problem in substituting the values of variables via 'sed'

[needee]

I need to substitute the value of a variable with the value of another variable

I did a test

# cat file
this is needee
# cp file file.old

# echo $STRING1
this is needee

# echo $STRING2
this is NOT needee

# cat file.old | sed 's/$STRING1/$STRING2/' > file

# cat file
this is needee

i.e no change in the 'file'

please help me in substituting the values of variables via 'sed'

[stokilo]

# cat file.old | sed "s/$STRING1/$STRING2/" > file

replace ' with " for variables

[needee]

Hi thanks stokilo, yes double quotes works

another relevant question
the following code has to do two task
1, if there is no TIMEZONE set, then set it in /etc/sysconfig/clock
2, if there is wrong TIMEZONE in /etc/sysconfig/clock, then fix it

ok, now
the proper value of timezone in /etc/sysconfig/clock should be
TIMEZONE="Asia/Tashkent"

right now the value of timezone(wrong) in /etc/sysconfig/clock is
TIMEZONE="aa/tshknt"

#!/bin/bash
CLOCK=/etc/sysconfig/clock
TIMEZONE_STRING='TIMEZONE="Asia/Tashkent"'
CHECK_TIMEZONE=$(grep ^TIMEZONE $CLOCK)

if [ -z "$CHECK_TIMEZONE" ]; then echo "$TIMEZONE_STRING" >> $CLOCK
else cp "$CLOCK" /tmp/clock
cat /tmp/clock | sed "s/$CHECK_TIMEZONE/$TIMEZONE_STRING/" > $CLOCK
fi
exit

upon execution of the above code, I got the following error

sed: -e expression #1, char 36: unknown option to `s'

[stokilo]

sed -e "s/$CHECK_TIMEZONE/$TIMEZONE_STRING/"

Did you forget -e option ? I didn't check this script but try with -e switch.

Best regards

Slawek