how to replace a known substring value in a string variable

Question

I have a string that looks like the following:

prop_1=val1 prop_2=val2 prop_3=val3

I want to replace the entirety of one of the tokens so the string looks like:

prop_1=val1 prop_X=valX prop_3=val3

I have tried

$params =~ s/prop_2=val2/prop_X=valX/ig;

but nothing is changing. What am I doing wrong?

ikegami · Accepted Answer · 2013-04-29 19:58:41Z

2

You are mistaken.

$params = 'prop_1=val1 prop_2=val2 prop_3=val3';
$params =~ s/prop_2=val2/prop_X=valX/ig;
print "$params\n";

gives

prop_1=val1 prop_X=valX prop_3=val3

You have not demonstrated the problem. The only thing I can think of that could cause the above to fail is if pos($params) isn't zero, such as if you perhaps did use /.../g in scalar context earlier, as in

if ($params =~ /.../g)   # Bad

Note that s/(?<!\S)prop_2=val(?!\S)/prop_X=valX/ig would be better as it won't match aprop_2=vals.

answered Apr 29, 2013 at 19:58

ikegami

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8 Comments

Jon Purdy Over a year ago

(?<!\S) and (?!\S) would be simpler as \b.

ikegami Over a year ago

@Jon Purdy, How can something that gives the wrong result be simpler? (e.g. foo.prod_2=val)

Jon Purdy Over a year ago

Just giving an alternative. There’s no indication in the question of whether it’s right or wrong, as OP’s examples used only word characters. But “word boundary” is certainly simpler than “not preceded/followed by non-whitespace”.

ikegami Over a year ago

@Jon Purdy, I don't see how you can call (?<!\S)\b and \b(?!\S), simpler than (?<!\S) and (?!\S).

Jon Purdy Over a year ago

I didn’t. Don’t worry about it.

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