I have below code to replace the variable value and store it in new variable and leave the original variable intact.
#!/usr/bin/perl
$hdisk="hdisk361";
($newdisk) = ($hdisk =~ s/(hdisk\D*)(\d+)/(($1 eq "hdiskpower"?"prw":"dsk").$2)/ei);
print "hdisk: $hdisk"."\n";
print "newdisk: $newdisk"."\n";
It gives this output:
hdisk: dsk361
newdisk: 1
I want the output like this:
hdisk: hdisk361
newdisk: dsk361
Please help me to fix this code?
Or a bit shorter:
#!/usr/bin/perl
$hdisk="hdisk361";
($newdisk = $hdisk) =~ s/(hdisk\D*)(\d+)/(($1 eq "hdiskpower"?"prw":"dsk").$2)/ei;
Otherwise, as you saw, you get 1, meaning a successful operation. In the code you provided you captured the return value instead of the result.
But don't forget to use use strict and use warnings ;)
Use
$hdisk = "hdisk361"
$newdisk = $hdisk;
$newdisk =~ s/(hdisk\D*)(\d+)/(($1 eq "hdiskpower"?"prw":"dsk").$2/
The substitution s/// works by side-effect. Its return value is hardly ever what you want. You especially don't want to use s here, when you seem not to want $hdisk to change.
Capture the pieces of $hdisk with m instead of s.
use strict;
use warnings;
my $hdisk="hdisk361";
my ($word, $number) = $hdisk =~ m/(hdisk\D*)(\d+)/i;
my $newdisk = ($word eq "hdiskpower"?"prw":"dsk").$number;
print "hdisk: $hdisk"."\n";
print "newdisk: $newdisk"."\n";
This answer is a bit superflous, but anyway, if you are using a fairly modern version of Perl (5.13+), you could have the original code working by just adding the r flag:
use 5.013;
($newdisk) = ($hdisk =~ s/(hdisk\D*)(\d+)/(($1 eq "hdiskpower"?"prw":"dsk").$2)/rei);
You could even let go of the parens:
use 5.013;
my $newdisk = $hdisk =~ s/(hdisk\D*)(\d+)/(($1 eq "hdiskpower"?"prw":"dsk").$2)/rei;
You can read more on the /r flag at Use the /r substitution flag to work on a copy.
Related
I'm experiencing a rather odd problem while using Data::Dumper to try and check on my importing of a large list of data into a hash.
My Data looks like this in another file.
##Product ID => Market for product
ABC => Euro
XYZ => USA
PQR => India
Then in my script, I'm trying to read in my list of data into a hash like so:
open(CONFIG_DAT_H, "<", $config_data);
while(my $line = <CONFIG_DAT_H>) {
if($line !~ /^\#/) {
chomp($line);
my #words = split(/\s*\=\>\s/, $line);
%product_names->{$words[0]} = $words[1];
}
}
close(CONFIG_DAT_H);
print Dumper (%product_names);
My parsing is working for the most part that I can find all of my data in the hash, but when I print it using the Data::Dumper it doesn't print it properly. This is my output.
$VAR1 = 'ABC';
';AR2 = 'Euro
$VAR3 = 'XYZ';
';AR4 = 'USA
$VAR5 = 'PQR';
';AR6 = 'India
Does anybody know why the Dumper is printing the '; characters over the first two letters on my second column of data?
There is one unclear thing in the code: is *product_names a hash or a hashref?
If it is a hash, you should use %product_names{key} syntax, not %product_names->{key}, and need to pass a reference to Data::Dumper, so Dumper(\%product_names).
If it is a hashref then it should be labelled with a correct sigil, so $product_names->{key} and Dumper($product_names}.
As noted by mob if your input has anything other than \n it need be cleaned up more explicitly, say with s/\s*$// per comment. See the answer by ikegami.
I'd also like to add, the loop can be simplified by loosing the if branch
open my $config_dat_h, "<", $config_data or die "Can't open $config_data: $!";
while (my $line = <$config_dat_h>)
{
next if $line =~ /^\#/; # or /^\s*\#/ to account for possible spaces
# ...
}
I have changed to the lexical filehandle, the recommended practice with many advantages. I have also added a check for open, which should always be in place.
