Replace a word in a line except for the last char of the word?

Ronnie Doorzon has asked for the wisdom of the Perl Monks concerning the following question:

I have a line that contains spaces which I want to replace
except for the last space.
eg: replace

$line = '<br>  2<br>   3<br>   4';
[download]

into

$line = '<br>&nbsp 2<br>&nbsp&nbsp 3<br>&nbsp&nbsp&nbsp 4';
[download]

The number of spaces can range from 0 to 50
Currently I'm doing this with many s/// lines but there should be a better way.

sub convert_line_1
{
my $line_convert = shift;
$line_convert =~ s/(\s{5})/\&nbsp\&nbsp\&nbsp\&nbsp /g;
$line_convert =~ s/(\s{4})/\&nbsp\&nbsp\&nbsp /g;
$line_convert =~ s/(\s{3})/\&nbsp\&nbsp /g;
$line_convert =~ s/(\s{2})/\&nbsp /g;
return($line_convert);
}

sub convert_line_2
{
my $line_convert = shift;
$line_convert =~ s/&nbsp&nbsp&nbsp&nbsp /     /g;
$line_convert =~ s/&nbsp&nbsp&nbsp /    /g;
$line_convert =~ s/&nbsp&nbsp /   /g;
$line_convert =~ s/&nbsp /  /g;
return($line_convert);
}
[download]

Comment on Replace a word in a line except for the last char of the word? Select or Download Code

Replies are listed 'Best First'.
Re: Replace a word in a line except for the last char of the word? by Roy Johnson (Monsignor) on Mar 11, 2005 at 12:03 UTC
`s/ (?= )/ /g;` [download] Caution: Contents may have been coded under pressure.	[reply] [d/l]
Re^2: Replace a word in a line except for the last char of the word? by cog (Parson) on Mar 11, 2005 at 12:20 UTC
Allow me to explain :-) `(?=)` matches only if what's between the "=" and the ")" is found to the right of the expression. This means the expression matches a space only if it is followed by another space.	[reply] [d/l]
Re: Replace a word in a line except for the last char of the word? by cog (Parson) on Mar 11, 2005 at 12:26 UTC
A few notes on your code: In HTML, it's ` `, and not `&nbsp`; you are missing the semicolons Your substitutions have unnecessary brackets. `(\s{5})` will match the same as `\s{5}`. The difference is that the brackets will capture those spaces into $1, but you don't seem to need that here On the right hand side of substitutions, you don't need to escape the ampersand character On your subroutines (or code blocks, for that matter), try to indent things (not sure if you just missed that because of pasting the code here) Hope you live a happier life because of this :-)	[reply] [d/l] [select]
Re: Replace a word in a line except for the last char of the word? by gube (Parson) on Mar 11, 2005 at 12:19 UTC
Hi try this, `$line = '<br> 2<br> 3<br> 4'; $line =~ s#(?= )#&nbsp#gsi; while ($line =~ s#(&nbsp) (&nbsp)#$1$2#gsi) { } print $line; o/p: <br>&nbsp&nbsp 2<br>&nbsp&nbsp&nbsp 3<br>&nbsp&nbsp&nbsp 4` [download] gube	[reply] [d/l]
Re: Replace a word in a line except for the last char of the word? by dmorelli (Scribe) on Mar 11, 2005 at 14:11 UTC
Hm. Some of these are so long. This seems to do the job. `s/ +( )/ $1/g` As mentioned by cog, be mindful that it must be ` ` with the semicolon.	[reply] [d/l] [select]
Re: Replace a word in a line except for the last char of the word? by perlsen (Chaplain) on Mar 11, 2005 at 12:58 UTC
Hi, just try the below code (is this meet your requirement?) `my $line = '<br> 2<br> 3<br> 4'; my $temp = $line; while ($temp =~m#(<br>)(.+?)( \d+)#gsi) { my $full= $&; my $text=$1; my $space = "$2"; my $digit = "$3"; my @arr=(); (@arr) = $space =~s#( )# #gsi; my $nbsptext = " " x "@arr"; $line =~s#$full#${text}${nbsptext}${digit}#gsi; $full=""; } print $line; output: <br>  2<br>   3<br>   4` [download]	[reply] [d/l]
Re: Replace a word in a line except for the last char of the word? by Ronnie Doorzon (Initiate) on Mar 11, 2005 at 14:38 UTC
Thanks Monks for the great and quick replies. It is working now! Thnx again.	[reply]
Re: Replace a word in a line except for the last char of the word? by PhilHibbs (Hermit) on Mar 11, 2005 at 15:40 UTC
You might be better replacing alternate spaces with   since this has the same effect in HTML, takes up less space in the output, and is simpler code: `s/ /  /g;`	[reply] [d/l]