I need (in PHP) to split a sententse by the word that cannot be the first or the last one in the sentence. Say the word is "pression" and here is my regex
/^.+?(?=[\s\.\,\:\;])pression(?=[\s\.\,\:\;]).+$/i
Live here: https://regex101.com/r/CHAhKj/1/
First, it doesn't match.
Next, I think - it is at all possible to split that way? I tryed simplified example
print_r(preg_split('/^.+pizza.+$/', 'my pizza is cool'));
live here http://sandbox.onlinephpfunctions.com/code/10b674900fc1ef44ec79bfaf80e83fe1f4248d02
and it prints an array of 2 empty strings, when I expect
['my ', ' is cool']
I need (in PHP) to split a sentence by the word that cannot be the first or the last one in the sentence
You may use this regex:
(?<=[^\s.?]\h)pression(?=\h[^\s.?])
RegEx Demo
RegEx Details:
(?<=[^\s.?]\h): Lookbehind to assert that ahead of current position we have a space and a character that not a whitespace, not a dot and not a ?.
pression: Match word pression
(?=\h[^\s.?]): Lookahead to assert that before current position we have a space and a character that not a whitespace, not a dot and not a ?
First, ^.+?(?=[\s\.\,\:\;])pression(?=[\s\.\,\:\;]).+$ can't match any string at all because the (?=[\s\.\,\:\;])p part requires p to be also either a whitespace char, or a ., ,, : or ;, which invalidates the whole match at once.
Second, ^.+pizza.+$ pattern does not ensure the pizza matched is not the first or last word in a sentence as . matches whitespace, too. It does not return anything meaningful, because preg_split uses the match to break string into chunks, and the two empty values are 1) start of string and 2) empty string positions.
That said, all you need is:
preg_match('~^(.*?\w\W+)pression(\W+\w.*)$~is', $text, $m)
See the regex demo. Details:
^ - start of string
(.*?\w\W+) - Capturing group 1: any zero or more chars, as few as possible, then a word char and then one or more non-word chars
pression - a word
(\W+\w.*) - Capturing group 2: one or more non-word chars, a word char, and then any zero or more chars as many as possible
$ - end of string.
s makes the . match across lines and i flag makes the pattern match in a case insensitive way.
See the PHP demo:
$text = "You can use any regular expression pression inside the lookahead ";
if (preg_match('~^(.*?\w\W+)pression(\W+\w.*)$~is', $text, $m)) {
echo $m[1] . " << | >> " . $m[2];
}
// => You can use any regular expression << | >> inside the lookahead
Related
regarding to this post "https://stackoverflow.com/questions/35413960/regular-expression-match-all-except-first-occurence" I'm wondering how to find the first occurence on a string only if it start's with a specfic character in PHP.
I would like to sanitize phonenumbers. Example bad phone number:
+49+12423#23492#aosd#+dasd
Regex to remove all "+" except first occurence.
\G(?:\A[^\+]*\+)?+[^\+]*\K\+
Problem: it should remove every "+" only if it starts with "+" not if the first occurence-position is greater than 1.
The regex to remove everything except numbers is easy:
[^0-9]*
But I don't know how to combine those two within one regex. I would just use preg_replace() twice.
Of course I would be able to use a workaround like if ($str[0] === '+') {...} but I prefer to learn some new stuff (regex :)
Thanks for helping.
You can use
(?:\G(?!\A)|^\+)[^+]*\K\+
See the regex demo. Details:
(?:\G(?!\A)|^\+) - either the end of the preceding successful match or a + at the start of string
[^+]* - zero or more chars other than +
\K - match reset operator discarding the text matched so far
\+ - a + char.
See the PHP demo:
$re = '/(?:\G(?!\A)|^\+)[^+]*\K\+/m';
$str = '+49+12423#23492#aosd#+dasd';
echo preg_replace($re, '', $str);
// => +4912423#23492#aosd#dasd
You seem to want to combine the two queries:
A regex to remove everything except numbers
A regex to remove all "+" except first occurence
Here is my two cents:
(?:^\+|\d)(*SKIP)(*F)|.
Replace what is matched with nothing. Here is an online demo
(?:^\+|\d) - A non-capture group to match a starting literal plus or any digit in the range from 0-9.
(*SKIP)(*F) - Consume the previous matched characters and fail them in the rest of the matching result.
| - Or:
. - Any single character other than newline.
I'd like to think that this is a slight adaptation of what some consider "The best regex trick ever" where one would first try to match what you don't want, then use an alternation to match what you do want. With the use of the backtracking control verbs (*SKIP)(*F) we reverse the logic. We first match what we do want, exclude it from the results and then match what we don't want.
My regfu has declined... and I'm having trouble getting expected matches.
Here's example of what needs to match and what not:
NLNL LL
LNLN LL LL
NNLL LL LL LL
LNLN LLL LL
LLNN LL LLL <-- skip because:
Only need:
1 to 3 Pairs of letters separated by one space
Which are consecutive to end of string
\s{1}([A-Z]{2}) is close, but also grabbing part of the skip above.
Why? I need to grab what are name initials from strings. There are either 1,2,or 3 persons initials appended to the strings. I will be grabbing those with PHP to store them.
