I want to remove the same repeated non-word character.
My code looks like this:
<?php
$name = 'Malines - Blockbuster (prod. Malines) ***** (((((( &%^$';
echo preg_replace('/[^\pL\pN\s]{2,}/u', '', $name);
?>
Malines - Blockbuster (prod. Malines) ***** (((((( &%^$ should be Malines - Blockbuster (prod. Malines) &%^$
Only repeated non-word character should be removed. Could you help me?
You must use back-reference to say that the second character is the same as the one before :
/([^\pL\pN\s])\1+/u
Using the same idea than Arlaud Agbe Pierre (that is the way to do it), the pattern can be shorten using the Xan character class. \p{Xan} is the union of \p{L} and \p{N}. \P{Xan} (with an uppercase "p") is the negation of this class (i.e. all that is not a letter or a number).
$str = 'Malines - Blockbuster (prod. Malines) ààà ***** (((((( &%^$';
echo preg_replace('~(\P{Xan})\1+~u', '$1', $str);
Note: in this pattern, consecutive white characters are removed too.
an other way:
echo preg_replace('~(\P{Xan})\K\1+~u', '', $str);
where \K resets the begining of the match from the match result. (note that you can however define capturing groups before that you can use after)
Related
I have a string with chars inside and I would like to match only the chars around a string.
"This is a [1]test[/1] string. And [2]test[/2]"
Rubular http://rubular.com/r/f2Xwe3zPzo
Currently, the code in the link matches the text inside the special chars, how can I change it?
Update
To clarify my question. It should only match if the opening and closing has the same number.
"[2]first[/2] [1]second[/2]"
In the code above, only first should match and not second. The text inside the special chars (first), should be ignored.
Try this:
(\[[0-9]\]).+?(\[\/[0-9]\])
Permalink to the example on Rubular.
Update
Since you want to remove the 'special' characters, try this instead:
foo = "This is a [1]test[/1] string. And [2]test[/2]"
foo.gsub /\[\/?\d\]/, ""
# => "This is a test string. And test"
Update, Part II
You only want to remove the 'special' characters when the surrounding tags match, so what about this:
foo = "This is a [1]test[/1] string. And [2]test[/2], but not [3]test[/2]"
foo.gsub /(?:\[(?<number>\d)\])(?<content>.+?)(?:\[\/\k<number>\])/, '\k<content>'
# => "This is a test string. And test, but not [3]test[/2]"
\[([0-9])\].+?\[\/\1\]
([0-9]) is a capture since it is surrounded with parentheses. The \1 tells it to use the result of that capture. If you had more than one capture, you could reference them as well, \2, \3, etc.
Rubular
You can also use a named capture, rather than \1 to make it a little less cryptic. As in: \[(?<number>[0-9])\].+?\[\/\k<number>\]
Here's a way to do it that uses the form of String#gsub that takes a block. The idea is to pull strings such as "[1]test[/1]" into the block, and there remove the unwanted bits.
str = "This is a [1]test[/1] string. And [2]test[/2], plus [3]test[/99]"
r = /
\[ # match a left bracket
(\d+) # capture one or more digits in capture group 1
\] # match a right bracket
.+? # match one or more characters lazily
\[\/ # match a left bracket and forward slash
\1 # match the contents of capture group 1
\] # match a right bracket
/x
str.gsub(r) { |s| s[/(?<=\]).*?(?=\[)/] }
#=> "This is a test string. And test, plus [3]test[/99]"
Aside: When I first heard of named capture groups, they seemed like a great idea, but now I wonder if they really make regexes easier to read than \1, \2....
Can someone tell me, what's wrong in this code:
if ((!preg_match("[a-zA-Z0-9 \.\s]", $username)) || (!preg_match("[a-zA-Z0-9 \.\s]", $password)));
exit("result_message=Error: invalid characters");
}
??
