How can I add 2 words in front and behind of a regex matched string?
Example:
hi1,hi2,6d371555e08ba2b2397fd44a0db31605e7def831585c4c11dbb21c70d89e3b3551350e36d2cef84097077a4f5f12e5ee359625ec0f776403895039c4442860fa9968827ab119c8e8362c8a5cbef4389c2c36a08eda30ce091fe9a8e19f9eec0d,hi3
Regex to match string: \b[A-Fa-f0-9]{64}\b
String: 6d371555e08ba2b2397fd44a0db31605e7def831585c4c11dbb21c70d89e3b3551350e36d2cef84097077a4f5f12e5ee359625ec0f776403895039c4442860fa9968827ab119c8e8362c8a5cbef4389c2c36a08eda30ce091fe9a8e19f9eec0d
I want to add: hi1, hi2, hi3.
Use $& to reference the match in the replacement string:
$s = '6d37...ec0d'
$s -replace '\b[a-f0-9]{64}\b', 'hi1,hi2,$&,hi3'
Uppercase characters in the match expression are not required because PowerShell operators (-replace in this case) are case-insensitive by default.
Without knowing what to match, here is an example:
$str = 'klpo6d371555e08ba2b2397fd44a0db31605e7def831585c4c11dbb21c70d89e3b3551350e36d2cef84097077a4f5f12e5ee359625ec0f776403895039c4442860fa9968827ab119c8e8362c8a5cbef4389c2c36a08eda30ce091fe9a8e19f9eec0dputy'
if ($str -match '\b[A-Fa-f0-9]{64}\b'){
'hi1,hi2,{0},hi3' -f $matches[0]
}
Related
For example, I have a below string starting with two front slashes. Now I want to get the string "foo_foo". How do I do that? Thanks in advance.
my $str = "// filename : foo_foo";
if ($_ =~ m/^filename\s+:\s+(.+)/) {print "regex $1 \n";}
You populate $str but bind the match against $_.
Use a different delimiter so you don't have to escape the slashes.
my $str = "// filename : foo_foo";
if ($str =~ m{^/+\s+filename\s+:\s+(.+)}) {
print "regex: '$1'\n";
}
You can use
my $str = "// filename : foo_foo";
if ($str =~ m{^//\h*filename\s*:\s*(.+)}) {
print "regex $1 \n";
}
See the online Perl demo. Here, I used {...} regex delimiters instead of /.../ and the pattern looks like ^//\h*filename\s*:\s*(.+) now, matching
^ - start of string
// - a // substring
\h* - zero or more horizontal whitespaces
filename - some fixed string
\s*:\s* - a : char enclosed with zero or more whitespaces
(.+) - Group 1: one or more chars other than line break chars as many as possible (greedy dot).
Something in line with following sample code should produce desired result.
use strict;
use warnings;
use feature 'say';
my $str = "// filename : foo_foo";
my($fname) = $str =~ m|// filename : (.*)\z|;
say $fname;
Output
foo_foo
I'd like to be able to use regex in Perl to insert characters into words.
So that the word "TABLE" would become "T%A%B%L%E%"
Can I ask for the syntax for such a feat?
Many thanks
Break the string into characters then join them with what you want in between; also append that
my $res = ( join '%', split //, $string ) . '%';
A simple-minded way with regex
$string =~ s/(.)/$1%/g;
where with /r modifier you can preserve $string and return the changed string instead
my $res = $string =~ s/(.)/$1%/gr;
You can use this command,
echo TABLE|perl -pe 's/\w/$&%/g'
This outputs T%A%B%L%E%
OR (in case your data is contained in a file)
perl -pe 's/\w/$&%/g' test.pl
You may replace \w with [a-zA-Z] if you just want to replace with alphabets as \w matchs alphabets numbers and underscore.
You can use look-behind also
my $s = "table";
$s=~s/(?<=.)/%/g;
print $s;
If your version >5.14 you can use \K
$s=~s/.\K/%/g;
I have string which should to be change from numbers to text in my case variable is:
$string = '18.3.0-31290741.41742-1'
I want to replace everything after '-' to be "-SNAPSHOT" and when perform echo $string to show information below. I tried with LastIndexOf(), Trim() and other things but seems not able to manage how to do it.
Expected result:
PS> echo $string
18.3.0-SNAPSHOT
Maybe that can be the light of the correct way, but when have two '-' is going to replace the last one not the first which can see:
$string = "18.3.0-31290741.41742-1" -replace '(.*)-(.*)', '$1-SNAPSHOT'
.* is a greedy match, meaning it will produce the longest matching (sub)string. In your case that would be everything up to the last hyphen. You need either a non-greedy match (.*?) or a pattern that won't match hyphens (^[^-]*).
Demonstration:
PS C:\> '18.3.0-31290741.41742-1' -replace '(^.*?)-.*', '$1-SNAPSHOT'
18.3.0-SNAPSHOT
PS C:\> '18.3.0-31290741.41742-1' -replace '(^[^-]*)-.*', '$1-SNAPSHOT'
18.3.0-SNAPSHOT
By using a positive lookbehind assertion ((?<=...)) you could eliminate the need for a capturing group and backreference:
PS C:\> "18.3.0-31290741.41742-1" -replace '(?<=^.*?-).*', 'SNAPSHOT'
18.3.0-SNAPSHOT
You could use Select-String and an regular expression to match the pattern, then pass the match to ForEach-Object (commonly shorthanded with alias %) to construct the final string:
$string = "18.3.0-31290741.41742-1" | Select-String -pattern ".*-.*-" | %{ "$($_.Matches.value)SNAPSHOT" }
$string
There are numerous questions on how to do a multiline regex in Perl. Most of them mention the s switch that makes a dot match a newline. However, I want to match an exact phrase (so, not a pattern) and I don't know where the newlines will be. So the question is: can you ignore newlines, instead of matching them with .?
