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could someone please explain, why perl doesn't replace the regexp:

root@machine08:~# VERSIONK8S='${VERSION_KUBERNETES:-1.23.3-00}'
# with sed as i want it
root@machine08:~# sed -e "s/^VERSIONK8S=.*/VERSIONK8S=${VERSIONK8S}/g" /root/coding/k8s-setup/allnodes_basic_setup.sh | grep ^VERSIONK8S=
VERSIONK8S=${VERSION_KUBERNETES:-1.23.3-00}
# with perl not exactly as i want it
root@machine08:~# perl -pe "s/^VERSIONK8S=.*/VERSIONK8S=${VERSIONK8S}/g" /root/coding/k8s-setup/allnodes_basic_setup.sh | grep ^VERSIONK8S=
VERSIONK8S=-e
Wiktor Stribiżew
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Dschungel Ratte
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1 Answers1

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From Perldoc:

If the delimiter chosen is a single quote, no variable interpolation is done on either the PATTERN or the REPLACEMENT. Otherwise, if the PATTERN contains a $ that looks like a variable rather than an end-of-string test, the variable will be interpolated into the pattern at run-time.

Perl tries to expand the substitution string ${VERSION_KUBERNETES:-1.23.3-00} as a perl variable starting with $ and fails. To avoid the expansion, please try:

perl -pe "s'^VERSIONK8S=.*'VERSIONK8S=${VERSIONK8S}'g" /root/coding/k8s-setup/allnodes_basic_setup.sh | grep ^VERSIONK8S=

by using a single quote as a delimiter instead of a slash such as s'pattern'replacement'.

tshiono
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    Don't build Perl programs from the shell. It's complicated and prone to [code injection bugs](https://en.wikipedia.org/wiki/Code_injection). There are [much better alternatives](https://stackoverflow.com/q/53524699/589924) – ikegami Mar 09 '22 at 09:01