Humm... this appears wrong to me, even you're using Perl6:
%product_names->{$words[0]} = $words[1];
I don't know Perl6 very well, but in Perl5 the reference should be like bellow considering that %product_names exists and is declared:
$product_names{...} = ... ;
If you could expose the full code, I can help to solve this problem.
The file uses CR LF as line endings. This would become evident by adding the following to your code:
local $Data::Dumper::Useqq = 1;
You could convert the file to use unix line endings (seeing as you are on a unix system). This can be achieved using the dos2unix utility.
dos2unix config.dat
Alternatively, replace
chomp($line);
with the more flexible
$line =~ s/\s+\z//;
Note: %product_names->{$words[0]} makes no sense. It happens to do what you want in old versions of Perl, but it rightfully throws an error in newer versions. $product_names{$words[0]} is the proper syntax for accessing the value of an element of a hash.
Tip: You should be using print Dumper(\%product_names); instead of print Dumper(%product_names);.
Tip: You might also find local $Data::Dumper::Sortkeys = 1; useful. Data::Dumper has such bad defaults :(
Tip: Using split(/\s*=>\s*/, $line, 2) instead of split(/\s*=>\s*/, $line) would permit the value to contain =>.
Tip: You shouldn't use global variable without reason. Use open(my $CONFIG_DAT_H, ...) instead of open(CONFIG_DAT_H, ...), and replace other instances of CONFIG_DAT_H with $CONFIG_DAT_H.
Tip: Using next if $line =~ /^#/; would avoid a lot of indenting.
I am working on a program in perl and I am trying to combine more than one regex in a binding operator. I have tried using the syntax below but it's not working. I will like to know if there is any other way to go with this.
$in =~ (s/pattern/replacement/)||(s/pattern/replacement/)||...
You can often get a clue about what the Perl makes of some code using B::Deparse.
$ perl -MO=Deparse -E'$in =~ (s/pattern1/replacement1/)||(s/pattern2/replacement2/)'
[ ... snip ... ]
s/pattern2/replacement2/u unless $in =~ s/pattern1/replacement1/u;
-e syntax OK
So it's attempting your first substitution on $in. And if that fails, it is then trying your second substitution. But it's not using $in for the second substitution, it's using $_ instead.
You're running up against precedence issues here. Perl interprets your code as:
($in =~ s/pattern1/replacement1/) or (s/pattern2/replacement2/)
Notice that the opening parenthesis has moved before $in.
As others have pointed out, it's best to use a loop approach here. But I thought it might be useful to explain why your version didn't work.
Update: To be clear, if you wanted to use syntax like this, then you would need:
($in =~ s/pattern1/replacement1/) or
($in =~ s/pattern2/replacement2/);
Note that I've included $in =~ in each expression. At this point, it becomes obvious (I hope) why the looping solution is better.
However, because or is a short-circuiting operator, this statement will stop after the first successful substitution. I assumed that's what you wanted from your use of it in your original code. If that's not what you want, then you need to either switch to using and or (better, in my opinion) break them out into separate statements.
$in =~ s/pattern1/replacement1/;
$in =~ s/pattern2/replacement2/;
The closest you could get with a syntax looking similar to that would be
s/one/ONE/ or
s/two/TWO/ or
...
s/ten/TEN/ for $str;
This will attempt each substitution in turn, once only, stopping after the first successful one.