Several things are wrong. I assume that the code you are looking for is:
if (preg_match('~[^a-z0-9\h.]~i', $username) || preg_match('~[^a-z0-9\h.]~i', $password))
exit('result_message=Error: invalid characters');
What is wrong in your code?
the pattern [a-zA-Z0-9 \.\s] is false for multiple reasons:
a regex pattern in PHP must by enclosed by delimiters, the most used is /, but as you can see, I have choosen ~. Example: /[a-zA-Z \.\s]/
the character class is strange because it contains a space and the character class \s that contains the space too. IMO, to check a username or a password, you only need the space and why not the tab, but not the carriage return or the line feed character! You can remove \s and let the space, or you can use the \h character class that matches all horizontal white spaces. /[a-zA-Z\h\.]/ (if you don't want to allow tabs, replace the \h by a space)
the dot has no special meaning inside a character class and doesn't need to be escaped: /[a-zA-Z\h.]/
you are trying to verify a whole string, but your pattern matches a single character! In other words, the pattern checks only if the string contains at least an alnum, a space or a dot. If you want to check all the string you must use a quantifier + and anchors for the start ^ and the end $ of the string. Example ∕^[a-zA-Z0-9\h.]+$/
in fine, you can shorten the character class by using the case-insensitive modifier i: /^[a-z0-9\h.]+$/i
But there is a faster way, instead of negate with ! your preg_match assertion and test if all characters are in the character range you want, you can only test if there is one character you don't want in the string. To do this you only need to negate the character class by inserting a ^ at the first place:
preg_match('/[^a-z0-9\h.]/i', ...
(Note that the ^ has a different meaning inside and outside a character class. If ^ isn't at the begining of a character class, it is a simple literal character.)
I've got strings like these:
+996999966966AA
-996999966966AA
I am using this code:
"+996999966966AA".gsub!(/\D/, "")
to get rid of any character except digits, but the sign + also being stripped. How can my code retain the +?
Use:
[^+\d]
to match anything that isn't + or a digit.
You can also use \W, "non-word character" which matches any character that is not a word character (alphanumeric & underscore)).
(\W\d+)\w+
Need parse a lot of text files and replace any quoted strings containing cyrillic symbols. They are may contains new lines, non-alphabetic characters and special symbols (for example '$' or escaped quote).
Can anyone help with regex?
From comments:
for example php code
function hello($word) {
$word2 = "ха-ха!";
echo "Привет, $word $word2\n";
}
hello('Мир');
I need match "ха-ха!", "Привет, $word $word2\n" and 'Мир'
This should work:
str = 'The cat is under the "таблица"'
regex = /"\p{Cyrillic}+.*?\.?"/ui
str.match(regex){|s| do_stuff_with_each_matching s}
# or...
str.gsub!(regex){|s| method_that_translates_russian s}
Check it out on live at http://rubular.com/r/0Mwbfinjvp.
http://www.ruby-doc.org/core-1.9.3/Regexp.html
".*[^a-zA-Z\d]+.*" matches any quoted character sequence containing at least one non-alphanumeric character.
i.e. it matches "aa$bb" and "a1$b1"
It doesn't match "aabb" or a$b.
Hope that this is what you want (Add required escaping).
I want to remove any leading and trailing non-alphabetic character in my string.
for eg. ":----- pt-br:-" , i want "pt-br"
Thanks
result = subject.gsub(/\A[\d_\W]+|[\d_\W]+\Z/, '')
will remove non-letters from the start and end of the string.
\A and \Z anchor the regex at the start/end of the string (^/$ would also match after/before a newline which is probably not what you want - but that might not matter in this case);
[\d_\W]+ matches one or more digits, the underscore or anything else that is not an alphanumeric character, leaving only letters.
| is the alternation operator.
In ruby 1.9.1 :
":----- pt-br:-".partition( /[a-zA-Z](...)[a-zA-Z]/ )[1]
partition searches the pattern in the string and returns the part before it, the match, and the part after it.
result = subject.gsub(/^[^a-zA-Z]+/, '').gsub(/[^a-zA-Z]+$/, '')