Why is the order of execution inversed in this bash script?

Question

I have this script :

ssh -T user@$123.456.789.123 <<EOF

    cd www

    var=$(tail index.htm)

    echo $var

EOF

What I thought it should do is :

1. Connect to the server through SSH ,
2. then change to the folder www ,
3. then store the tail of index.htm into the variable var
4. and finally echo the result.

Instead it seems that tail is executed before the change of folder, and thus doesn't find the index.htm file.

I've tried with different commands, and each time it seems the result from command substitution I'm trying to store into a variable is executed right after the SSH connexion is opened, before any other piece of script.

What am I missing here ?

Answer 1

The $(...) is being expanded locally, before the contents of the here document are passed to ssh . To send literal text to the remote server, quote the here document delimiter.

ssh -T user@$123.456.789.123 <<'EOF'
    cd www
    var=$(tail index.htm)
    echo "$var"
EOF

(Also, quote the expansion of $var to protect any embedded spacing from the shell.)

Answer 2

The tail is running in the bash script on your local machine, not on the remote host. The substitution is getting made before you even execute the ssh command.

Your script can be replaced with simply:

ssh -T user@$123.456.789.123 tail www/index.htm

Answer 3

If you want to send those commands to the remote server, you can write

ssh -T user@$123.456.789.123 'cd www && var=$(tail index.htm) && echo $var'

Note that conditioning the next command on the result of the previous allows SSH to return a meaningful return code. In your heredoc, whatever happens (eg tail fails), SSH will return with $?=0 because echo will not fail.

Another option is to create a script there and launch it with ssh.