Bash variable in rename regex

Question

I'm writing a bash script to rename files names and using command

rename 's/\d+/sprintf("%02d", $&)/e' *

But would like to replace "%02d" with something like "%0"$var"d" thus replacing '2' with $var but it seems like it doesn't work and I'm not well versed in regex and how to use escape \\ and would like help

Answer 1

Sure, your whole expression is inside single quotes where no expansion is performed. To expand a part of the command, move it out of the quotes:

rename 's/\d+/sprintf("%0'"$var"'d", $&)/e' *

Or put the whole expression in double quotes, but then you'll need to escape more:

rename "s/\\d+/sprintf(\"%0${var}d\", \$&)/e" *

Answer 2

您也可以通过环境传递值

env var=5 rename 's/\d+/sprintf("%0$ENV{var}d", $&)/e' *