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What is wrong with this script?

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Old Unix and Linux 05-12-2011   -   Original Discussion by Oman_Member
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What is wrong with this script?

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

Script Name: shell_script1

while [ “$looptrack” = 1 ]
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

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Old Unix and Linux 05-12-2011   -   Original Discussion by Oman_Member
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- 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

Last edited by Scrutinizer; 05-12-2011 at 05:18 AM..
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Old Unix and Linux 05-12-2011   -   Original Discussion by Oman_Member
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while condition; do

and your double quotes look strange...

if looptrack = 'y' then looptrack != '1' !!! because 'y' != '1'
Old Unix and Linux 05-12-2011   -   Original Discussion by Oman_Member
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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.
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