Recently I tried passing a bash variable to perl command in bash script, it didn’t end well.
Troy Engel from http://tacticalvim.wordpress.com/ was nice enough to point out the issue:
use sed instead of perl for what you need; it’s simpler, faster and uses the bash variables easily.
I set up a test script /home/someuser/test.sh to show:
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=== # cat test.conf A=domain B=PATH_DEST === # cat test.sh domain="test" PATH_DEST_APACHE="/test" PATH_DEST_CONFIG="/home/someuser/${domain}.conf" sed -i "s|domain|${domain}|g" "${PATH_DEST_CONFIG}" sed -i "s|PATH_DEST|${PATH_DEST_APACHE}|g" "${PATH_DEST_CONFIG}" === # sh -x ./test.sh + domain=test + PATH_DEST_APACHE=/test + PATH_DEST_CONFIG=/home/someuser/test.conf + sed -i 's|domain|test|g' /home/someuser/test.conf + sed -i 's|PATH_DEST|/test|g' /home/someuser/test.conf === # cat test.conf A=test B=/test === |
What’s important to note is that A, you *must* use double-quotes in the sed expression so that variable expansion works; single-quotes inhibits expansion. B, as you found out, you need to use something other than “/” as your expression separator since you have a replacement variable with that character in it. I used pipes in sed, but you could use almost anything (“#”, etc.) for the same effect.
