What regex would I need to use to achieve the following;
Check if string does not contain one or more options... I tried a lot expressions.
I think this is closest to be the correct one.
/^[^(256K)]$|^[^(2M)]$/
I would like preg_match to tell me if there is anything other than 256K or 2M, and I cant negate preg_match (!preg_match) for reasons that take to long to explain ;)
You can not place whole words or capturing groups inside of Character Classes. A character class matches any one character from a set of characters.
Your regular expression matches the beginning of the string, any character except: (, 2, 5, 6, K, ), followed by the end of the string, OR the beginning of the string, any character except: (, 2, M, ), followed by the end of string.
I believe you are wanting a Negative Lookahead here instead.
/^((?!256K|2M).)*$/i
Regular expression:
^ # the beginning of the string
( # group and capture to \1 (0 or more times)
(?! # look ahead to see if there is not:
256K # '256K'
| # OR
2M # '2M'
) # end of look-ahead
. # any character except \n
)* # end of \1
$ # before an optional \n, and the end of the string
Related
I 'm tring to do a php regexp to mask a email so that
example#gmail.com turn to e*****e#gmail.com.
$maskedEmail=preg_replace('/^*#/', '*', $email);
You may use
preg_replace('~((?!^)\G|^[^#])[^#](?=[^#]+#)~', '$1*', $s)
See the regex demo
Details
((?!^)\G|^[^#]) - Group 1: the end of the previous match or start of string and any char other than a #
[^#] - a char other than #
(?=[^#]+#) - a positive lookahead that requires 1+ chars (that + is important here, you can't use *) other than # followed with a # immediately to the right of the current location.
The replacement is the value captured in Group 1 (so that the first char is kept in the string, and then all but the last char before a # are replaced with *.
To not mask the first character you could assert what is directly to the left is not the start of the string.
To not mask the character directly before the #, you could assert that what is on the right is always one character that is not an # before matching it.
(?<!^).(?=[^#]+#)
In the replacement use:
*
Explanation
(?<!^) Negative lookbehind, assert what is directly on the left is not the start of the string
. Match any character except a newline
(?= Positive lookahead, assert what is directly on the right is
[^#]+# Match 1+ times any char except # using a negated character class, and match #
) Close positive lookahead
Regex demo | Php demo
For example
$email = "example#gmail.com";
$maskedEmail=preg_replace('/(?<!^).(?=[^#]+#)/', '*', $email);
echo $maskedEmail;
Result
e*****e#gmail.com
I need to match a series of strings that:
Contain at least 3 numbers
0 or more letters
0 or 1 - (not more)
0 or 1 \ (not more)
These characters can be in any position in the string.
The regular expression I have so far is:
([A-Z0-9]*[0-9]{3,}[\/]?[\-]?[0-9]*[A-Z]*)
This matches the following data in the following cases. The only one that does not match is the first one:
02ABU-D9435
013DFC
1123451
03323456782
ADS7124536768
03SDFA9433/0
03SDFA9433/
03SDFA9433/1
A41B03423523
O4AGFC4430
I think perhaps I am being too prescriptive about positioning. How can I update this regex to match all possibilities?
PHP PCRE
The following would not match:
01/01/2018 [multiple / or -]
AA-AA [no numbers]
Thanks
One option could be using lookaheads to assert 3 digits, not 2 backslashes and not 2 times a hyphen.
(?<!\S)(?=(?:[^\d\s]*\d){3})(?!(?:[^\s-]*-){2})(?!(?:[^\s\\]*\\){2})[A-Z0-9/\\-]+(?!\S)
About the pattern
(?<!\S) Assert what is on the left is not a non whitespace char
(?=(?:[^\d\s]*\d){3}) Assert wat is on the right is 3 times a whitespace char or digit
(?!(?:[^\s-]*-){2}) Assert what is on the right is not 2 times a whitespace char a hyphen
(?!(?:[^\s\\]*\\){2}) Assert what is on the right is not 2 times a whitespace char a backslash
[A-Z0-9/\\-]+ Match any of the listed 1+ times
(?!\S) Assert what is on the right is not a non whitespace char
Regex demo
Your patterns can be checked with positive/negative lookaheads anchored at the start of the string:
at least 3 digits -> find (not necessarily consecutive) 3 digits
no more than 1 '-' -> assert absence of (not necessarily consecutive) 2 '-' characters
no more than 1 '/' -> assert absence of (not necessarily consecutive) 2 '/' characters
0 or more letters -> no check needed.
If these conditions are met, any content is permitted.
The regex implementing this:
^(?=(([^0-9\r\n]*\d){3}))(?!(.*-){2})(?!(.*\/){2}).*$
Check out this Regex101 demo.
Remark
This solution assumes that each string tested resides on its own line, ie. not just being separated by whitespace.