Use for to "topicalize" (alias $_ to your variable).
for ($in) {
s/pattern/replacement/;
s/pattern/replacement/;
}
A simpler way might be to create an array of all such patterns and replacements, then simply iterate through your array applying the substitution one pattern at a time.
my $in = "some string you want to modify";
my #patterns = (
['pattern to match', 'replacement string'],
# ...
);
$in = replace_many($in, \#patterns);
sub replace_many {
my ($in, $replacements) = #_;
foreach my $replacement ( #$replacements ) {
my ($pattern, $replace_string) = #$replacement;
$in =~ s/$pattern/$replace_string/;
}
return $in;
}
It's not at all clear what you need, and it's not at all clear that you can accomplish what you appear to want by the means you suggest. The OR operator is a short circuit operator, and you may not want this behavior. Please give an example of the input you expect, and the output you desire, hopefully several examples of each. Meanwhile, here is a test script.
use warnings;
use strict;
my $in1 = 'George Walker Bush';
my $in2 = 'George Walker Bush';
my $in3 = 'George Walker Bush';
my $in4 = 'George Walker Bush';
(my $out1 = $in1) =~ s/e/*/g;
print "out1 = $out1 \n";
(my $out2 = $in2) =~ s/Bush/Obama/;
print "out2 = $out2 \n";
(my $out3 = $in3) =~ s/(George)|(Bush)/Obama/g;
print "out3 = $out3\n";
$in4 =~ /(George)|(Walker)|(Bush)/g;
print "$1 - $2 - $3\n";
exit(0);
You will notice in the last case that only the first OR operator matches in the regular expression. If you wanted to replace 'George Walker Bush' with Barack Hussein Obama', you could do that easily enough, but you would also replace 'George Washington'with 'Barack Washington' - is this what you want? Here is the output of the script:
out1 = G*org* Walk*r Bush
out2 = George Walker Obama
out3 = Obama Walker Obama
Use of uninitialized value $2 in concatenation (.) or string at pq_151111a.plx line 19.
Use of uninitialized value $3 in concatenation (.) or string at pq_151111a.plx line 19.
George - -
I'm trying for substitution in which a condition will allow or disallow substitution.
I have a string
$string = "There is <tag1>you can do for it. that dosen't mean <tag2>you are fool.There <tag3>you got it.";
Here are two hashes which are used to check condition.
my %tag = ('tag1' => '<you>', 'tag2'=>'<do>', 'tag3'=>'<no>');
my %data = ('you'=>'<you>');
Here is actual substitution in which substitution is allowed for hash tag values not matched.
$string =~ s{(?<=<(.*?)>)(you)}{
if($tag{"$1"} eq $data{"$2"}){next;}
"I"
}sixe;
in this code I want to substitute 'you' with something with the condition that it is not equal to the hash value given in tag.
Can I use next in substitution?
Problem is that I can't use \g modifier. And after using next I cant go for next substitution.
Also I can't modify expression while matching and using next it dosen't go for second match, it stops there.
You can't use a variable length look behind assertion. The only one that is allowed is the special \K marker.
With that in mind, one way to perform this test is the following:
use strict;
use warnings;
while (my $string = <DATA>) {
$string =~ s{<([^>]*)>\K(?!\1)\w+}{I}s;
print $string;
}
__DATA__
There is <you>you can do for it. that dosen't mean <notyou>you are fool.
There is <you>you can do for it. that dosen't mean <do>you are fool.There <no>you got it.
Output:
There is <you>you can do for it. that dosen't mean <notyou>I are fool.
There is <you>you can do for it. that dosen't mean <do>I are fool.There <no>you got it.
It was simple but got my two days to think about it. I just written another substitution where it ignores previous tag which is cancelled by next;
$string = "There is <tag1>you can do for it. that dosen't mean <tag2>you are fool.There <tag3>you got it.";
my %tag = ('tag1' => '<you>', 'tag2'=>'<do>', 'tag3'=>'<no>');
my %data = ('you'=>'<you>');
my $notag;
$string =~ s{(?<=<(.*?)>)(you)}{
$notag = $2;
if($tag{"$1"} eq $data{"$2"}){next;}
"I"
}sie;
$string =~ s{(?<=<(.*?)>)(?<!$notag)(you)}{
"I"
}sie;
Why do I get a syntax error in the following script?
print "Enter Sequence:";
$a = <STDIN>;
if ($a=="A")|| ($a== "T")|| ( $a == "C")|| ($a== "G")
{
print $a;
}
else
{
print "Error";
}
First, you have a syntax error: The condition expression of an if statement must be in parens.
