如何将bash / awk命令放入变量（不输出命令）

Question

I am trying to just put bash commands into a variable and later run with exit code function.

command1="$(awk -v previous_date=$reply -f $SCRIPT_HOME/tes.awk <(gzip -dc $logDir/*) > $OUTFILE)" # Step 1

check_exit $command1 # Step2

Here it runs the command in step 1 and always returns 0 exit code.

How Can I put commands in a variable and later run with exit function.

Answer 1

You can declare your function this way:

function command1 { 
    awk -v "previous_date=$reply" -f "$SCRIPT_HOME/tes.awk" <(gzip -dc "$logDir"/*) > "$OUTFILE"
}

Or the older form:

command1() { 
    awk -v "previous_date=$reply" -f "$SCRIPT_HOME/tes.awk" <(gzip -dc "$logDir"/*) > "$OUTFILE"
}

Take attention on placing variables inside double quotes to prevent word splitting and pathname expansion.

Some POSIXists would prefer that you use the old version for compatibility but if you're just running your code with bash, everything can just be a preference.

And you can also have another function like checkexit which would check the exit code of the command:

function checkexit {
    "$@"
    echo "Command returned $?"
}

Example run:

checkexit command1

Answer 2

Try using a function:

do_stuff() { echo "$var"; echo that; }

You can later call this function and check the error code:

$ var="this"
$ do_stuff
this
that
$ echo $?
0

The variables defined in the global scope of your script will be visible to the function.

So for your example:

command1() { awk -v previous_date="$reply" -f "$SCRIPT_HOME/tes.awk" <(gzip -dc "$logDir/*") > "$OUTFILE"); }

check_exit command1

By the way, I put some quotes around the variables in your function, as it's typically a good idea in case their values contain spaces.