In case the strings are separated by whitespace, choose the solution of user #TheFourthBird (which essentially is the same as this one but caters for the whitespace separation)
You can test the condition for both the hyphen and the slash into a same lookahead using a capture group and a backreference:
~\A(?!.*([-/]).*\1)(?:[A-Z/-]*\d){3,}[A-Z/-]*\z~
demo
detailled:
~ # using the tild as pattern delimiter avoids to escape all slashes in the pattern
\A # start of the string
(?! .* ([-/]) .* \1 ) # negative lookahead:
# check that there's no more than one hyphen and one slash
(?: [A-Z/-]* \d ){3,} # at least 3 digits
[A-Z/-]* # eventual other characters until the end of the string
\z # end of the string.
~
To better understand (if you are not familiar with): these three subpatterns start from the same position (in this case the beginning of the string):
\A
(?! .* ([-/]) .* \1 )
(?: [A-Z/-]* \d ){3,}
This is possible only because the two first are zero-width assertions that are simple tests and don't consume any character.
This question already has an answer here:
Reference - What does this regex mean?
(1 answer)
Closed 7 years ago.
The preg_replace() function has so many possible values, like:
<?php
$patterns = array('/(19|20)(\d{2})-(\d{1,2})-(\d{1,2})/', '/^\s*{(\w+)}\s*=/');
$replace = array('\3/\4/\1\2', '$\1 =');
echo preg_replace($patterns, $replace, '{startDate} = 1999-5-27');
What does:
\3/\4/\1\2
And:
/(19|20)(\d{2})-(\d{1,2})-(\d{1,2})/','/^\s*{(\w+)}\s*=/
mean?
Is there any information available to help understand the meanings at one place? Any help or documents would be appreciated! Thanks in Advance.
Take a look at http://www.tutorialspoint.com/php/php_regular_expression.htm
\3 is the captured group 3
\4 is the captured group 4
...an so on...
\w means any word character.
\d means any digit.
\s means any white space.
+ means match the preceding pattern at least once or more.
* means match the preceding pattern 0 times or more.
{n,m} means match the preceding pattern at least n times to m times max.
{n} means match the preceding pattern exactly n times.
(n,} means match the preceding pattern at least n times or more.
(...) is a captured group.
So, the first thing to point out, is that we have an array of patterns ($patterns), and an array of replacements ($replace). Let's take each pattern and replacement and break it down:
Pattern:
/(19|20)(\d{2})-(\d{1,2})-(\d{1,2})/
Replacement:
\3/\4/\1\2
This takes a date and converts it from a YYYY-M-D format to a M/D/YYYY format. Let's break down it's components:
/ ... / # The starting and trailing slash mark the beginning and end of the expression.
(19|20) # Matches either 19 or 20, capturing the result as \1.
# \1 will be 19 or 20.
(\d{2}) # Matches any two digits (must be two digits), capturing the result as \2.
# \2 will be the two digits captured here.
- # Literal "-" character, not captured.
(\d{2}) # Either 1 or 2 digits, capturing the result as \3.
# \3 will be the one or two digits captured here.
- # Literal "-" character, not captured.
(\d{2}) # Either 1 or 2 digits, capturing the result as \4.
# \4 will be the one or two digits captured here.
This match is replaced by \3/\4/\1\2, which means:
\3 # The two digits captured in the 3rd set of `()`s, representing the month.
/ # A literal '/'.
\4 # The two digits captured in the 4rd set of `()`s, representing the day.
/ # A literal '/'.
\1 # Either '19' or '20'; the first two digits captured (first `()`s).
\2 # The two digits captured in the 2nd set of `()`s, representing the last two digits of the year.
Pattern:
/^\s*{(\w+)}\s*=/
Replacement:
$\1 =
This takes a variable name encoded as {variable} and converts it to $variable = <date>. Let's break it down:
/ ... / # The starting and trailing slash mark the beginning and end of the expression.
^ # Matches the beginning of the string, anchoring the match.
# If the following character isn't matched exactly at the beginning of the string, the expression won't match.
\s* # Any whitespace character. This can include spaces, tabs, etc.
# The '*' means "zero or more occurrences".
# So, the whitespace is optional, but there can be any amount of it at the beginning of the line.
{ # A literal '{' character.
(\w+) # Any 'word' character (a-z, A-Z, 0-9, _). This is captured in \1.
# \1 will be the text contained between the { and }, and is the only thing "captured" in this expression.
} # A literal '}' character.
\s* # Any whitespace character. This can include spaces, tabs, etc.
= # A literal '=' character.
This match is replaced by $\1 =, which means:
$ # A literal '$' character.
\1 # The text captured in the 1st and only set of `()`s, representing the variable name.
# A literal space.
= # A literal '=' character.
Lastly, I wanted to show you a couple of resources. The regex-format you're using is called "PCRE", or Perl-Compatible Regular Expressions. Here is a quick cheat-sheet on PCRE for PHP. Over the last few years, several tools have been popping up to help you visualize, explain, and test regular expressions. One is Regex 101 (just Google "regex tester" or "regex visualizer"). If you look here, this is an explanation of the first RegEx, and here is an explanation of the second. There are others as well, like Debuggex, Regex Tester, etc. But I find the detailed match breakdown on Regex 101 to be pretty useful.
