I tried to convert strings from lower case to upper case. While I achieved that, I found my double quotes were removed from the output. I want to be clear why the substitution fails here.
Perl code:
#!/usr/bin/perl
use strict;
use warnings;
my #ar = <DATA>;
my $str = join '', #ar;
#print $str;
if ( $str =~ s/\"([^"]*)\"/uc($1)/eg ) {
print $str;
}
__DATA__
output = " as (10) print "it if achieved. print" # This is comment.NUMBERS"
Obtained output:
output = AS (10) PRINT it if achieved. print # THIS IS COMMENT.NUMBERS
Expected output:
output = " AS (10) PRINT "it if achieved. print" # THIS IS COMMENT.NUMBERS"
The quotes are all disappearing because they're outside the capture in your regex. Changing it to s/("[^"]*")/uc($1)/eg gets us your desired output:
output = " AS (10) PRINT "it if achieved. print" # THIS IS COMMENT.NUMBERS"
Note that the inner section isn't mapped to all caps. This is a hint to how the regex is actually matching things in your string - it finds two matches: " as (10) print " and " # This is comment.NUMBERS". The text inside the inner quotes isn't matched by the regex at all. This is important, because it means that this solution is fragile and it will break on data which doesn't contain two nested sets of quotes and you want to capitalize only the text which is inside the outer quotes, but outside the inner quotes.
s/\"([^"]*)\"/uc($1)/eg - $1 represents match inside brackets (). You have include double quotes in match to be replaced but outside brackets.
Possible fix:
s/(\"[^"]*\")/uc($1)/eg
Related
I need a regex to match \' <---- literally backslash apostrophe.
my $line = '\'this';
$line =~ s/(\o{134})(\o{047})/\\\\'/g;
$line =~ s/\\'/\\\\'/g;
$line =~ s/[\\][']/\\\\'/g;
printf('%s',$line);
print "\n";
All I get out of this is
'this
When what I want is
\\'this
This occurs whether the string is declared using ' or ". This was a test script for tracking down a file parsing bug. I wanted to confirm that the regex was working as expected.
I don't know if when the backslash apostrophe is parsed by the regex it is not treated as 2 characters, but is instead treated as an escaped apostrophe.
Either way. what is the best way to match \' and print out \\'? I don't want to escape any other back slashes or apostrophes and I can't change the text I am parsing, just the way it is handled and outputted.
s/\\'/\\\\'/g
All three of your patterns match a backslash followed by a quote, the above being the simplest.
Your testing was in vain because your string doesn't contain any backslashes. Both string literals "\'this" (from earlier edit) and '\'this' (from later edit) produce the string 'this.
say "\'this"; # 'this
say '\'this'; # 'this
To produce the string \'this, you could use either of the following string literals (among others):
"\\'this"
'\\\'this'
say "\\'this"; # \'this
say '\\\'this'; # \'this
The answer is, of course
s/[\\][']/\\\\'/g
This will match
\'this
And substitute with this
\\'this
This was the only way I could get it to work.
Perl
Too much "regexing" in your snippet. Try:
my $line = '\'this';
$line =~ s/'/\\\\\'/g;
printf('%s',$line);
print "\n";
# \\'this
or... if you want another mode:
my $line = '\'this';
$line =~ s/'/\\'/g;
printf('%s',$line);
print "\n";
# \'this
I'm trying to find the index of white space in a string in Perl.
For example, if I have the string
stuff/more stuffhere
I'd like to select the word "more" with a substring method. I can find the index of "/" but haven't figured out how to find the index of white space. The length of the substring I'm trying to select will vary, so I can't hard code the index. There will only be one white space in the string (other than those after the end of the string).
Also, if anybody has any better ideas of how to do this, I'd appreciate hearing them. I'm fairly new to programming so I'm open to advice. Thanks.
Just use index:
#!/usr/bin/perl
use warnings;
use strict;
use feature qw{ say };
my $string = 'stuff/more stuffhere';
my $index_of_slash = index $string, '/';
my $index_of_space = index $string, ' ';
say "Between $index_of_slash and $index_of_space.";
The output is
Between 5 and 10.
Which is correct:
0 1
01234567890123456789
stuff/more stuffhere
If by "whitespace" you also mean tabs or whatever, you can use a regular expression match and the special variables #- and #+:
#!/usr/bin/perl
use warnings;
use strict;
use feature qw{ say };
my $string = "stuff/more\tstuffhere";
if ($string =~ m{/.*(?=\s)}) {
say "Between $-[0] and $+[0]";
}
The (?=\s) means is followed by a whitespace character, but the character itself is not part of the match, so you don't need to do any maths on the returned values.
