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Bash script - variable content as a command to run

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bash

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How do you execute a command stored in a variable in a shell script?

Here, the first line of the script i.e. “#!/bin/bash” shows that this file is in fact a Bash file. Then we have created a variable named “test” and have assigned it the value “$(echo “Hi there!”)”. Whenever you want to store the command in a variable, you have to type that command preceded by a “$” symbol.

How do you save a command output of a variable in a shell script?

Here are the different ways to store the output of a command in shell script. You can also use these commands on terminal to store command outputs in shell variables. variable_name=$(command) variable_name=$(command [option ...] arg1 arg2 ...) OR variable_name=`command` variable_name=`command [option ...]


You just need to do:

#!/bin/bash
count=$(cat last_queries.txt | wc -l)
$(perl test.pl test2 $count)

However, if you want to call your Perl command later, and that's why you want to assign it to a variable, then:

#!/bin/bash
count=$(cat last_queries.txt | wc -l)
var="perl test.pl test2 $count" # You need double quotes to get your $count value substituted.

...stuff...

eval $var

As per Bash's help:

~$ help eval
eval: eval [arg ...]
    Execute arguments as a shell command.

    Combine ARGs into a single string, use the result as input to the shell,
    and execute the resulting commands.

    Exit Status:
    Returns exit status of command or success if command is null.

You're are probably looking for eval $var.


line=$((${RANDOM} % $(wc -l < /etc/passwd)))
sed -n "${line}p" /etc/passwd

just with your file instead.

In this example I used the file /etc/password, using the special variable ${RANDOM} (about which I learned here), and the sed expression you had, only difference is that I am using double quotes instead of single to allow the variable expansion.


In the case where you have multiple variables containing the arguments for a command you're running, and not just a single string, you should not use eval directly, as it will fail in the following case:

function echo_arguments() {
  echo "Argument 1: $1"
  echo "Argument 2: $2"
  echo "Argument 3: $3"
  echo "Argument 4: $4"
}

# Note we are passing 3 arguments to `echo_arguments`, not 4
eval echo_arguments arg1 arg2 "Some arg"

Result:

Argument 1: arg1
Argument 2: arg2
Argument 3: Some
Argument 4: arg

Note that even though "Some arg" was passed as a single argument, eval read it as two.

Instead, you can just use the string as the command itself:

# The regular bash eval works by jamming all its arguments into a string then
# evaluating the string. This function treats its arguments as individual
# arguments to be passed to the command being run.
function eval_command() {
  "$@";
}

Note the difference between the output of eval and the new eval_command function:

eval_command echo_arguments arg1 arg2 "Some arg"

Result:

Argument 1: arg1
Argument 2: arg2
Argument 3: Some arg
Argument 4: