I am after some help trying to convert the following log I have to plain text.
This is a URL so there maybe %20 = 'space' and other but the main bit I am trying convert is the char(1,2,3,4,5,6) to text.
Below is an example of what I am trying to convert.
select%20char(45,120,49,45,81,45),char(45,120,50,45,81,45),char(45,120,51,45,81,45)
What I have tried so far is the following while trying to added into the char(in here) to convert with the chr($2)
perl -pe "s/(char())/chr($2)/ge"
All this has manage to do is remove the char but now I am trying to convert the number to text and remove the commas and brackets.
I maybe way off with how I am doing as I am fairly new to to perl.
perl -pe "s/word to remove/word to change it to/ge"
"s/(char(what goes in here))/chr($2)/ge"
Output try to achieve is
select -x1-Q-,-x2-Q-,-x3-Q-
Or
select%20-x1-Q-,-x2-Q-,-x3-Q-
Thanks for any help
There's too much to do here for a reasonable one-liner. Also, a script is easier to adjust later
use warnings;
use strict;
use feature 'say';
use URI::Escape 'uri_unescape';
my $string = q{select%20}
. q{char(45,120,49,45,81,45),char(45,120,50,45,81,45),}
. q{char(45,120,51,45,81,45)};
my $new_string = uri_unescape($string); # convert %20 and such
my #parts = $new_string =~ /(.*?)(char.*)/;
$parts[1] = join ',', map { chr( (/([0-9]+)/)[0] ) } split /,/, $parts[1];
$new_string = join '', #parts;
say $new_string;
this prints
select -x1-Q-,-x2-Q-,-x3-Q-
Comments
Module URI::Escape is used to convert percent-encoded characters, per RFC 3986
It is unspecified whether anything can follow the part with char(...)s, and what that might be. If there can be more after last char(...) adjust the splitting into #parts, or clarify
In the part with char(...)s only the numbers are needed, what regex in map uses
If you are going to use regex you should read up on it. See
perlretut, a tutorial
perlrequick, a quick-start introduction
perlre, the full account of syntax
perlreref, a quick reference (its See Also section is useful on its own)
Alright, this is going to be a messy "one-liner". Assuming your text is in a variable called $text.
$text =~ s{char\( ( (?: (?:\d+,)* \d+ )? ) \)}{
my #arr = split /,/, $1;
my $temp = join('', map { chr($_) } #arr);
$temp =~ s/^|$/"/g;
$temp
}xeg;
The regular expression matches char(, followed by a comma-separated list of sequences of digits, followed by ). We capture the digits in capture group $1. In the substitution, we split $1 on the comma (since chr only works on one character, not a whole list of them). Then we map chr over each number and concatenate the result into a string. The next line simply puts quotation marks at the start and end of the string (presumably you want the output quoted) and then returns the new string.
Input:
select%20char(45,120,49,45,81,45),char(45,120,50,45,81,45),char(45,120,51,45,81,45)
Output:
select%20"-x1-Q-","-x2-Q-","-x3-Q-"
If you want to replace the % escape sequences as well, I suggest doing that in a separate line. Trying to integrate both substitutions into one statement is going to get very hairy.
This will do as you ask. It performs the decoding in two stages: first the URI-encoding is decoded using chr hex $1, and then each char() function is translated to the string corresponding to the character equivalents of its decimal parameters
use strict;
use warnings 'all';
use feature 'say';
my $s = 'select%20char(45,120,49,45,81,45),char(45,120,50,45,81,45),char(45,120,51,45,81,45)';
$s =~ s/%(\d+)/ chr hex $1 /eg;
$s =~ s{ char \s* \( ( [^()]+ ) \) }{ join '', map chr, $1 =~ /\d+/g }xge;
say $s;
output
select -x1-Q-,-x2-Q-,-x3-Q-
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
I want to remove date identifier and * from string .
$string = "*102015 Supplied air hood";
$output = "Supplied air hood";
i have used
$string =~ s/[#\%&\""*+]//g;
$string =~ s/^\s+//;
what should i used to get string value = "Supplied air hood";
Thanks in advance
To remove everything from the string up to the first space, you can write
$str =~ s/^\S*\s+//;
Your pattern doesn't contain numbers. It would remove the *, but nothing else. If you want to remove a * followed by six digits and a blank at the beginning of the string, do it like this:
$string =~ s/^\*\d{6} //;
However, if that string always contains a pattern like this, you don't need a regular expression substitution. You can simply take a substring.
my $output = substr $string, 8;
That will assign the content of $string starting from the 9th character
The script below does what you want, assuming that the date always appears at the beginning the line, and that it is follow by exactly one space.
use strict;
use warnings;
while (<DATA>)
{
# skip one or more characters not a space
# then skip exactly one space
# then capture all remaining characters
# and assign them to $s
my ($s) = $_ =~ /[^ ]+ (.*)/;
print $s, "\n";
}
__DATA__
*110115 first date
*110115 second date
*110315 third date
Output is:
first date
second date
third date
I have a two strings, XXXXXXnumber and XXXXXXdate and I want to strip all the XXXXXX from each string. The actual number of character represented by XXXXXX can vary. The suffixes 'number' and 'date' are constant. XXXXXXnumber and XXXXXXXdate should become XXXXXX.
my ($prefix) = ($string =~ /\A (.+?) (?:date|number) \z/x);
Alternatively:
$string =~ s/ (?:date|number) \z//x;
I would use a regular expression like $line =~ s/(number|date)$// for that task, where $line can be either line.
If your line has additional characters after number or date, they must be filtered out, too. An alternative approach would be using an expression like ($num) = ($line =~ /^(.*)(number|date).*$/);
use regexes:
($newvar = $oldvar) =~ s/^(.*)(number|date)$/$1/;
if you have no mor euse for $oldvar's original value (including the Xes) this simplifies to
$oldvar =~ s/^(.*)(number|date)$/$1/;
A simple substitution takes care of it:
$str =~ s/(?:number|date)\z/;
my code is as follows
my $string = $cells[71];
print $string;
this prints the string but where spaces should be there is a box with 01 10 in it. I opened it in Notepad++ and the box turned into a black GS (which i am assuming is group separator).
I looked online and it said to use:
s/[^[:print:]]+//g
but when i set the string to:
my $string =~s/[^[:print:]]+//g
and I run the program i get:
4294967295
How do i resolve this?
I did what HOBBS said and it worked... thanks :)
Is there anyway I could print an enter where each of these characters are ( the box with 1001)?
When doing a regex match, you need to be careful to write $var =~ /pattern/, not $var = ~ /pattern/. When you use the second one, you're doing /pattern/, which is a regex match against $_, returning a number in scalar context. Then you do ~, which takes the bitwise inverse of that number, then ($var =) you assign that result to $var. Not what you wanted at all.
You have to assign the variable first, then do the substitution:
my $string = $cells[71];
$string =~ s/[^[:print:]]+//g;