I'm using this regex to get house number of a street adress.
[a-zA-ZßäöüÄÖÜ .]*(?=[0-9])
Usually, the street is something like "Ohmstraße 2a" or something. At regexpal.com my pattern matches, but I guess preg_replace() isn't identical with it's regex engine.
$num = preg_replace("/[a-zA-ZßäöüÄÖÜ .]*(?=[0-9])/", "", $num);
Update:
It seems that my pattern matches, but I've got some encoding problems with the special chars like äöü
Update #2:
Turns out to be a encoding problem with mysqli.
First of all if you want to get the house number then you should not replace it. So instead of preg_replace use preg_match.
I modified your regex a little bit to match better:
$street = 'Öhmsträße 2a';
if(preg_match('/\s+(\d+[a-z]?)$/i', trim($street), $matches) !== 0) {
var_dump($matches);
} else {
echo 'no house number';
}
\s+ matches one or more space chars (blanks, tabs, etc.)
(...) defines a capture group which can be accesses in $matches
\d+ matches one or more digits (2, 23, 235, ...)
[a-z] matches one character from a to z
? means it's optional (not every house number has a letter in it)
$ means end of string, so it makes sure the house number is at the end of the string
Make sure you strip any spaces after the end of the house number with trim().
The u modifier can help sometimes for handling "extra" characters.
I feel this may be a character set or UTF-8 issue.
It would be a good idea to find out what version of PHP you're running too. If I recall correctly, full Unicode support came in around 5.1.x
Related
I have this regex that matches strings that I want to check on validity.
However recently I want to use this same regex to replace every character that is not valid to the regex with a character (let's say x).
My regex to match these types of strings is: '#^[\pL\'\’\d][\pL\.\-\ \'\/\,\’\d]*$#iu'
Which allows for the first character to be of any language or any digit and some determined special chars. And all the following letters to be slightly the same but slightly more special characters.
This is what I do (nothing special).
preg_replace($regex, 'x', $string);
Things I tried include trying to negate the regex:
'(?![\pL\'\’\d][\pL\.\-\ \'\/\,\’\d]*)'
'[^\pL\'\’\d][^\pL\.\-\ \'\/\,\’\d]*'
I've also tried splitting up the string into the firstchar and the rest of the string and split the regex in 2.
$validationRegex1 = '[^\pL\'\’\d]';
$validationRegex2 = '[^\pL\.\-\ \'\/\,\’\d]*';
$fixedStr1 = (string) preg_replace($validationRegex1, 'x', $firstChar)
. (string) preg_replace($validationRegex2, 'x', $theRest);
But this also did not seemed to work.
I've experimented a bit with this online tool: https://www.functions-online.com/preg_replace.html
Does anyone know what I am overlooking?
Examples of strings and their expected results
'-' should become 'x'.
'Random-morestuff' stays 'Random-morestuff'
'Random%morestuff' should become 'Randomxmorestuff'
'Rândôm' stays 'Rândôm'
Just an idea but if I got you right, you could use
(?(DEFINE)
(?<first>[\pL\d'’])
(?<other>[-\ \pL\d.'/,’])
)
\b(?&first)(?&other)+\b(*SKIP)(*FAIL)|.
This needs to be replaced by x. You do not have to escape everything in a character class, I changed this accordingly.
See a demo on regex101.com.
A bit more explanation: The (?(DEFINE)...) thingy lets you define subroutines that can be used afterwards and is just syntactic sugar in this case (maybe a bit showing off, really). As you have stated that other characters are allowed depending on theirs positions, I just called them first and other. The \b marks a word boundary, that is a boundary between \w (usually [a-zA-Z0-9_]) and \W (not \w). All of these "words" are allowed, so we let the engine "forget" what has been matched with the (*SKIP)(*FAIL) mechanism and match any other character on the right side of the alternation (|). See how (*SKIP)(*FAIL) works here on SO.
Use
$fixedStr1 = preg_replace('/[\p{L}\'\’\d][\p{L}\.\ \'\/\,\’\d-]*(*SKIP)(*FAIL)|./u', 'x', $input_string);
See regex proof.