I am a serious newbie with regular expression so please disregard my mistakes. I need to be sure that several criteria in a string are met.
Requirements:
Have at most 5 words
Max of 256 characters
Word is considered 1 or more characters - no spaces
Shouldn't contain two consecutive spaces
Example:
Tree blows in the wind
1-Tree falls over
Failure Example:
Tree blows in the night sky
Tree breaks 2 limbs during night
Can this be done in one single expression or should it be broken up?
Validating for 2 spaces:
- /^\s\s$/
Max of 256 characters:
- /^[a-zA-Z0-9]{,256}$/
I am not sure how to test case for the 5 words and combine the other criteria that I impose. Can anyone help?
Test for word:
- /^\w{1,5}$
You can try this:
(?s)\A(?!.{257}|.*\s\s)\W*\w*(?:\W+\w+){0,4}\W*\z
pattern details:
(?s) # turn on the singleline mode: allow the dot to match newlines
\A # start of the string anchor
(?! # open a negative lookahead assertion: means not followed by
.{257} # 257 characters
| # OR
.*\s\s # two consecutive whitespaces
) # close the negative lookahead
\W* # optional non-word characters
\w* # optional word characters (nothing in your requirements forbids to have a string without words or an empty string)
(?: # open a non-capturing group
\W+ # non-word characters: words are obviously separated with non-word characters
\w+ # an other word
){0,4} # repeat the non-capturing group between zero and 4 times
\W* # optional non-word characters
\z # anchor for the end of the string
I know this regex divides a text into sentences. Can someone help me understand how?
/(?<!\..)([\?\!\.])\s(?!.\.)/
You can use YAPE::Regex::Explain to decipher Perl regular expressions:
use strict;
use warnings;
use YAPE::Regex::Explain;
my $re = qr/(?<!\..)([\?\!\.])\s(?!.\.)/;
print YAPE::Regex::Explain->new($re)->explain();
__END__
The regular expression:
(?-imsx:(?<!\..)([\?\!\.])\s(?!.\.))
matches as follows:
NODE EXPLANATION
----------------------------------------------------------------------
(?-imsx: group, but do not capture (case-sensitive)
(with ^ and $ matching normally) (with . not
matching \n) (matching whitespace and #
normally):
----------------------------------------------------------------------
(?<! look behind to see if there is not:
----------------------------------------------------------------------
\. '.'
----------------------------------------------------------------------
. any character except \n
----------------------------------------------------------------------
) end of look-behind
----------------------------------------------------------------------
( group and capture to \1:
----------------------------------------------------------------------
[\?\!\.] any character of: '\?', '\!', '\.'
----------------------------------------------------------------------
) end of \1
----------------------------------------------------------------------
\s whitespace (\n, \r, \t, \f, and " ")
----------------------------------------------------------------------
(?! look ahead to see if there is not:
----------------------------------------------------------------------
. any character except \n
----------------------------------------------------------------------
\. '.'
----------------------------------------------------------------------
) end of look-ahead
----------------------------------------------------------------------
) end of grouping
----------------------------------------------------------------------
There is the Regular Expression Analyzer which will do quite the same as toolic already suggested - but completely webbased.
(? # Find a group (don't capture)
< # before the following regular expression
! # that does not match
\. # a literal "."
. # followed by 1 character
) # (End look-behind group)
( # Start a group (capture it to $1)
[\?\!\.] # Containing any one of the characters in the following set "?!."
) # End group $1
\s # followed by a whitespace character " ", \t, etc.
(? # Followed by a group (don't capture)
# after the preceding regular expression
! # that does not have
. # 1 character
\. # followed by a literal "."
) # (End look-ahead group)
The first part (?<!\..) is a negative look-behind. It specifies a pattern which invalidates the match. In this case it's looking for two characters--the first a period and the other one any character.
The second part is a standard capture/group, which could be better expressed: ([?!.]) (you don't need the escapes in the class brackets), that is a sentence ending punctuation character.
The next part is a single (??) white-space character: \s
And the last part is a negative look-ahead: (?!.\.). Again it is guarding against the case of a single character followed by a period.
This should work, relatively well. But I don't think I would recommend it. I don't see what the coder was getting at trying to make sure that just a period wasn't the second most recent character, or that it wasn't the second one to come.
I mean if you are looking to split on terminal punctuation, why don't you want to guard against the same class being two-back or two-ahead? Instead it relies on periods not being there. Thus a more regular expression would be:
/(?<![?!.].)([?!.])\s(?!.[?!.])/
Portions:
([\?\!\.])\s: split by ending character (.,!,or ?) which is followed by a whitespace character (space, tab, newline)
(?<!\..) where the characters before this 'ending character' arent a .+anything
(?!.\.) after the whitespace character any character directly followed by any . isn't allowed.
Those look-ahead ((?!) & look-behind ((?<!) assertions mainly seem to prevent splitting on (whitespaced?) abbreviations (q. e. d. etc.).