As you stated, you want to select the word between the first /
and the first space following it.
If this is the case, you maybe don't need any index (you need just
the word).
A perfect tool to find something in a text is regex.
Look at the following code:
$txt = 'stuff/more stuffxx here';
if ($txt =~ /\/(.+?) /) {
print "Match: $1.\n";
}
The regex used tries to match:
a slash,
a non-empty sequence of any chars (note ? - reluctant
version), enclosed in a capturing group,
a space.
So after the match $1 contains what was captured by the first
capturing group, i.e. "your" word.
But if for any reason you were interested in starting and ending
offsets to this word, you can read them from $-[1]
and $+[1] (starting / ending indices of the first capturing group).
The arrays #- (#LAST_MATCH_START) and #+ (#LAST_MATCH_END) give offsets of the start and end of last successful submatches. See Regex related variables in perlvar.
You can capture your real target, and then read off the offset right after it with $+[0]
#+
This array holds the offsets of the ends of the last successful submatches in the currently active dynamic scope. $+[0] is the offset into the string of the end of the entire match. This is the same value as what the pos function returns when called on the variable that was matched against.
Example
my $str = 'target and target with spaces';
while ($str =~ /(target)\s/g)
{
say "Position after match: $+[0]"
}
prints
Position after match: 7
Position after match: 18
These are positions right after 'target', so of spaces that come after it.
Or you can capture \s instead and use $-[1] + 1 (first position of the match, the space).
You can use
my $str = "stuff/more stuffhere";
if ($str =~ m{/\K\S+}) {
... substr($str, $-[0], $+[0] - $-[0]) ...
}
But why substr? That's very weird there. Maybe if you told us what you actually wanted to do, we could provide a better alternatives. Here are three cases:
Data extraction:
my $str = "stuff/more stuffhere";
if ( my ($word) = $str =~ m{/(\S+)} ) {
say $word; # more
}
Data replacement:
my $str = "stuff/more stuffhere";
$str =~ s{/\K\S+}{REPLACED};
say $str; # stuff/REPLACED stuffhere
Data replacement (dynamic):
my $str = "stuff/more stuffhere";
$str =~ s{/\K(\S+)}{ uc($1) }e;
say $str; # stuff/MORE stuffhere
How can I print a string (single-quoted) containing double-backslash \\ characters as is without making Perl somehow interpolating it to single-slash \? I don't want to alter the string by adding more escape characters also.
my $string1 = 'a\\\b';
print $string1; #prints 'a\b'
my $string1 = 'a\\\\b';
#I know I can alter the string to escape each backslash
#but I want to keep string as is.
print $string1; #prints 'a\\b'
#I can also use single-quoted here document
#but unfortunately this would make my code syntactically look horrible.
my $string1 = <<'EOF';
a\\b
EOF
print $string1; #prints a\\b, with newline that could be removed with chomp
The only quoting construct in Perl that doesn't interpret backslashes at all is the single-quoted here document:
my $string1 = <<'EOF';
a\\\b
EOF
print $string1; # Prints a\\\b, with newline
Because here-docs are line-based, it's unavoidable that you will get a newline at the end of your string, but you can remove it with chomp.
Other techniques are simply to live with it and backslash your strings correctly (for small amounts of data), or to put them in a __DATA__ section or an external file (for large amounts of data).
If you are mildly crazy, and like the idea of using experimental software that mucks about with perl's internals to improve the aesthetics of your code, you can use the Syntax::Keyword::RawQuote module, on CPAN since this morning.
use syntax 'raw_quote';
my $string1 = r'a\\\b';
print $string1; # prints 'a\\\b'
Thanks to #melpomene for the inspiration.
Since the backslash interpolation happens in string literals, perhaps you could declare your literals using some other arbitrary symbol, then substitute them for something else later.
my $string = 'a!!!b';
$string =~ s{!}{\\}g;
print $string; #prints 'a\\\b'
Of course it doesn't have to be !, any symbol that does not conflict with a normal character in the string will do. You said you need to make a number of strings, so you could put the substitution in a function
sub bs {
$_[0] =~ s{!}{\\}gr
}
my $string = 'a!!!b';
print bs($string); #prints 'a\\\b'
P.S.
