Loop Help

[joec369]

Hello,

I'm trying to write a simple script. Here is what I have so far:
#! /bin/bash
echo "$USERNAME Please Login With Password!!"
COUNT=5
while [ COUNT > 0 ] ; do
echo "Login Name: $USERNAME "
echo "Password:"
read Psword
if [ $PASSWORD = $PSWORD ] ; then
echo "Login Successful"
let COUNT=COUNT-5
else
if [ $PASSWORD != $PSWORD ] ; then
echo "Wrong Password!!"
let COUNT=COUNT-1
fi
fi
done

Basically what I'm trying to do is to compare variables $PASSWORD to $PSWORD. If they are equal the just echo "Login Good" if they are not the same, then echo "Wrong Password." Now the part I can't figure out. I want to limit the number of failed attempts to five. So basically limit the endless while loop to only five go arounds. I've been trying to mess with this COUNT thing, but I don't know if it's what I need or not.

Any Help Greatly apreciated!!!!!

[vsemaska]

The '>' usually redirects output. If you look in your directory you'll find a file named 0. To make bash use '>' as 'greater than', 'escape' it by preceeding it with a backslash(\).

while [ $COUNT \> 0 ] ; do

or use -gt

while [ $COUNT -gt 0 ] ; do

Note that I preceeded COUNT with a'$' in order to get the value of COUNT.

Also, variables are case sensitive so Psword in your 'read' command isn't the same as PSWORD is your 'if' statements.

Finally, when comparing strings in 'if' statements enclose them in double quotes.

if [ "$PASSWORD" = "$PSWORD" ] ; then

[joec369]

Thanks so much for the help. It works and does exactly what I was trying to get it to do. Here is the final result:

#! /bin/bash
echo "$USERNAME Please Login With Password!!"
COUNT=5
while [ COUNT -gt 0 ] ; do
echo "Login Name: $USERNAME "
echo "Password:"
read PSWORD
if [ $PASSWORD = $PSWORD ] ; then
let COUNT=COUNT-5
echo "Login Successful"
else
if [ $PASSWORD != $PSWORD ] ; then
let COUNT=COUNT-1
echo "Wrong Password!!"
fi
fi
done