bash insert backslash for every slash in string

Question

I have a string that is a path:

/tmp/something

I need to escape the forward slashes with backslashes:

\/tmp\/something

How can I do this? Maybe sed?

Please point me in the right direction.

Answer 1

You can do it with Bash's own variable manipulation methods (parameter expansion), though the syntax is fairly monstrous:

$ string='/tmp/something'
$ escapedstring="${string//\//\\\/}"
$ printf '%s\n' "$escapedstring"     
\/tmp\/something

To explain this further, in bash ${var//xxx/yyy} means "take variable var and replace all occurrences of xxx with yyy." In this case, you need to escape the slashes in specifying the substituend and substitute; the former is \/ for a single forward slash and latter is \\\/ for a backslash plus forward slash, so you end up with the ridiculous looking ${string//\//\\\/}.

Remember not to use echo to display its contents as depending on the implementation or environment, echo may do it's own processing of backslash characters.

Answer 2

sed is the right direction! Follow a tutorial like https://www.grymoire.com/Unix/Sed.html#uh-1 ; the very basic method is

echo $string | sed 's;old;replacement;g' 

where s is the command, i.e. "search and replace", g is the flag that means "repeat on each line until you're done", old is what you want to replace (/ in your case) and replacement is what you want to replace it with. (You might need to escape the backslash, so \\/.)

Have fun!

Answer 3

You can do something along these lines.

# let
str='/tmp/something'
str_esc=$(printf '%s\n' "$str" | sed 's:/:\\&:g')