MWE:
$pattern = "Match this exact phrase across newlines";
$text1 = "Match\nthis exact\nphrase across newlines";
$text2 = "Match this\nexact phra\nse across\nnewlines";
$text3 = "Keep any newlines\nMatch this exact\nphrase across newlines\noutside\nof the match";
$text1 =~ s/$pattern/replacement text/s;
$text2 =~ s/$pattern/replacement text/s;
$text3 =~ s/$pattern/replacement text/s;
print "$text1\n---\n$text2\n---\n$text3\n";
I can put dots in the pattern instead of spaces ("Match.this.exact.phrase") but that does not work for the second example. I can delete all newlines as preprocessing but I would like to keep newlines that are not part of the match (as in the third example).
Desired output:
replacement text
---
replacement text
---
Keep any newlines
replacement text
outside
of the match
Just replace the literal spaces with a character class that matches a space or a newline:
$pattern = "Match[ \n]this[ \n]exact[ \n]phrase[ \n]across[ \n]newlines";
Or, if you want to be more lenient, use \s or \s+ instead, since \s also matches newlines.
Most of the time, you are treating newlines as spaces. If that's all you wanted to do, all you'd need is
$text =~ s/\n/ /g;
$text =~ /\Q$text_to_find/ # or $text =~ /$regex_pattern_to_match/
Then there's the one time you want to ignore it. If that's all you wanted to do, all you'd need is
$text =~ s/\n//g;
$text =~ /\Q$text_to_find/ # or $text =~ /$regex_pattern_to_match/
Doing both is next to impossible if you have a regex pattern to match. But you seem to want to match literal text, so that opens up some possibilities.
( my $pattern = $text_to_find )
=~ s/(.)/ $1 eq " " ? "[ \\n]" : "\\n?" . quotemeta($1) /seg;
$pattern =~ s/^\\n\?//;
$text =~ /$pattern/
It sounds like you want to change your "exact" pattern to match newlines anywhere, and also to allow newlines instead of spaces. So change your pattern to do so:
$pattern = "Match this exact phrase across newlines";
$pattern =~ s/\S\K\B/\n?/g;
$pattern =~ s/ /[ \n]/g;
It certainly is ugly, but it works:
M\n?a\n?t\n?c\n?h\st\n?h\n?i\n?s\se\n?x\n?a\n?ct\sp\n?h\n?r\n?a\n?s\n?e\sa\n?c\n?r\n?o\n?s\n?s\sn\n?e\n?w\n?l\n?i\n?n\n?e\n?s
For every pair of letters inside a word, allow a newline between them with \n?. And replace each space in your regex with \s.
May not be usable, but it gets the job done ;)
Check it out at regex101.
How can I find the first substring until I find the first digit?
Example:
my $string = 'AAAA_BBBB_12_13_14' ;
Result expected: 'AAAA_BBBB_'
Judging from the tags you want to use a regular expression. So let's build this up.
We want to match from the beginning of the string so we anchor with a ^ metacharacter at the beginning
We want to match anything but digits so we look at the character classes and find out this is \D
We want 1 or more of these so we use the + quantifier which means 1 or more of the previous part of the pattern.
This gives us the following regular expression:
^\D+
Which we can use in code like so:
my $string = 'AAAA_BBBB_12_13_14';
$string =~ /^\D+/;
my $result = $&;
Most people got half of the answer right, but they missed several key points.
You can only trust the match variables after a successful match. Don't use them unless you know you had a successful match.
The $&, $``, and$'` have well known performance penalties across all regexes in your program.
You need to anchor the match to the beginning of the string. Since Perl now has user-settable default match flags, you want to stay away from the ^ beginning of line anchor. The \A beginning of string anchor won't change what it does even with default flags.
This would work:
my $substring = $string =~ m/\A(\D+)/ ? $1 : undef;
If you really wanted to use something like $&, use Perl 5.10's per-match version instead. The /p switch provides non-global-perfomance-sucking versions:
my $substring = $string =~ m/\A\D+/p ? ${^MATCH} : undef;
If you're worried about what might be in \D, you can specify the character class yourself instead of using the shortcut:
my $substring = $string =~ m/\A[^0-9]+/p ? ${^MATCH} : undef;
I don't particularly like the conditional operator here, so I would probably use the match in list context:
my( $substring ) = $string =~ m/\A([^0-9]+)/;
If there must be a number in the string (so, you don't match an entire string that has no digits, you can throw in a lookahead, which won't be part of the capture:
my( $substring ) = $string =~ m/\A([^0-9]+)(?=[0-9])/;
$str =~ /(\d)/; print $`;
This code print string, which stand before matching
perl -le '$string=q(AAAA_BBBB_12_13_14);$string=~m{(\D+)} and print $1'
AAAA_BBBB_