The second error is found by using use strict; use warnings;, something you should always do. The error is the use of numerical comparison (==) where string comparison (eq) is called for.
The final problem is that $a will almost surely contain a string ending with a newline, so a chomp is in order.
The immediate problem is that he entire logical expression for an if must be in parentheses.
In addition
You must use eq instead of == for comparing strings
Your input string will have a trailing newline, so it will look like "C\n" and will not match a simple one-character string. You need to chomp the input before you compare it
It is generally better to read from STDIN using <> rather than <STDIN>. That way you can specify an input file on the command line, or read from the STDIN if no input was provided
You must always put use strict and use warnings at the top of your program. That will catch many simple errors that you may otherwise overlook
You shouldn't use $a as a variable name. It is a symbol reserved by Perl itself, and says nothing about the purpose of the variable
It is best to use a regular expression for simple comparisons like this. It makes your code much easier to read and will usually make the execution very much faster
Please take a look at this program, which I think does what you want.
use strict;
use warnings;
print "Enter Sequence: ";
my $input = <>;
chomp $input;
if ( $input =~ /^[ATCG]$/i ) {
print $input, "\n";
}
else {
print "Error";
}
I have a Perl application that takes from command line an input as:
application --fields 1-6,8
I am required to display the fields as requested by the user on command line.
I thought of substituting '-' with '..' so that I can store them in array e.g.
$str = "1..15,16" ;
#arr2 = ( $str ) ;
#arr = ( 1..15,16 ) ;
print "#arr\n" ;
print "#arr2\n" ;
The problem here is that #arr works fine ( as it should ) but in #arr2 the entire string is not expanded as array elements.
I have tried using escape sequences but no luck.
Can it be done this way?
If this is user input, don't use string eval on it if you have any security concerns at all.
Try using Number::Range instead:
use Number::Range;
$str = "1..15,16" ;
#arr2 = Number::Range->new( $str )->range;
print for #arr2;
To avoid dying on an invalid range, do:
eval { #arr2 = Number::Range->new( $str )->range; 1 } or your_error_handling
There's also Set::IntSpan, which uses - instead of ..:
use Set::IntSpan;
$str = "1-15,16";
#arr2 = Set::IntSpan->new( $str )->elements;
but it requires the ranges to be in order and non-overlapping (it was written for use on .newsrc files, if anyone remembers what those are). It also allows infinite ranges (where the string starts -number or ends number-), which the elements method will croak on.
You're thinking of #arr2 = eval($str);
Since you're taking input and evaluating that, you need to be careful.
You should probably #arr2 = eval($str) if ($str =~ m/^[0-9.,]+$/)
P.S. I didn't know about the Number::Range package, but it's awesome. Number::Range ftw.
I had the same problem in dealing with the output of Bit::Vector::to_Enum. I solved it by doing:
$range_string =~ s/\b(\d+)-(\d+)\b/expand_range($1,$2)/eg;
then also in my file:
sub expand_range
{
return join(",",($_[0] .. $_[1]));
}
So "1,3,5-7,9,12-15" turns into "1,3,5,6,7,9,12,13,14,15".
I tried really hard to put that expansion in the 2nd part of the s/// so I wouldn't need that extra function, but I couldn't get it to work. I like this because while Number::Range would work, this way I don't have to pull in another module for something that should be trivial.
#arr2 = ( eval $str ) ;
Works, though of course you have to be very careful with eval().
You could use eval:
$str = "1..15,16" ;
#arr2 = ( eval $str ) ;
#arr = ( 1..15,16 ) ;
print "#arr\n" ;
print "#arr2\n" ;
Although if this is user input, you'll probably want to do some validation on the input string first, to make sure they haven't input anything dodgy.
Use split:
#parts = split(/\,/, $fields);
print $parts[0];
1-6
print $parts[1];
8
You can't just put a string containing ',' in an array, and expect it to turn to elements (except if you use some Perl black magic, but we won't go into that here)
But Regex and split are your friends.