Bash script print commands, but do not print echo command

You could use trap DEBUG instead of set -v as one option.

For example

#!/bin/bash


# Makes the bash script to print out every command before it is executed except echo
trap '[[ $BASH_COMMAND != echo* ]] && echo $BASH_COMMAND' DEBUG

echo "Cleaning test database"
RAILS_ENV=test bundle exec rake db:drop
echo "************************************************************"
echo ""

echo "Setting up the test database"
RAILS_ENV=test bundle exec rake db:setup
echo "************************************************************"
echo ""

Debug is executed after every command.
$BASH_COMMAND is currently running command.

BASH_COMMAND The command currently being executed or about to be executed, unless the shell is executing a command as the result of a trap, in which case it is the command executing at the time of the trap.

So the trap just checks if the last command did not start with echo and prints it.

Thanks @123 !

To filter multiple strings from the DEBUG trap (not just echo commands), we can use regexp match too.

For example, to filter out any command that starts with "echo" or "read" or "if", I use:

trap '! [[ "$BASH_COMMAND" =~ ^(echo|read|if) ]] && echo $PS4$BASH_COMMAND' DEBUG

Update:

In order to show bash ${variable} values expended, I now use:

trap '! [[ "$BASH_COMMAND" =~ ^(echo|read|if) ]] && \
cmd=`eval echo "$BASH_COMMAND" 2>/dev/null` && echo "$cmd"' DEBUG