Perl Shell脚本转义的替换模式不起作用

Question

I have this NOT working issue.

echo "aabbccdd" | perl -w -pe "s/(?<Naa>aa)/\g{Naa}-$1/;"

it outputs

Unrecognized escape \\g passed through at -e line 1.

my perl version is "subversion 4 (v5.18.4)"

my bash version is "version 4.2.53(1)-release (x86_64-redhat-linux-gnu)"

my OS distro is "Fedora release 20 (Heisenbug)"

my locale is "C"

how to solve ?

update:

how to display same labeled group by number 1 ?

Answer 1

You can only use \\g{name} (or \\k<name> ) on the matching side of the regular expression. You can use $+{name} on the replacement side:

echo "aabbaaccdd" | perl -w -pe "s/(?<Naa>aa)(?=bb\g{Naa}).*/$+{Naa}/;"

prints out

aa

This is because named captures are placed in the hash table %+ .

EDIT: Wiktor Stribiżew beat me to it while I was writing this, in the comments to the question.