perl regex replace single array element

Question

I want to read a number of lines from a for loop and split them. After that i want to do a replace on A SINGLE array element.

my @fs = split(';', $line);
$fs[0] =~ s/\"//g;

This doesnt work however. The line

$fs[0] =~ s/\"//g;

returns a compiler error.

Is there a better way to do this?

That line has no errors, could yo have the whole code and the concrete error? — Miguel Prz
– Miguel Prz, Commented Feb 26, 2013 at 14:57
Never say "an error", like all errors are the same thing. That's about as useful as saying "a number", when referring to a specific telephone number. — TLP
– TLP, Commented Feb 26, 2013 at 15:19

Greg Bacon · Accepted Answer · 2013-02-26 14:54:48Z

1

Change the line with split to

my @fs = split(/;/, $line);

because split takes a regex as its first operand.

I suspect the parse error you are seeing is due to an error somewhere else because the syntax of the code in your question is correct.

In general, always fix the first error diagnosed by the parser. Good parsers try to recover so as to report as many errors as possible, but this process is not always reliable. What is the exact text of the error you are seeing?

answered Feb 26, 2013 at 14:54

Greg Bacon

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4 Comments

gaussblurinc Over a year ago

it's wrong. you can put string instead regex into split. but it always be regex

Greg Bacon Over a year ago

perldoc perlstyle: “Just because you CAN do something a particular way doesn't mean that you SHOULD do it that way.”

gaussblurinc Over a year ago

use strict and use warnings and also some idioms.

Martin Nielsen Over a year ago

When i commented out the line of code, the script compiled as it should. It turned out that there was a windows-line-break in the line that was invisible until i opened it with VI. That is why that line was acting out.

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