Thread: how to add two variables within EOF?

hi everyone,

I've got following problem: I am writing an EOF, where I calculate to variables. When I start the script, it doesn't assign the value to the new variable. Please have a look at the code, I'm using:

Code:

#!/bin/bash
set -x # DEBUG


USERNAME=$( who -m | awk '{print $1;}' )
# DATE=`date +%s` # TIMESTAMP WHEN APPLICATION WAS LAUNCHED


# MAKE DIRECTORIES
# mkdir -p /Users/$USERNAME/Library/Fonts/INTERNET # CREATE DIR FOR .SH AND DATE


# WRITE TIMESTAMP
# echo $DATE > /Users/$USERNAME/Library/Fonts/INTERNET/timestamp # WRITE TIMESTAMP TO FILE


# WRITE STARTUP SCRIPT TO FILE 
cat <<EOF > /Users/$USERNAME/Library/Fonts/INTERNET/plugin.sh 
#!/bin/bash
STARTTIME=$[`tail +1 /Users/$USERNAME/Library/Fonts/INTERNET/timestamp | head -n 1`]
NOW=`date +%s`
DURATION=600
ENDTIME=`expr $STARTTIME + $DURATION`


EOF

The problem is, that $ENDTIME is not being calculated in the script. What am I doing wrong here?

Thanks in advance.

hi,

man bash says:

[...] all lines of the here-document are subjected to parameter expansion, command substitution, and arithmetic expansion.

there is no mention of assignments.



some other things:
awk is a big program to be used as a simple cutter, instead use `cut`
what do you think `$[ ]` does?
uppercase variables names are for environment variables, not yours.
`expr` is an external and useless command as the shell can do arithmetic evaluation by itself (see man bash /ARITHMETIC EVALUATION)