Shell Programming and Scripting

Could someone tell me how to replace varibles using SED inside Korn Shell?

e.g. I have a ksh file program.ksh below:
------------------------------------
#!/bin/ksh

sed -n '/ABC/p' $1 > output.txt

if [[ $2 = AAA ]]
then
status=new
elif [[ $2 = BBB ]]
then
status=old
fi

sed -n '/$status/p' $1 >> output.txt
------------------------------------


I have a text file called filename.txt with the contents below:
------------------------------------
DDD old
CCC new
ABC old
------------------------------------

When I run program.ksh with filename.txt and BBB as input I want output.txt to contain the two lines below but with the current program it only outputs "ABC new".

DDD old
ABC new


I guess $status is not being recognized as a varible but how do I get this work? I tried putting "\" in front of $status but it didn't work.

Any help will be greatly appreciated.

i wonder how you would be arriving at
DDD old
ABC new <-- this line

there is no such file entry as ABC new

> cat filename.txt
DDD old
CCC new
ABC old

here's the modification

Code:

> cat program.ksh

#!/bin/ksh

sed -n '/ABC/p' $1 > output.txt
# ABC old
if [[ $2 = AAA ]]
then
status=new
elif [[ $2 = BBB ]]
then
status=old
fi

sed -n "/$status/p" $1 >> output.txt
#DDD old
#ABC old

>cat output.txt
ABC old
DDD old
ABC old

Thanks matrixmadhan!
It worked!
Yes, you are right about "ABC new" line. It should have been "ABC old".
Thank you for your quich response!

cheers
Steve

I have a different question on sed.
How do you delete everything after a particular string using sed?

e.g. I have a file with three lines below:

A100 ;
B2000 ; B2000
C30 ; C30 ; C30

I want to delete everything after ";" to become like below:

A100
B2000
C30

Any help will be greatly appreciated.

Code:

[~/temp]$ echo "A100 ; 
B2000 ; B2000
C30 ; C30 ; C30" |  sed -e 's/;.*//g'
A100 
B2000 
C30 
[~/temp]$

Thanks vino!