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regex - Why does my Perl one-liner report "Bareword found where operator expected"?

I want to convert block size into MB. I am using the /e option in my substitution. When I add starting MB in substitution part it giving me the error.

e.g:

This works.

 echo "16777216 SELECT" |perl -lane 's#(d+)(s+SELECT)#$1/(1024*1024*2)#e; print'
8

This giving me the error.

echo "16777216 SELECT" |perl -lane 's#(d+)(s+SELECT)#$1/(1024*1024*2) MB $2#e; print'
Bareword found where operator expected at -e line 1, near ") MB"
        (Missing operator before MB?)
syntax error at -e line 1, near ") MB "
Execution of -e aborted due to compilation errors.

Any help fixing the second one?

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1 Answer

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by (71.8m points)

Change

(1024*1024*2) MB $2

to

(1024*1024*2)."MB".$2

The /e modifier tells the engine to treat the substitution field as Perl code.


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