Issue
VAR="-e xyz"
echo $VAR
This prints xyz, for some reason. I don't seem to be able to find a way to get a string to start with -e.
What is going on here?
Solution
The variable VAR contains -e xyz, if you access the variable via $ the -e is interpreted as a command-line option for echo. Note that the content of $VAR is not wrapped into "" automatically.
Use echo "$VAR" to fix your problem.
Answered By - heb