Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
628 views
in Technique[技术] by (71.8m points)

bash - echo "-e" doesn't print anything

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").

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Answer

0 votes
by (71.8m points)

The one true way to print any arbitrary string:

printf "%s" "$vars"

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome to OStack Knowledge Sharing Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

2.1m questions

2.1m answers

60 comments

57.0k users

...