How to extract number between two words?

[sjit]

Hi There,
I am trying to extract the number between two words.
Below is the code which am using but not getting any output.
Could you please help me to get the output.

$x = "The start text always precedes 5 times
              of the end text.";
 
if ($x=~ /The(.*)times(\d+)/) {
 
print "X:$1."
}
 

Thanks
Jit

[numberwhun]

You referenced $1 in your code, but you should actually be referencing $2. Try changing it and see if that prints what you want.

[sjit]

Thanks numberwhun for your response.
If I don't use (\d+)then am gettign the right output for $1 that is "start text always precedes 5"
anyway I used $2 and still not getting the output.

[Rabbit]

The . symbol is greedy. It will eat up your digit along with everything else.

Aren't you actually looking for /\b(\d)+\b/

[sjit]

My Intention is to capture the number between two words. In this example between The and times. That is why am using
/The(.*)times(\d+)/. I also tried with \b but still no output. if I remove \d, them am getting the text.

[Rabbit]

I'm not sure why (\d+) is working unless it's seeing the period as a numeric character. But the reason it doesn't work without it is because you didn't account for the period at the end of the sentence so it's not matching. So really it should be /The(.*)times\./

But that .* is too greedy. Try /The([A-Za-z\s])*\b(\d+)\b([A-Za-z\s])*times\./

[sjit]

Thanks Rabbit. it is working now. but I have to remove "\." and $2 is responding but not $1.(any reason?)
The modified code is:
if ($x=~ /The([A-Za-z\s])*\b(\d+)\b([A-Za-z\s])*times/) {
To have more understanding of the code I put another number (6) in 2nd line and trying to extract (6) by $3 but its not giving any output.
**************
Could you help me on this please.
$x = "The start text always precedes 5 times
of the 6 end text.";
if ($x=~ /The([A-Za-z\s])*\b(\d+)\b([A-Za-z\s])*text/) {
print "\n X:$3\n"
}

[numberwhun]

My mistake. Had to wake up and rethink what was going on. $2 would not have it either because the number was contained in $1. You are grabbing way to much. You just want the number. So why not either use what Rabbit has given you, or try this sample, which does grab the number for you:

#!/usr/bin/perl
 
use strict;
use warnings;
 
my $x = "The start text always precedes 5 times of the end text.";
 
if ($x=~ /(\d+)/) {
    print "X: $1\n";
}
 

Regards,

Jeff

[sjit]

I got it.
I have to put /s.
Here is the code:
$x = "The start text always precedes 5 times
of the 6 end text.";
if ($x=~ /The([A-Za-z\s])*\b(\d+)\b([A-Za-z\s])*text/s){
print "\n X:$3\n"
}

[sjit]

Sorry I was too excited to jump into conclusion.
$3 is not fetching anything. but$2 is fetching 5.
Am still not able to get $3=6

[Rabbit]

I don't know how perl implements regex so I don't know what $3 and what /s does but if you're trying to return all number matches, how come you can't use m/\b\d+\b/g

However, if you wish to use the more precise expression and return all number matches, you would need to get the submatches from your expression wouldn't you? Unfortunately I don't know how you go about doing that in perl though.

[sjit]

I can use m/\b\d+\b/g to get all the matching numbers but need to use between two words from multiple lines.
That is why am using this code to minimize the search area between two words:
$x = "The start text always precedes 5 times
of the 6 end text.";
if ($x=~ /The([A-Za-z\s])*\b(\d+)\b([A-Za-z\s])*text/){
print "\n X:$2\nY:$3\n"
}
Any suggestion to get the values for X & Y

[numberwhun]

Sure, you could always match the first number and then use a lookahead assertion to match the next number.

[Rabbit]

So you just want numbers that aren't at the start or end of the string? What about m/[^\A]\b(\d+)\b[^\Z]/g

[sjit]

Thanks Rabit for your patience with me and response to my queries.
I got only one output. Please see below;
***********
$x = "The start text always precedes 5 times
of the 6 end text.";

#if ($x=~ /The([A-Za-z\s])*\b(\d+)\b([A-Za-z\s])*text/g) {
if ($x=~ m/[^\A]\b(\d+)\b[^\Z]/g) {
print "\nZ=$1 X:$2\nY:$3\n"
}
*************
Output:
Z=5 X:
Y:
---------
This time got a response for $1 but still didn't get number 6.
Thanks

[Rabbit]

Is global mode enabled in your regex settings? I read that just putting /g will turn on global mode but if it's global, it should be matching.

[Rabbit]

I was reading up on the perl regex engine and it turns out that to get all matches, you have to do something like this

while ($string =~ m/regex/g) {
  print "Found '$&'.  Next attempt at character " . pos($string)+1 . "\n";
}

[RonB]

#!/usr/bin/perl
 
use strict;
use warnings;
use 5.010;
 
my $x = "The start text always precedes 5 times
of the 6 end text.";
 
while ( $x =~ /.+?(\d+)./g ) {
    say $1;
}

[sjit]

Hi RonB,
Thanks for your response.
I tried but getting only 5:
z:5
X:
Y:
Am not getting number 6.

[RonB]

You didn't run the test script I provided.

E:\TEMP\test>type perl-2.pl

#!/usr/bin/perl
 
use strict;
use warnings;
use 5.010;
 
my $x = "The start text always precedes 5 times
of the 6 end text.";
 
while ( $x =~ /.+?(\d+).+/g ) {
    say $1;
}
 
 

E:\TEMP\test>perl-2.pl
5
6