Perl while loops not working - perl

Question

I'm quite new to perl and apologies if this has already been answered in a previous discussion. I have a script that needs to use the declared variables outside the loops, but only one loop is working, even though I have declared the variables outside of the loop, the code is:
my $sample;
open(IN, 'ls /*_R1_*.gz |');
while (my $sample = <IN>) {
chomp $sample;
print "sample = $sample\n";
my $fastq1="${sample}"; #need to use fastq1 later on hence it's declared here
my $sample2;
open(IN, 'ls /*_R2_*.gz |');
while (my $sample2 = <IN>) {
chomp $sample2;
print "sample2 = $sample2\n";
my $fastq2="${sample2}"; #need to use fastq2 later on hence it's declared here
}
}
Sample2 works but sample1 does not, only the first sample is output and then the loop goes onto sample2, the output is:
sample =/sample1_R1_001.fastq.gz
sample2 =/sample1_R2_001.fastq.gz
sample2 =/sample2_R2_001.fastq.gz
sample2 =/sample3_R2_001.fastq.gz
etc..
Can anyone figure this out?
Thanks

Answer 1

From your comments, I assume that your problem is probably that you declare $fastq1 and $fastq2 inside the loop. That means they will be out of scope outside the loops, and not accessible. You need something like:
my ($fastq1, $fastq2);
while ( ... ) {
....
$fastq1 = $sample;
}
Note that this will only save the last value in the loop of that variable. The others will of course be overwritten each loop iteration. If you have more values to save, use an array or hash.
Some other notes on your code.
You should always use
use strict;
use warnings;
Not doing so is a very bad idea, as it will only hide the errors and warnings, not solve them.
my $sample;
You declare this variable twice.
open(IN, 'ls /*_R1_*.gz |');
This is just bad on all possible levels:
System calls are always the least desirable option, unless no alternatives exist
Perl has many ways of reading file names
Parsing the output of ls is fragile and not portable
Piping the result of the system command through open is compounding the other flaws with this approach.
Recommended solution: Use either opendir + readdir or glob:
for my $files (</*_R1_*.gz>) { ... }
# or
opendir my $dh, "/" or die $!;
while (my file = readdir $dh) {
next unless $file =~ /_R1_.*\.gz$/;
...
}
my $fastq1 = "${sample}";
You do not need to quote a variable. Nor use support curly braces.
When declaring the variable with my inside a loop, it only retains its value that single loop iteration. Since you never use this variable, I assume you meant to use it outside the loop. But it will be out of scope there.
This can be written
my $fastq1 = $sample;
But you probably want to declare those variables outside your while loops, or they will be out of scope there. You should know that this will only save the last value for these variables, of course.
Also, as Rohit says, your loops are nested, which I assume is not what you wanted. This is most likely because you do not use a proper text editor to write your code, so your indentation is all messed up, and it is hard to see where one loop ends. Follow Rohit's advice there.

Answer 2

You are closing the first while loop after the end of 2nd while loop. Because of that, your 2nd while loop become a part of your 1st while loop, wherein, you are re-assigning the file handler - IN to a different file. And since you are exhausting it in the inner while loop, your outer while loop never run again.
You should close the brace before starting the next while:
while(my $sample = <IN>){
chomp $sample;
print "sample = $sample\n";
my $fastq1="${sample}";
} # You need this
my $sample2;
open(IN, 'ls /data_n2/vmistry/Fluidigm_Exome/300bp_fastq/*_R2_*.gz |');
while(my $sample2 = <IN>){
chomp $sample2;
print "sample2 = $sample2\n";
my $fastq2="${sample2}";
}
# } # Remove this