Bash script - variable content as a command to run - bash

Question

I have a Perl script that gives me a defined list of random numbers that correspond to the lines of a file. Next I want to extract those lines from the file using sed.
#!/bin/bash
count=$(cat last_queries.txt | wc -l)
var=$(perl test.pl test2 $count)
The variable var returns an output like: cat last_queries.txt | sed -n '12p;500p;700p'. The problem is that I can't run this last command. I tried with $var, but the output is not correct (if I run manually the command it works fine, so no problem there). What is the correct way to do this?
P.S: Sure I could do all the work in Perl, but I'm trying to learn this way, because it could help me in other situations.

Answer 1

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.

Answer 2

You're are probably looking for eval $var.

Answer 3

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.

Answer 4

There're 2 basic ways of executing a string command in a shell script whether it's given as parameter or not here's.
COMMAND="ls -lah"
$(echo $COMMAND)
or
COMMAND="ls -lah"
bash -c $COMMAND

Answer 5

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:

Answer 6

Better ways to do it
Using a function:
# define it
myls() {
ls -l "/tmp/test/my dir"
}
# run it
myls
Using an array:
# define the array
mycmd=(ls -l "/tmp/test/my dir")
# run the command
"${mycmd[#]}"

Answer 7

cmd="ls -atr ${HOME} | tail -1" <br/>
echo "$cmd" <br/>
VAR_FIRST_FILE=$( eval "${cmd}" ) <br/>
or
cmd=("ls -atr ${HOME} | tail -1") <br/>
echo "$cmd" <br/>
VAR_FIRST_FILE=$( eval "${cmd[#]}" )