I was wondering, why .* and .*? is not the same in PCRE regular expressions (for example in PHP's preg_match(). Dot . is symbol for any possible character and * is symbol for 0 to infinity repetition. Why is there symbol ? which means 0 to 1 repetition? However it is not obviously the same, because .*? is not interchangeable with .*, but I can't see logic difference, I have to always try what works and what does not work in certain case. I suppose that .* should match nothing to anything and ? is redundant, because it specify that .* can be 0 or 1 times - but zero times is empty string and empty string should be matched by .* too.
Can anyone explain me what is the exact difference and show me short example?
Thanks
i love wantons because they are tasty snacks
In the above string, let's say you try to match it with i.*s. The result would be the entire string, because this is called a greedy match. It matches from the first instance of i until the last instance of s.
If you were to use the non-greedy modifier ?, like i.*?s, then you would result in the following:
i love wantons
This is because the non-greedy ? modifier only matches until the first instance of s.
* is a greedy match - in other words, match zero to many times, as many times as possible. *? is a minimal match - in other words, match zero to many times, as few times as possible for the rest of the pattern to make sense. Similarly, +? is a minimally-matching version of +.
Consider the string this is "quoted" and this is "also quoted". The regular expression ".*" would match one result, "quoted" and this is "also quoted"; ".*?" would match twice, "quoted" and "also quoted".
Related
I would like to build a regular expression for replacing a sentence with "per" when it should be (a readable version of a sentence with quantities).
That is:
"3/unit" must match
"unit/3" must match
"feet/second" must match
"05/07" must not match
I know how to create something like "\D+/\D+".
But how can I build a regex saying "not both right and left expressions match \D+" ?
You can use
^(?![0-9]+/[0-9]+$)[^/]+/[^/]+$
See the regex demo. Details:
^ - start of string
(?![0-9]+/[0-9]+$) - a negative lookahead that fails the match if there are one or more digits, /, one or more digits and end of string position immediately to the right of the current location
[^/]+/[^/]+ - one or more chars other than /, a / char, and then one or more chars other than /
$ - end of string.
I am trying to do a regex string to find all cases of force unwrapping in swift. This will search all words with exclamation points in the entire code base. However, the regex that I already have has included implicit declaration of variable which I am trying to exclude.
This is the regex that I already have.
(:\s)?\w+(?<!as)\)*!
And it works fine. It searches for "variableName!", "(variableName)!", "hello.hello!". The exclusion of force casting also works. It avoids cases like "hello as! UIView", But I am trying also to exclude another cases such as "var hello: UIView!" which has an exclamation point. That's the problem I am having. I tried negative lookahead and negative lookbehind and nothing solved this kind of case.
This is the sample regex I am working on
(:\s)?\w+(?<!as)\)*!
And this is the result
testing.(**test)))!**
Details lists capture **groups!**
hello as! hello
**Hello!**
**testing!**
testing**.test!**
Hello != World
var noNetworkBanner**: StatusBarNotificationBanner!** <-- need to exclude
"var noNetworkBanner**: StatusBarNotificationBanner!**" <-- need to exclude
You may use
(?<!:\s)\b\w+(?<!\bas)\b\)*!
I added \b word boundaries to match whole words only, and changed the (:\s)? optional group to a negative lookbehind, (?<!:\s), that disallows a : + space before the word you need to match.
See the regex demo and the regex graph:
Details
(?<!:\s) - a negative lookbehind that fails the match if, immediately to the left of the current location, there is a : and a whitespace
\b - word boundary
\w+ - 1+ word chars
(?<!\bas) - a negative lookbehind that fails the match if, immediately to the left of the current location, there is a whole word as
\b - word boundary
\)* - 0 or more ) chars
! - a ! char.
I'm trying to increment:
Text_1_string(0)
to
Text_1_string(1)
and so on.
Note that I only want to increment the number in the parenthesis.
I've used:
name =~ s/\(([0-9]+)\)/$1 + 1/e;
but it turns out as:
Text_1_string1
and I don't understand why. The group captured is the number, it shouldn't replace the parenthesis.
It replaces the whole pattern that it matched, not only what is also captured. So you do need to put back the parens
$name =~ s/\(([0-9]+)\)/'('.($1 + 1).')'/e;
Since the replacement part is evaluated as code it need be normal Perl code, thus the quotes and concatenation, and parenthesis for precedence.
To add, there are patterns that need not be put back in the replacement part: lookahead and lookbehind assertions. Like common anchors, these are zero width assertions, so they do not consume what they match -- you only "look"
$name =~ s/(?<=\() ([0-9]+) (?=\))/$1 + 1/xe;
The lookbehind can't be of variable length (like \w+); it takes only a fixed string pattern.
The (?<=...) asserts that the (fixed length) pattern in parenthesis (which do not capture!) must precede the number while (?=...) asserts that the pattern in its parens must follow, for the whole pattern to match.
Often very useful is the lookbehind-type construct \K, which makes the engine keep in the string what it had matched up to that point (instead of "consuming" it); so it "drops" previous matches, much like the (?<=...) form
$name =~ s/\(\K ([0-9]+) (?=\))/$1 + 1/xe;
This is also more efficient. While it is also termed a "lookbehind" in documentation, there are in fact distinct differences in behavior. See this post and comments. Thanks to ikegami for a comment.
All these are positive lookarounds; there are also negative ones, asserting that given patterns must not be there for the whole thing to match.
A bit of an overkill in this case but a true gift in some other cases.
I want to make all a.b.c.top*.gz mentions to new-word/new-table.
Something like -->
es.fr.en.top20.gz becomes binarised-model/phrase-table
I did this :
sed -i 's/es\.fr\.en\.top*\.gz/binarised-model\/phrase-table/g' top*/mert-work/moses.ini
I had initially not used backslash before periods, but, once it did not work, I thought maybe period is tricky.
