Generate code with own arguments properly quoted

snipplet:
arguments, quoting, escaping, wrapper
LastUpdate:
2010-07-31
Contributor:
Jan Schampera
type:
snipplet
Keywords: arguments,escape,quote,wrapper,generate
Contributor: self

There are situation where Bash code needs to generate Bash code. A script that writes out another script the user or cron may start, for example.

The general issue is easy, just write out text to the file.

A specific detail of it is tricky: If the generated script needs to call a command using the arguments the first original script got, you have problem in writing out the correct code.

I.e. if you run your generator script like

./myscript "give me 'some' water"
then this script should generate code that looks like
echo give me 'some' water"
you need correct escapes or quotes to not generate shell special characters out of normal text (like embedded dollar signs $).

Solution:

A loop over the own arguments that writes out properly quoted/escaped code to the generated script file

There are two (maybe more) easy options:

  • writing out singlequoted strings and handle the embedded singlequotes
  • the printf command knows the %q format specification, which will print a string (like %s does), but with all shell special characters escaped

Using singlequoted string

#!/bin/bash

# first option:
# generate singlequoted strings out of your own arguments and handle embedded singlequotes
# here to call 'echo' in the generated script

{
printf "#!/bin/bash\n\n"
printf "echo "
for arg; do
  arg=${arg/\'/\'\\\'\'}
  printf "'%s' " "${arg}"
done

printf "\n"
} >s2

The generated script will look like:

#!/bin/bash

echo 'fir$t' 'seco "ond"' 'thir'\''d' 

Using printf

The second method is easier, though more or less Bash-only (due to the %q in printf):

#!/bin/bash

{
printf "#!/bin/bash\n\n"
printf "echo "
for arg; do
  printf '%q ' "$arg"
done

printf "\n"
} >s2

The generated script will look like:

#!/bin/bash

echo fir\$t seco\ \"ond\" thir\'d 

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