How to ignore last newline in multiline line with Perl
I need to turn foo,bar
into
BAZ(foo) \
BAZ(bar)
\
there is as-is.
I tried with echo 'foo,bar' | tr , '\n' | perl -pe 's/(.*)\n/BAZ($1) \\\n/g'
but created
BAZ(foo) \
BAZ(bar) \
So this is either a completely wrong approach, or I need a way to ignore the last newline in a multi-line string from Perl.
+3
Marcel Stör
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4 answers
Using eof
to determine if you are on the last line:
echo 'foo,bar' | tr , '\n' | perl -pe 'my $break = (eof) ? "" : "\\"; s/(.*)\n/BAZ($1) $break\n/g'
0
RobEarl
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You can use join
with map
for example:
$ echo 'foo,bar' | perl -F, -lape '$_ = join(" \\\n", map { "BAZ(" . $_ . ")" } @F)'
BAZ(foo) \
BAZ(bar)
-
-a
enables automatic split mode by splitting the input with a separator,
and assigning it@F
-
map
takes each element of the array and wraps it -
join
adds backslash and newline between each element
+3
Tom fenech
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echo 'foo,bar' | perl -ne'print join " \\\n", map "BAZ($_)", /\w+/g'
Output
BAZ(foo) \
BAZ(bar)
+3
Borodin
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Using awk
, you can:
s='foo,bar'
awk -F, '{for (i=1; i<=NF; i++) printf "BAZ(%s) %s\n", $i, (i<NF)? "\\" : ""}' <<< "$s"
BAZ(foo) \
BAZ(bar)
Or using sed
:
sed -r 's/([^,]+)$/BAZ(\1)/; s/([^,]+),/BAZ(\1) \\\n/g' <<< "$s"
BAZ(foo) \
BAZ(bar)
+1
anubhava
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