I need to replace string SALT in a file with content of another file. Problem is that the input file has multilines. I tried something like this in my bash script:

SALT=`cat salt.txt`;
sed "s/SALT/$SALT/" wp-config.php > result.txt

It work's fine when the salt.txt is single line, but if there are more lines it fails. I've read that it could do PERL. But I don't know how. Could you help me?

4 Answers 4


Another perl way:

perl -pe 's/SALT/`cat salt.txt`/e' wp-config.php > result.txt

The key here is the /e regexp option allowing us to use a perl command result as a substitution string.


If you want to stick to bash, choose a character that doesn't appear in neither your string nor in your file, let's say @; then:

SALT=`< salt.txt tr '\n' '@'`
sed "s/SALT/$SALT/" wp-config.php | tr '@' '\n' > result.txt

This way before the replacement the newline characters in your string are changed to @ and after the replacement the @ characters are changed back to newlines, so that SALT is not treated as an array anymore but just as a variable containing a long string.


You can do this:

sed -e "/SALT/{r salt.txt" -e "d}" wp-config.php  > result.txt

Where salt.txt is the salt, wp-config.php is the input file and SALT is the string to replace

  • 2
    This will replace whole line with content of file salt.txt if containing a SALT in it not SALT word itself!! Commented Mar 30, 2015 at 13:49
  • is there anyway to fix this? instead of replacing the whole line, just the keyword Commented Apr 27, 2020 at 9:38
perl -e 'open my $S, "<", "salt.txt" or die $!;
         $salt = do { local $/ ; <$S> };
         s/SALT/$salt/, print while <>;
        ' wp-config.php > result.txt

The first line opens the salt.txt.
The second line reads it contents into the $salt variable.
The third takes the command line arguments as file names, reads the files line by line and replaces the string.

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