I execute the following bash script:
#!/bin/bash
version=$1
echo $version
sed 's/\${version.number}/$version/' template.txt > readme.txt
I'm expecting to replace all instances of ${version.number} with the contents of the variable "version". Instead the literal text $version is being inserted.
What do I need to do to make sed use the current value of $version instead?
sed "s/\${version.number}/$version/" template.txt > readme.txt
Only double quotes do dollar-sign replacement. That also means single quotes don't require the dollar sign to be escaped.
sed. So I think it should be \\$.
$ sign is only an anchor when it's the last character of the regex (or possibly subexpression). But I admit I got lucky. ;) You can optionally add an escaped slash, but then it's \\\$.You could also simply unquote the variables
sed 's/'${version.number}'/'$version'/' template.txt > readme.txt
It's a bit old, but it might still be helpful... For me it worked using a double escape as Philip said (and escaping parenthesis, if you use them...):
#!/bin/bash
LIVE_DB_NAME='wordpress';
STAGING='staging';
sed -r "s/define\('DB_NAME', '[a-zA-Z0-9]+'\);/define('DB_NAME', '\\${LIVE_DB_NAME}');/" ${STAGING}/wp-config.php >tmp1;
