Question

I’m writing a bash script that will receive a password via STDIN and go through a few different checks.

I’m getting a problem when I use a password that contain symbols.

The plain password grassy is OK, but a complex combination such as gra$$y will expand to gra3308y.

This is the script I am using:

read INPUT

if [ $(echo -n $INPUT|wc -m) -ge 6 ]

then exit 0
else exit 1

fi

I’ve tried all different kinds of quotations but I can’t stop the STDIN password from expanding special characters.

Here is an example, without putting single quotes around the original value what can be done to solve this?

david@hostname ~ $ echo gra$$| { read INPUT; echo ${INPUT}; }
gra2598

Answer 1

Use double quotes around variable expansions inside your script:

read INPUT
echo "Password was: $INPUT"
if [ $(echo -n "$INPUT" | wc -m) -ge 6 ]
then echo "OK"; exit 0
else echo "Bogus"; exit 1
fi

While you’re developing a script, use echo to show what you’re working with, but also use double quotes.

If you type a line at this program, there will be no shell expansion done on the characters you type. If you simulate a user typing with echo, then you need to prevent the shell that executes the echo from expanding any metacharacters, thus:

echo 'gra$$' | bash yourscript

Omitting the single quotes means that the shell will replace the $$ with a PID (probably the PID of the parent shell rather than the shell in the pipeline, but with some PID). This is different from you typing 5 characters plus newline after running:

bash yourscript

You could also use:

$ cat > password-file
gra$$
<control-D>
$ cat password-file
gra$$
$ bash yourscript < password-file

The <control-D> is the EOF indication; it flushes all zero characters typed since the last newline was entered, and cat interprets zero bytes available as EOF. The file therefore contains 6 characters: the letters ‘g‘, ‘r‘, ‘a‘, two ‘$‘ signs and a newline. This will be read by your script verbatim; there will be no expansion on the data.