Shell Programming and Scripting

I write this script and do not no whrer the wrong!!
Can anyone hlep me?

#=========================================
Script Name: shell_script1
#=========================================

Code:

looptrack=y
while [ “$looptrack” = 1 ]
do
echo -n “Type in the account number:” read account
echo -n “Type the first and last name:” ; read full_name
echo -n “Type the age:” red age
echo -n “Enter another record?” ; read looptrack
finish

- looptrack is initialised to y instead of 1
- Use done instead of finish.
- Use regular " instead of “ and ”
- Use read instead of red
- Separate commands by newline or semicolon

Code:

while condition; do
    ...
done

and your double quotes look strange...

if looptrack = 'y' then looptrack != '1' !!! because 'y' != '1'

The slandy quotes “ ” look like they have come from Microsoft Word and are not valid in a unix script. They should be standard double quotes " " . You need to be using a unix editor to create a unix script file in the correct character set.

After taking into account the advice from various posters, please post the new version and show what happens when you execute the script. It is hard to suggest changes without knowing what the script is designed to do.