Reading password in Unix shell

admin — Tue, 02 Feb 2010 21:18:51 +0000

print -n "Enter Your password:"
stty_orig=`stty -g`
trap "stty ${stty_orig}; exit" 1 2 3 15
stty -echo >&- 2>&-
read PASS
stty ${stty_orig} >&- 2>&-
trap 1 2 3 15
print

trap :catches interruptions. I.e. if the user presses Ctrl+C, the normal stty mode is set before stopping the program
stty -echo :switches off the display echo
>&- 2>&- :helps to avoid “stty: Not a typewriter” message for non-interactive scripts.

Unix shell: workaround for loop problem

admin — Fri, 15 Jan 2010 23:04:40 +0000

It’s not possible to get the value of the loop variables in some versions of ksh.

Example:

#!/bin/ksh

num=0
cat $0 | while read line ; do
let num=num+1
done

echo "Number=$num"

This script will return “Number=0″ as the result on some Linux machines.

Here is the workaround for the problem: You should change the redirection method for the input file.

#!/bin/ksh

l=0
while read line ; do
let l=l+1
done < $0

echo "Number=$l"

The last script will return the correct result: “Number=9″

Another possibility is to use the named pipe:

#!/bin/ksh

TMPPIPE=/tmp/thepipe.$$
mkfifo $TMPPIPE
cat $0 > $TMPPIPE &
num=0
while read line ; do
let num=num+1
done < $TMPPIPE

echo "Number=$num"

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Reading password in Unix shell

Unix shell: workaround for loop problem