You may use
if (preg_match('~(?: [A-Z]{2})+$~', $s, $match)) {
print_r(explode(" ", trim($match[0])));
}
Here, (?: [A-Z]{2})+$ matches one or more sequences of a space and then two uppercase ASCII letters till the end of string, and then explode(" ", trim($match[0])) splits the trimmed match with a space into chunks.
Or, if you want to match all occurrences with one regex call:
if (preg_match_all('~(?:\G(?!\A)|(?=(?:\s[A-Z]{2})+$))\s\K[A-Z]{2}~', $s, $matches)) {
print_r($matches[0]);
}
Here, the regex matches:
(?:\G(?!\A)|(?=(?:\s[A-Z]{2})+$)) - end of previous match (\G(?!\A)) or (|) a location immediately followed with one or more sequences of a space and then two uppercase ASCII letters till the end of string
\s - a whitespace
\K - match reset operator
[A-Z]{2} - two uppercase ASCII letters.
See the PHP demo.
We are extracting text from PDF files, and there is a high frequency of results that contain malformed text. Specifically adding spaces between the characters of a word. e.g. SEATTLE is being returned as S E A T T L E.
Is there a RegEx expression for preg_replace that can remove any spaces in the case of n number of single character "words"? Specifically, remove spaces from any occurrence of a string that is more than 3 single alpha characters and is separated by spaces?
If googled this for awhile, but can't even imagine how to construct the expression. As expressed in a comment, I don't want ALL spaces removed, but only when there is an occurrence of >3 single alpha characters, e.g. Welcome to the Greater S E A T T L E area should become Welcome to the Greater SEATTLE area. The result is to be used in full text searching, so case sensitivity is not a concern.
You may use a simple approach with a preg_replace_callback. Match '~\b[A-Za-z](?: [A-Za-z]){2,}\b~' and str_replace spaces in the anonymous function:
$regex = '~\b[A-Za-z](?: [A-Za-z]){2,}\b~';
$result = preg_replace_callback($regex, function($m) {
return str_replace(" ", "", $m[0]);
}, $s);
See the regex demo.
To only match sequences of uppercase letters, remove a-z from the pattern:
$regex = '~\b[A-Z](?: [A-Z]){2,}\b~';
And another thing: there may be soft/hard spaces, tabs, other kind of whitespace. Then, use
$regex = '~\b[A-Za-z](?:\h[A-Za-z]){2,}\b~u';
^^ ^
Finally, to match any Unicode letter, use \p{L} (to only match uppercase ones, \p{Lu}) instead of [a-zA-Z]:
$regex = '~\b\p{L}(?:\h\p{L}){2,}\b~u';
NOTE: It will most probably fail to work in some cases, e.g. when there are one-letter words. You will have to handle those cases separately/manually. Anyway, there is no safe regex-only way to fix OCR issues.
Pattern details
\b - a word boundary
[A-Za-z] - a single letter
(?: [A-Za-z]){2,} - 2 or more occurrences of
- a space (\h matches any kind of horizontal whitespace)
[A-Za-z] - a single letter
\b - a word boundary
When usign u modifier, \h becomes Unicode-aware.
You could do this in one go:
(?i:(?<!\S)([a-z]) +((?1))|\G(?!\A) +((?1))\b)
See live demo here
Explanation:
(?i: # Start of non-capturing group with case-insensitive modifier on
(?<!\S) # Negative lookbehind to ensure there is no leading non-whitespace character
([a-z]) + # Capture one letter and at least one space
((?1)) # Capture one letter in 2nd capturing group
| # Or
\G(?!\A) + # Start match from where previous match ends
# with matching spaces
((?1))\b # Match a letter at word boundary
) # End of non-capturing group
PHP code:
$str = preg_replace('~(?i:(?<!\S)([a-z]) +((?1))|\G(?!\A) +((?1))\b)~', '$1$2$3', $str);
You may use this pure regex approach with lookarounds and \G:
$re = '~\b(?:(?=(?:\pL\h+){3}\pL\b)|(?<!^)\G)(\pL)\h+(?=\pL\b)~';
$repl = preg_replace($re, '$1', $str);
RegEx Demo
RegEx Details:
\b: Match word boundary
(?:: Start non-capture group
(?=(?:\pL\h+){3}\pL\b): Lookahead to assert we have 3+ single letters separated by 1+ spaces
|: OR
(?<!^)\G: \G asserts position at the end of the previous match. (?<!^) ensures we don't match start of the string for the first match
): End non-capture group
(\pL): Match a single letter and capture it
\h+: Followed by 1+ horizontal whitespace
(?=\pL\b): Assert that we only have a single letter ahead
In the replacement we use $1 which is the letter left of whitespace we capture
im looking for a regex that matches words that repeat a letter(s) more than once and that are next to each other.
Here's an example:
This is an exxxmaple oooonnnnllllyyyyy!
By far I havent found anything that can exactly match:
exxxmaple and oooonnnnllllyyyyy
I need to find it and place them in an array, like this:
preg_match_all('/\b(???)\b/', $str, $arr) );
Can somebody explain what regexp i have to use?
