I use the split function by two ways. First way (string argument to split):
my $string = "chr1.txt";
my #array1 = split(".", $string);
print $array1[0];
I get this error:
Use of uninitialized value in print
When I do split by the second way (regular expression argument to split), I don't get any errors.
my #array1 = split(/\./, $string); print $array1[0];
My first way of splitting is not working only for dot.
What is the reason behind this?
"\." is just ., careful with escape sequences.
If you want a backslash and a dot in a double-quoted string, you need "\\.". Or use single quotes: '\.'
If you just want to parse files and get their suffixes, better use the fileparse() method from File::Basename.
Additional details to the information provided by Mat:
In split "\.", ... the first parameter to split is first interpreted as a double-quoted string before being passed to the regex engine. As Mat said, inside a double-quoted string, a \ is the escape character, meaning "take the next character literally", e.g. for things like putting double quotes inside a double-quoted string: "\""
So your split gets passed "." as the pattern. A single dot means "split on any character". As you know, the split pattern itself is not part of the results. So you have several empty strings as the result.
But why is the first element undefined instead of empty? The answer lies in the documentation for split: if you don't impose a limit on the number of elements returned by split (its third argument) then it will silently remove empty results from the end of the list. As all items are empty the list is empty, hence the first element doesn't exist and is undefined.
You can see the difference with this particular snippet:
my #p1 = split "\.", "thing";
my #p2 = split "\.", "thing", -1;
print scalar(#p1), ' ', scalar(#p2), "\n";
It outputs 0 6.
The "proper" way to deal with this, however, is what #soulSurfer2010 said in his post.
Related
What is difference between /.../ and m/.../?
use strict;
use warnings;
my $str = "This is a testing for modifier";
if ($str =~ /This/i) { print "Modifier...\n"; }
if ($str =~ m/This/i) { print "W/O Modifier...\n"; }
However, I checked with this site for Reference not clearly understand with the theory
There's no difference. If you just supply /PATTERN/ then it assumes m. However, if you're using an alternative delimiter, you need to supply the m. E.g. m|PATTERN| won't work as |PATTERN|.
In your example, i is the modifier as it's after the pattern. m is the operation. (as opposed to s, tr, y etc.)
Perhaps slightly confusingly - you can use m as a modifier, but only if you put if after the match.
m/PATTERN/m will cause ^ and $ to match differently than in m/PATTERN/, but it's the trailing m that does this, not the leading one.
Perl has a number of quote-like operators where you can choose the delimiter to suit the data you're passing to the operator.
q(...) creates a single-quoted string
qq(...) creates a double-quoted string
qw(...) creates a list by splitting its arguments on white-space
qx(...) executes a command and returns the output
qr(...) compiles a regular expression
m(...) matches its argument as a regular expression
(There's also s(...)(...) but I've left that off the list as it has two arguments)
For some of these, you can omit the letter at the start of the operator if you choose the default delimiter.
You can omit q if you use single quote characters ('...').
You can omit qq if you use double quote characters ("...").
You can omit qx if you use backticks (`...`).
You can omit m if you use slashes (/.../).
So, to answer your original question, m/.../ and /.../ are the same, but because slashes are the default delimitor for the match operator, you can omit the m.
I am trying to split a string with "." but getting nothing in the array. File name is "Head-First-Java-2nd-edition.pdf" After splitting I want to extract extension, but don't know why it is giving blank array.
my #fileInfo = split(/./, $filename);
&logMsg("Array is: #fileInfo");
The split is giving an empty list because you are splitting on a wildcard .. Period is a meta character, and if you want to split on a literal period, you need to escape it
my #fileInfo = split(/\./, $filename);
Also, the syntax for calling a subroutine is NAME(LIST). Using the & prefix has a certain hidden feature, in that it circumvents prototypes. Read more in perldoc perlsub.
. in a regular expression means any character except \n. To split on a literal ., you need to escape it:
split /\./, $filename;
I have one string
"8.53" I want my resulting string "853"
I have tried
the following code
tr|.||;
but its not replacing its giving 8.53 only .
I have tried another way using
tr|.|NULL|;
but its giving 8N53 can anyone please suggest me how to use tr to replace a character with NULL.
Thanks
You need to specify the d modifier to delete chars with no corresponding char:
tr/.//d;
Or you could use the (slower but more familiar) substitution operator:
s/\.//g;
You don't want tr because that transliterates characters from the 1st list with the corresponding character in the 2nd list (which was N in your example since that was the first character). You'll want the substitution operator.
my $var = "8.53";
$var =~ s/\.//;
print $var;
Add the g flag if there are multiple instances you want to replace (s/\.//g).
I am trying to get that Perl split working for more than 2 hours. I don't see an error. Maybe some other eyes can look at it and see the issue. I am sure its a silly one:
#versionsplit=split('.',"15.0.3");
print $versionsplit[0];
print $versionsplit[1];
print $versionsplit[2];
I just get an empty array. Any idea why?
You need:
#versionsplit=split(/\./,"15.0.3");
The first argument to split is a regular expression, not a string. And . is the regex symbol which means ‘match any character’. So all the characters in your input string were being treated as separators, and split wasn't finding anything between them to return.
the "." represents any character.You need to escape it for split function to recognise as a field separator.
change your line to
#versionsplit=split('\.',"15.0.3");
So I thought this would just be a simple issue however I'm getting the incorrect results. Basically I am trying to remove the quotes around a string. For example I have the string "01:00" and I want 01:00, below is the code on how I thought I would be able to do this:
$expected_start_time = $conditions =~ m/(\"[^\"])/;
Every time this runs it returns 1, so I'm guessing that it is just returning true and not actually extracting the string from the quotes. This happen no matter what is in the quotes "02:00", "02:20", "08:00", etc.
All you forgot was parens for the LHS to put the match into list context so it returns the submatch group(s). The normal way to do this is:
($expected_start_time) = $condition =~ /"([^"]*)"/;
It appears that you know that the first and last character are quotes. If that is the case, use
$expected_start_time = substr $conditions, 1, -1;
No need for a regexp.
The brute force way is:
$expected_start_time = $conditions;
$expected_start_time =~ s/"//g;
Note that the original regex:
m/(\"[^\"])/
would capture the opening quote and the following non-quote character. To capture the non-quote characters between double quotes, you'd need some variant on:
m/"([^"]*)"/;
This being Perl (and regexes), TMTOWTDI - There's More Than One Way Do It.
In scalar context a regex returns true if the regex matches the string. You can access the match with $1. See perlre.