44

Perhaps easiest to explain with an example:

$ echo '\&|'
\&|
$ echo '\&|' | while read in; do echo "$in"; done
&|

It seems that the read command is interpreting the backslashes in the input as escapes and is removing them. I need to process a file line by line without changing its contents and I'm not sure how to stop read from being smart here. Any ideas?

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2 Answers 2

Reset to default
65

Accrding to: http://www.vias.org/linux-knowhow/bbg_sect_08_02_01.html :

-r
If this option is given, backslash does not act as an escape character. The backslash is considered to be part of the line. In particular, a backslash-newline pair may not be used as a line continuation.

It works on my machine.

$ echo '\&|' | while read -r in; do echo "$in"; done
\&|
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5 Comments

Awesome, thanks! I feel dumb for not having found the documentation for this, but it's one of those things that's hard to search for because the terms are rather generic :(
Btw, what shell are you using that does not require the -r? I'm using bash.
Watch out that this is not exactly true for read on zsh: documentation of built-in commands. Difference: -r just prints '\' if it is at the end of the line. So read -r in && echo $in will echo two lines if you enter foo\nbar
@normanius the difference is not in read, it's in echo - the zsh built in echo converts escape sequences like \n by default, whereas the bash built in you have to opt in to escape processing with echo -e
Indeed, what Ian says. I propose to change the solution to use printf "%s\n" "$in" which make it work with non-bash shells - I ran into this with dash.
7

Use read -r, as per http://www.ss64.com/bash/read.html:

-r
If this option is given, backslash does not act as an escape character.

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