That function uses the non-destructive substitution modifier /r introduced in v5.14. If you are using an older version, then the function would need to be written like this
sub bs {
$_[0] =~ s{!}{\\}g;
return $_[0];
}
Or if you like something more readable
sub bs {
my $str = shift;
$str =~ s{!}{\\}g;
return $str;
}
I'm using Perl 5.16.2 to try to count the number of occurrences of a particular delimiter in the $_ string. The delimiter is passed to my Perl program via the #ARGV array. I verify that it is correct within the program. My instruction to count the number of delimiters in the string is:
$dlm_count = tr/$dlm//;
If I hardcode the delimiter, e.g. $dlm_count = tr/,//; the count comes out correctly. But when I use the variable $dlm, the count is wrong. I modified the instruction to say
$dlm_count = tr/$dlm/\t/;
and realized from how the tabs were inserted in the string that the operation was substituting every instance of any of the four characters "$", "d", "l", or "m" to \t — i.e. any of the four characters that made up my variable name $dlm.
Here is a sample program that illustrates the problem:
$_ = "abcdefghij,klm,nopqrstuvwxyz";
my $dlm = ",";
my $dlm_count = tr/$dlm/\t/;
print "The count is $dlm_count\n";
print "The modified string is $_\n";
There are only two commas in the $_ string, but this program prints the following:
The count is 3
The modified string is abc efghij,k ,nopqrstuvwxyz
Why is the $dlm token being treated as a literal string of four characters instead of as a variable name?
You cannot use tr that way, it doesn't interpolate variables. It runs strictly character by character replacement. So this
$string =~ tr/a$v/123/
is going to replace every a with 1, every $ with 2, and every v with 3. It is not a regex but a transliteration. From perlop
Because the transliteration table is built at compile time, neither the SEARCHLIST nor the REPLACEMENTLIST are subjected to double quote interpolation. That means that if you want to use variables, you must use an eval():
eval "tr/$oldlist/$newlist/";
die $# if $#;
eval "tr/$oldlist/$newlist/, 1" or die $#;
The above example from docs hints how to count. For $dlms in $string
$dlm_count = eval "\$string =~ tr/$dlm//";
The $string is escaped so to not be interpolated before it gets to eval. In your case
$dlm_count = eval "tr/$dlm//";
You can also use tools other than tr (or regex). For example, with string being in $_
my $dlm_count = grep { /$dlm/ } split //;
When split breaks $_ by the pattern that is empty string (//) it returns the list of all characters in it. Then the grep block tests each against $dlm so returning the list of as many $dlm characters as there were in $_. Since this is assigned to a scalar, $dlm_count is set to the length of that list, which is the count of all $dlm.
In the section of the docs on perlop 'Quote Like Operators', it states:
Because the transliteration table is built at compile time, neither
the SEARCHLIST nor the REPLACEMENTLIST are subjected to double quote
interpolation. That means that if you want to use variables, you must
use an eval():
As documented and as you discovered, tr/// doesn't interpolate. The simple solution is to use s/// instead.
my $dlm = ",";
$_ = "abcdefghij,klm,nopqrstuvwxyz";
my $dlm_count = s/\Q$dlm/\t/g;
If the transliteration is being performed in a loop, the following might speed things up noticeably:
my $dlm = ",";
my $tr = eval "sub { tr/\Q$dlm\E/\\t/ }";
for (...) {
my $dlm_count = $tr->();
...
}
Although several answers have hinted at the eval() idiom for tr///, none have the form that covers cases where the string has tr syntax characters in it, e.g.- (hyphen):
$_ = "abcdefghij,klm,nopqrstuvwxyz";
my $dlm = ",";
my $dlm_count = eval sprintf "tr/%s/%s/", map quotemeta, $dlm, "\t";
But as others have noted, there are lots of ways to count characters in Perl that avoid eval(), here's another:
my $dlm_count = () = m/$dlm/go;
Perl colon split show one character when using it in variables.
Code
#!/usr/bin/perl
my $serialnumber= "0123456789";
my $macaddr = "a1:b2:c3:d4:e5:f6";
my $macaddress = split /:/, $macaddr;
print $macaddress;
print "\n";
The result is only "6" or last character of the text.
See image...
Sean explained why you are getting 6. However, instead of using split, just substitute to get rid of colons:
my $macaddr = "a1:b2:c3:d4:e5:f6";
$macaddr =~ s/://g;
print $macaddr;
# a1b2c3d4e5f6
The first paragraph of perldoc -f split says:
Splits the string EXPR into a list of strings and returns the
list in list context, or the size of the list in scalar context.
Be assigning the result of split to the scalar variable $macaddress, you're providing scalar context, and so you're getting the size of the list back. It's just a coincidence that this happens to be the same as the last character of your input string.
You could use split in list context:
my #macaddress = split /:/, $macaddr;
print join "", #macaddress;