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I'm using GNU bash, version 3.00.15(1)-release (x86_64-redhat-linux-gnu). And this command:
echo "-e"
doesn't print anything. I guess this is because "-e" is one of a valid options of echo command because echo "-n" and echo "-E" (the other two options) also produce empty strings.
The question is how to escape the sequence "-e" for echo to get the natural output ("-e").
456
Printing Newline in Bash The most common way is to use the echo command. However, the printf command also works fine. Using the backslash character for newline “\n” is the conventional way. However, it's also possible to denote newlines using the “$” sign.
On the command line, press Shift + Enter to do the line break inside the string.
From the bash manual: The backslash character '\' may be used to remove any special meaning for the next character read and for line continuation.
The one true way to print any arbitrary string:
printf "%s" "$vars"
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