But, it does not seem to replace anything. What's going wrong ?
Thanks !
Using * as a wildcard is correct for bash globbing, but not if you work with regex, which is the case when using sed. Instead of *, try .*.
In regex, * means match the preceding character any number of times. The wildcard character is ., so .* matches any number of any characters.
If you know that the character you want to match is always a number, it's safer to use [0-9]*. If you even know how many characters this number will have, then you can even use e.g. [0-9]\{2\} to match exactly two numerals.
Sed uses regular expressions, not shell globbing. That means that (1) . matches any single character except a newline, so you are right to escape them to match a literal dot, and (2) * matches zero or more of the token preceding it, here that's p. You need
sed -i 's/es\.fr\.en\.top.*\.gz/binarised-model\/phrase-table/g' top*/mert-work/moses.ini
# ˆ
I've to match a regular-expression, stored in a variable:
#!/bin/env perl
use warnings;
use strict;
my $expr = qr/\s*(\w+(\[\d+\])?)\s+(\w+(\[\d+\])?)/sx;
$str = "abcd[3] xyzg[4:0]";
if ($str =~ m/$expr/) {
print "\n%%%%%%%%% $`-----$&-----$'\n";
}
else {
print "\n********* NOT MATCHED\n";
}
But I'm getting the outout in $& as
%%%%%%%%% -----abcd[3] xyzg-----[4:0]
But expecting, it shouldn't go inside the if clause.
What is intended is:
if $str = "abcd xyzg" => %%%%%%%%% -----abcd xyzg----- (CORRECT)
if $str = "abcd[2] xyzg" => %%%%%%%%% -----abcd[2] xyzg----- (CORRECT)
if $str = "abcd[2] xyzg[3] => %%%%%%%%% -----abcd[2] xyzg[3]----- (CORRECT)
if $str = "abcd[2:0] xyzg[3] => ********* NOT MATCHED (CORRECT)
if $str = "abcd[2:0] xyzg[3:0] => ********* NOT MATCHED (CORRECT)
if $str = "abcd[2] xyzg[3:0]" => ********* NOT MATCHED (CORRECT/INTENDED)
but output is %%%%%%%%% -----abcd[2] xyzg-----[3:0] (WRONG)
OR better to say this is not intended.
In this case, it should/my_expectation go to the else block.
Even I don't know, why $& take a portion of the string (abcd[2] xyzg), and $' having [3:0]?
HOW?
It should match the full, not a part like the above. If it didn't, it shouldn't go to the if clause.
Can anyone please help me to change my $expr pattern, so that I can have what is intended?
By default, Perl regexes only look for a matching substring of the given string. In order to force comparison against the entire string, you need to indicate that the regex begins at the beginning of the string and ends at the end by using ^ and $:
my $expr = qr/^\s*(\w+(\[\d+\])?)\s+(\w+(\[\d+\])?)$/;
(Also, there's no reason to have the /x modifier, as your regex doesn't include any literal whitespace or # characters, and there's no reason for the /s modifier, as you're not using ..)
EDIT: If you don't want the regex to match against the entire string, but you want it to reject anything in which the matching portion is followed by something like "[0:0]", the simplest way would be to use lookahead:
my $expr = qr/^\s*(\w+(\[\d+\])?)\s+(\w+(\[\d+\]|(?=[^[\w])|$ ))/x;
This will match anything that takes the following form:
beginning of the string (which your example in the comments seems to imply you want)
zero or more whitespace characters
one or more word characters
optional: [, one or more digits, ]
one or more whitespace characters
one or more word characters
one of the following, in descending order of preference:
[, one or more digits, ]
an empty string followed by (but not including!) a character that is neither [ nor a word character (The exclusion of word characters is to keep the regex engine from succeeding on "a[0] bc[1:2]" by only matching "a[0] b".)
end of string (A space is needed after the $ to keep it from merging with the following ) to form the name of a special variable, and this entails the reintroduction of the /x option.)
Do you have any more unstated requirements that need to be satisfied?
The short answer is your regexp is wrong.
We can't fix it for you without you explaining what you need exactly, and the community is not going to write a regexp exactly for your purpose because that's just too localized a question that only helps you this one time.
You need to ask something more general about regexps that we can explain to you, that will help you fix your regexp, and help others fix theirs.
Here's my general answer when you're having trouble testing your regexp. Use a regexp tool, like the regex buddy one.
So I'm going to give a specific answer about what you're overlooking here:
Let's make this example smaller:
Your pattern is a(bc+d)?. It will match: abcd abccd etc. While it will not match bcd nor bzd in the case of abzd it will match as matching only a because the whole group of bc+d is optional. Similarly it will match abcbcd as a dropping the whole optional group that couldn't be matched (at the second b).
Regexps will match as much of the string as they can and return a true match when they can match something and have satisfied the entire pattern. If you make something optional, they will leave it out when they have to including it only when it's present and matches.
Here's what you tried:
qr/\s*(\w+(\[\d+\])?)\s+(\w+(\[\d+\])?)/sx
First, s and x aren't needed modifiers here.
Second, this regex can match:
Any or no whitespace followed by
a word of at least one alpha character followed by
optionally a grouped square bracketed number with at least one digit (eg [0] or [9999]) followed by
at least one white space followed by
a word of at least one alpha character followed by
optionally a square bracketed number with at least one digit.
Clearly when you ask it to match abcd[0] xyzg[0:4] the colon ends the \d+ pattern but doesn't satisfy the \] so it backtracks the whole group, and then happily finds the group was optional. So by not matching the last optional group, your pattern has matched successfully.