You can use a very simple regex like
\S*(\w)(?=\1+)\S*
See how the regex matches at http://regex101.com/r/rF3pR7/3
\S matches anything other than a space
* quantifier, zero or more occurance of \S
(\w) matches a single character, captures in \1
(?=\1+) postive look ahead. Asserts that the captrued character is followed by itsef \1
+ quantifiers, one or more occurence of the repeated character
\S* matches anything other than space
EDIT
If the repeating must be more than once, a slight modification of the regex would do the trick
\S*(\w)(?=\1{2,})\S*
for example http://regex101.com/r/rF3pR7/5
Use this if you want discard words like apple etc .
\b\w*(\w)(?=\1\1+)\w*\b
or
\b(?=[^\s]*(\w)\1\1+)\w+\b
Try this.See demo.
http://regex101.com/r/kP8uF5/20
http://regex101.com/r/kP8uF5/21
You can use this pattern:
\b\w*?(\w)\1{2}\w*
The \w class and the word-boundary \b limit the search to words. Note that the word boundary can be removed, however, it reduces the number of steps to obtain a match (as the lazy quantifier). Note too, that if you are looking for words (in the common meaning), you need to remove the word boundary and to use [a-zA-Z] instead of \w.
(\w)\1{2} checks if a repeated character is present. A word character is captured in group 1 and must be followed with the content of the capture group (the backreference \1).
I've written the next regular expression
$pattern = "~\d+[.][\s]*[A-Z]{1}[A-Za-z0-9\s-']+~";
in order to match substrings as 2.bon jovi - it's my life
the problem is the only part that is recognized is - bon jovi
none " - " or " ' " are recognized by this regular expression.
I'd prefer to know what is wrong with the regular expression that I've wrote rather than getting a new one.
Your regular expressions states that after the period character (can be changed to \.), you will have zero or more white space characters which should then be followed by 1 upper case letter. In your string, you do not have any upper case letters.
Secondly, the - should be placed last when you want to match it. So, changing your regex to this: ~\d+[.][\s]*[A-Z]{1}[A-Za-z0-9\s'-]+~ will match something like so: 2.Bon jovi - it's my life.
On the other hand, you can change it to this: ~\d+[.][\s]*[A-Za-z0-9\s'-]+~ to match something like so: 2.bon jovi - it's my life.
EDIT: Ammended as per the comments of Marko D and aleation.
A better regular expression to handle that would be...
$pattern = "~\d+\.\s*[\pL\pP\s]+~";
CodePad.
This will match a number, followed by a ., followed by optional whitespace, followed by one or more Unicode letters, whitespace or punctuation marks.
$pattern = "~\d+\..*~";
$string = "2.bon jovi - it's my life";
preg_match($pattern, $string, $match);
print_r($match);
output: Array ( [0] => 2.bon jovi - it's my life )
So the way I understand this regular expression is:
\d+ // Match any digit, 1 or more times
[.] // Match a dot
[\s]* // Match 0 or more whitespace characters
[A-Z]{1} // Match characters between an UPPERCASE A-Z Range 1 time
[A-Za-z0-9\s-']+ // Match characters between A-Z, a-z, 0-9, whitespace, dashe and apostrophe
So straight away, your 'bon jovi' might not get matched as it's lower case and you're only looking for uppercase characters. 'bon jovi' also contains a space so perhaps changing that part of the regular expression to allow for lowercase characters and whitespace might help so you'd end up with:
$pattern = "~\d+[.][\s]*[A-Za-z\s]{1}[A-Za-z0-9\s-']+~";
Note: I quickly tested this on RegExr ( http://gskinner.com/RegExr/ ) and it appeared to match the string fine.
Your regrex is as follows.
~ // delimiter
\d+ // 1 or more numbers
[.] // a period
[\s]* // 0 or more whitespace characters
[A-Z]{1} // 1 upper case letter
[A-Za-z0-9\s-\']+ // 1 or more characters, from the character class
~ //delimiter
Comparing that to the string "2.bon jovi" You have:
~ //
\d+ // "2"
[.] // "."
[\s]* // ""
[A-Z]{1} // <- NO MATCH
[A-Za-z0-9\s-\']+ //
~ //
"bon" does not start with a captial letter, it therefore does not match [A-Z]{1}
Cleaner regex
There are a few simple things you can do to clean up your regex
don't use character-classes for one character
don't specify {1} it's the same as not being present
Applying the above to your existing regex you get:
$pattern = "~\d+\.\s*[A-Z][A-Za-z0-9\s-']+~";
Which is slightly easier to read.
Your [A-Z]{1} sub-pattern requires one capital letter, so "2.bon jovi - it's my life" will not match.
And you need to escape the - in the [A-Za-z0-9\s-'] character class, or put it at the start or end, otherwise it is specifying a range.
"~\d+\.[A-Za-z0-9\s'-]+~"
As pointed out in the comments, it is actually not necessary to escape the - in the character class in your regex. That is only because you happened to precede it with a metacharacter \s that cannot be part of a range. Normally, if you want to match a literal - and you have it in a character class, you must escape it or position it as described above.