How can I escape double quotes in Bash?

Question

I want to escape double quotes in Bash. I followed the following approach:

#!/bin/bash
this is a  \"number\"!

But is there another way?

score 7 · Answer 1 · edited Sep 16 '21 at 13:56

7

You can enclose the double quotes in single quotes:

echo '"'hola'"'

Or alternatively the whole text, including the double quotes:

echo '"hola"'

edited Sep 16 '21 at 13:56

Peter Mortensen

answered Feb 23 '16 at 19:41

Dirk Herrmann

thanks for the support this works well – neo33 Feb 23 '16 at 20:01
1

while this will work it won't interpolate vars – Marc Young Feb 23 '16 at 20:10
1

You wouldn't put the parameter expansion inside the single quotes, but you'd have to quote them as well. This approach quickly gets unreadable:, e.g., `echo '"'"$string"'"'` – chepner Feb 23 '16 at 23:39

score 5 · Accepted Answer · edited Sep 16 '21 at 13:57

5

With GNU Bash:

echo -e "this is a \x22number\x22"

Output:

this is a "number"

edited Sep 16 '21 at 13:57

Peter Mortensen

answered Feb 23 '16 at 19:47

Cyrus

2

Better yet, `printf "this is a \42number\42\n"`, which will work in any POSIX-compatible shell. (`\42` is the character whose ASCII value is 42 in *octal*.) – chepner Feb 23 '16 at 23:32

2 Answers2