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Old 04-18-2012, 04:46 PM   #1
tuts73
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Registered: Apr 2012
Posts: 7

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variables


Hi all, just wondering if you can help - i have a program that works but now i want to prompt for username and use the store value, but it prompts but doesnt use the variable as far as i can see....

#!/bin/bash

showMenu () {
echo "1) Display user account on the laptop"
echo "2) Display the bash history"
echo "3) Does the Mozilla Firefox history log file exist"
echo "4) quit"
}


var=username

while [ 1 ]
do
showMenu
read CHOICE
case "$CHOICE" in
"1")
cat /etc/group |cut -d: -f1
;;
"2")
cat /home/john403920394/.bash_history
;;

"3") echo "Please enter a username: "
read $username
FILE=$1/home/$username/.mozilla/firefox/b7wuc85h.default/places.sqlite

if [ -f $FILE ];
then
echo "File $FILE this file does exist."
else
echo "File $FILE does not exist."

fi

;;


"4")
exit
;;
esac
done


Would appriciate any help - thank you
 
Old 04-18-2012, 05:00 PM   #2
Kustom42
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Registered: Mar 2012
Distribution: Red Hat
Posts: 1,568

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remove the dollar sign from your read statement.

Code:
#!/bin/bash
echo "What is your username?"
read username
echo $username
 
1 members found this post helpful.
Old 04-19-2012, 02:07 AM   #3
grail
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Registered: Sep 2009
Location: Perth
Distribution: Manjaro
Posts: 7,516

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As demonstrated, please also use [code][/code] tags when displaying code to make it readable and keep formatting.
 
  


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