I would like to extract the string between "t=" and '&' from the string below, but '&2384' may not be present sometimes. I have tried the following, but I get the result "123455asdfgh&2384" instead of "123455asdfgh", what am I doing wrong here? thanks for help.
$string="t=123455asdfgh&2384";
$match=array();
preg_match('/t=(.*)(&.*)?/', $string, $match);
echo $match[1];
NOTE: I need to use regular exp...
Try this one instead
preg_match('/t=([^&]*)(&.*)?/', $string, $match);
This is better suited for parse_str(), and not a regex.
parse_str( "t=123455asdfgh&2384", $params);
echo $params['t'];
This prints:
123455asdfgh
This is what parse_str is for (parsing query strings):
$string="t=123455asdfgh&2384";
$args = array();
parse_str($string, $args);
echo $args['t']; // outputs '123455asdfgh'
Try a non greedy quantifier:
preg_match('/t=(.*?)(?=&|$)/', $string, $match);
The problem is that the first .* matches everything up till the end of the string, which is still a complete match because the latter group is optional.
How about using the strpos() to find the starting and ending point in the string, and then extract it?
Related
I have this code in perl where I want to extract the value of 'EUR_AF', in this case '0.39'.
Sometimes 'EUR_AF' ends with ';', sometimes it doesn't.
Alternatively, 'EUR_AF' may end with '=0' instead of '=0.39;' or '=0.39'.
How do I make the code handle that? Can't seem to find it online...I could of course wrap everything in an almost endless if-elsif-else statement, but that seems overkill.
Example text:
AVGPOST=0.9092;AN=2184;RSQ=0.5988;ERATE=0.0081;AC=144;VT=SNP;THETA=0.0045;AA=A;SNPSOURCE=LOWCOV;LDAF=0.0959;AF=0.07;ASN_AF=0.05;AMR_AF=0.10;AFR_AF=0.11;EUR_AF=0.039
Code: $INFO =~ m/\;EUR\_AF\=(.*?)(;)/
I did find that: $INFO =~ m/\;EUR\_AF\=(.*?0)/ handles the cases of EUR_AF=0, but how to handle alternative scenarios efficiently?
Extract one value:
my ($eur_af) = $s =~ /(?:^|;)EUR_AF=([^;]*)/;
my ($eur_af) = ";$s" =~ /;EUR_AF=([^;]*)/;
Extract all values:
my %rec = split(/[=;]/, $s);
my $eur_af = $rec{EUR_AF};
This regex should work for you: (?<=EUR_AF=)\d+(\.\d+)?
It means
(?<=EUR_AF=) - look for a string preceeded by EUR_AF=
\d+(\.\d+)? - consist of a digit, optionally a decimal digit
EDIT: I originally wanted the whole regex to return the correct result, not only the capture group. If you want the correct capture group edit it to (?<=EUR_AF=)(\d+(?:\.\d+)?)
I have found the answer. The code:
$INFO =~ m/(?:^|;)EUR_AF=([^;]*)/
seems to handle the cases where EUR_AF=0 and EUR_AF=0.39, ending with or without ;. The resulting $INFO will be 0 or 0.39.
This is my string: left_image_12.
I would like to leave out _12 and only display left_image. I am unable to figure this out.
You can do that with a capture group:
^(.*?)_\d+$
$1 (or \1, depending on your language) will contain the name without the number at the end.
Try this:
preg_match($string, "/^([A-Za-z_]+)_\d+$/", $match);
$output = $match[1];
I want to find a regular expression in Perl which matches a pattern such as this:
my $sumthing = "people say
for -->";
Over here after say there is a single newline character. So I need to find a regular expression which could match such a pattern which includes a newline within a pattern. Please help me to find this as I'm new to Perl & regular expression.
The possible methods I tried were these:
if (($sumthing !~ (/\n+$/)) && ($sumthing !~ (/^\n+/m)))
They kindly help me to find out an expression to match this kind of a pattern, but not getting the output as desired.
It's not clear what you want. Do you want match that string exactly? If so, you could use
$sumthing =~ /^people say\nfor -->\z/
or
$sumthing eq "people say\nfor -->"
Or maybe what you need to know is that . matches any character including newline when /s is used?
/people .* -->/s
The following will check for anything then new line then anything. Not sure if I totally understood your question.
if($sumthing =~ m/.*\n.*/)
Have a look at the /s modifier which causes .to match anything, including a newline.
my $str = "people say for\nsomething...";
$str =~ m{say(.*)}s and print "'$1'\n";
This would print:
' for
something...'
I have some strings like below
my-name-is-2547
this-is-stack-2012
hllo-how-2011
Now I want the above strings to be changed to something like the ones below using regex.
my-name-is-(2547)
this-is-stack-(2012)
hllo-how-(2011)
I don't want to use substr or other, only regex replace.
$pattern = '/(\d+)$/';
$replacement = '($1)';
echo preg_replace($pattern, $replacement, $string);
If you are sure that a numbers are only at the end:
regular expression:
(\d+)
using 1 capturing group. Replaced by: ($1).
so the outpu will be:
my-name-is-(2547)
this-is-stack-(2012)
hllo-how-(2011)
I have what I thought would be an easy problem to solve but I am not able to find the answer to this.
How can I find and replace the last occurrence of a character in a string?
I have a string: GE1/0/1 and I would like it to be: GE1/0:1 <- This can be variable length so no substrings please.
Clarification:
I am looking to replace the last / with a : no matter what comes before or after it.
use strict;
use warnings;
my $a = 'GE1/0/1';
(my $b = $a) =~ s{(.*)/}{$1:}xms;
print "$b\n";
I use the greedy behaviour of .*
Perhaps I have not understand the problem with variable length, but I would do the following :
You can match what you want with the regex :
(.+)/
So, this Perl script
my $text = 'GE1/0/1';
$text =~ s|(.+)/|$1:|;
print 'Result : '.$text;
will output :
Result : GE1/0:1
The '+' quantifier being 'greedy' by default, it will match only the last slash character.
Hope this is what you were asking.
This finds a slash and looks ahead to make sure there are no more slashes past it.:
Raw regex:
/(?=[^/]*$)
I think the code would look something like this, but perl isn't my language:
$string =~ s!/(?=[^/]*$)!\:!g;
"last occurrence in a string" is slightly ambiguous. The way I see it, you can mean either:
"Foo: 123, yada: GE1/0/1, Bar: null"
Meaning the last occurrence in the "word" GE1/0/1, or:
"GE1/0/1"
As a complete string.
In the latter case, it is a rather simple matter, you only have to decide how specific you can be in your regex.
$str =~ s{/(\d+)$}{:$1};
Is perfectly fine, assuming the last character(s) can only be digits.
In the former case, which I don't think you are referring to, but I'll include anyway, you'd need to be much more specific:
$str =~ s{(\byada:\s+\w+/\w+)/(\w+\b)}{$1:$2};