
2
$ eval sh ls try1.sh try1.sh.backup 1>out.txt 2>err.txt
$ echo $?
127
In the second case, the exit code is of sh shell. In first case, the exit code is of ls command. You need to make cautious choice depending on your needs.
I figure out one way but it's ugly:
echo script.sh $string | sh
I think you can just put the name into a string variable
and then use data redirection
file_name="file1"
outfile="$file_name"".log"
errorfile="$file_name"".err"
sh script.sh 1> $outfile 2> $errorfile
I've simplified my example to the following:
file1.sh:
#!/bin/bash
bash -c "./file2.sh $#"
file2.sh:
#!/bin/bash
echo "first $1"
echo "second $2"
I expect that if I call ./file1.sh a b to get:
first a
second b
but instead I get:
first a
second
In other words, my later arguments after the first one are not getting passed through to the command that I'm executing inside a new bash shell. I've tried many variations of removing and moving around the quotation marks in the file1.sh file, but haven't got this to work.
Why is this happening, and how do I get the behavior I want?
(UPDATE - I realize it seems pointless that I'm calling bash -c in this example, my actual file1.sh is a proxy script for a command that gets called locally to run in a docker container so it's actually docker exec -i mycontainer bash -c '')
Change file1.sh to this with different quoting:
#!/bin/bash
bash -c './file2.sh "$#"' - "$#"
- "$#" is passing hyphen to populate $0 and $# is being passed in to populate all other positional parameters in bash -c command line.
You can also make it:
bash -c './file2.sh "$#"' "$0" "$#"
However there is no real need to use bash -c here and you can just use:
./file2.sh "$#"
If you are passed in an argument in a KornShell script, and it's also a command like:
ksh argument.ksh "wc -l"
how would you execute this command inside the script? Do you store it in a variable, and then execute it? Also, is there a way to retrieve the standard output/standard error from executing the command inside the script?
Place this inside your argument.ksh script:
echo "Running command $1." ## optional message
eval "$1" ## evaluate "$1" as a whole new command
A better or safer way actually is to use "$#":
echo "Running command $*." ## optional message
"$#"
And pass your arguments like this:
ksh argument.ksh wc -l
I'm writing a bash script that starts the tcsh interpreter as a login shell and has it execute my_command. The tcsh man page says that there are two ways to start a login shell. The first is to use /bin/tcsh -l with no other arguments. Not an option, because I need the shell to execute my_command. The second is to specify a dash (-) as the zeroeth argument.
Now the bash exec command with the -l option does exactly this, and in fact the following works perfectly:
#!/bin/bash
exec -l /bin/tcsh -c my_command
Except... I can't use exec because I need the script to come back and do some other things afterwards! So how can I specify - as the zeroeth argument to /bin/tcsh without using exec?
You can enclose the exec command into a sub-shell of your script.
#!/bin/bash
(exec -l /bin/tcsh -c my_command)
# ... whatever else you need to do after the command is done
You can write a wrapper (w.sh) script that contains:
#!/bin/bash
exec -l /bin/tcsh -c my_command
and execute w.sh in your main script.
I am writing a bash script that takes a number of command line arguments (possibly including spaces) and passes all of them to a program (/bin/some_program) via a login shell. The login shell that is called from the bash script will depend on the user's login shell. Let's suppose the user uses /bin/bash as their login shell in this example... but it might be /bin/tcsh or anything else.
If I know how many arguments will be passed to some_program, I can put the following lines in my bash script:
#!/bin/bash
# ... (some lines where we determine that the user's login shell is bash) ...
/bin/bash --login -c "/bin/some_program \"$1\" \"$2\""
and then call the above script as follows:
my_script "this is too" cool
With the above example I can confirm that some_program receives two arguments, "this is too" and "cool".
My question is... what if I don't know how many arguments will be passed? I'd like to pass all the arguments that were sent to my_script along to some_program. The problem is I can't figure out how to do this. Here are some things that don't work:
/bin/bash --login -c "/bin/some_program $#" # --> 3 arguments: "this","is","too"
/bin/bash --login -c /bin/some_program "$#" # --> passes no arguments
Quoting the bash manual for -c:
If the -c option is present, then commands are read from string. If there are arguments after the string, they are assigned to the positional parameters, starting with $0.
Works for me:
$ cat x.sh
#!/bin/bash
/bin/bash --login -c 'echo 1:$1 2:$2 3:$3' echo "$#"
$ ./x.sh "foo bar" "baz" "argh blargh quargh"
1:foo bar 2:baz 3:argh blargh quargh
I don't know how you arrived at the "passes no arguments" conclusion, maybe you missed the $0 bit?
Avoid embedding variables into other scripts, pass them on as arguments instead. In this case:
bash --login -c 'some_program "$#"' some_program "$#"
The first argument after -c '...' is taken as $0, so I just put in some_program there.
On a side note, it's an odd requirement to require a login shell. Doesn't the user log in?