Fail matches that match valid symbol words and replace every character appearing in other places.
I have been looking around and googling about regex and I came up with this to make sure some variable has letters in it (and nothing else).
/^[a-zA-Z]*$/
In my mind ^ denotes the start of the string, the token [a-zA-Z] should in my mind make sure only letters are allowed. The star should match everything in the token, and the $-sign denotes the end of the string I'm trying to match.
But it doesn't work, when I try it on regexr it doesn't work sadly. What's the correct way to do it then? I would also like to allow hyphen and spaces, but figured just starting with letters are good enough to start with and expand.
Short answer : this is what you are looking for.
/^[a-zA-Z]+$/
The star * quantifier means "zero or more", meaning your regexp will match everytime even with an empty string as subject. You need the + quantifier instead (meaning "one or more") to achieve what you need.
If you also want to match at least one character which could also be a whitespace or a hyphen you could add those to your character class ^[A-Za-z -]+$ using the plus + sign for the repetition.
If you want to use preg_match to match at least one character which can contain an upper or lowercase character, you could shorten your pattern to ^[a-z]+$ and use the i modifier to make the regex case insensitive. To also match a hyphen and a whitespace, this could look like ^[a-z -]+$
For example:
$strings = [
"TeSt",
"Te s-t",
"",
"Te4St"
];
foreach ($strings as $string) {
if(preg_match('#^[a-z -]+$#i', $string, $matches)){
echo $matches[0] . PHP_EOL;
}
}
That would result in:
TeSt
Te s-t
Output php
I am trying to grab the text after the last number in the string and grab the whole string if it doesn't contain numbers.
The best regex I could come up with is:
([^\d\s]*)$
However I found that \s and \d aren't supported in mysql regexp rather [[:space:]] and not sure what \d is equivalent too.
This is what I'm trying to accomplish:
'1/2 Oz' returns 'Oz'
'2 3/4 Oz' returns 'Oz'
'As needed' returns 'As needed'
This is the regex you will need:
/^.*?(\d+(?=\D*$)\s*)/
And just replace matched text with empty string ""
PHP code:
$s = preg_replace('/^.*?(\d+(?=\D*$)\s*)/', '', 'Foo Oz');
//=> Foo Oz
$s = preg_replace('/^.*?(\d+(?=\D*$)\s*)/', '', '1/2 Oz');
//=> Oz
Live Demo: http://ideone.com/u887D7
First of all, you could simply avoid the class, and use a range instead:
[^0-9[:space:]]*$
But there is one for digits as well (which may actually include non-ASCII digits). The documentation has a list of these. They are called POSIX bracket expressions by the way.
[^[:digit:][:space:]]*$
However, the general problem with this approach is that it doesn't allow for spaces later on in the string (like the one between As and needed. To get those, but still avoid capturing trailing spaces after digits, make sure, the first character is neither space nor digit, then match the rest of the string as non-digits. In addition, make the whole thing optional, to ensure that it still works with strings ending in a digit.
([^[:digit:][:space:]][^:digit:]*)?$
How would I go about removing numbers and a space from the start of a string?
For example, from '13 Adam Court, Cannock' remove '13 '
Because everyone else is going the \d+\s route I'll give you the brain-dead answer
$str = preg_replace("#([0-9]+ )#","",$str);
Word to the wise, don't use / as your delimiter in regex, you will experience the dreaded leaning-toothpick-problem when trying to do file paths or something like http://
:)
Use the same regex I gave in my JavaScript answer, but apply it using preg_replace():
preg_replace('/^\d+\s+/', '', $str);
Try this one :
^\d+ (.*)$
Like this :
preg_replace ("^\d+ (.*)$", "$1" , $string);
Resources :
preg_replace
regular-expressions.info
On the same topic :
Regular expression to remove number, then a space?
regular expression for matching number and spaces.
I'd use
/^\d+\s+/
It looks for a number of any size in the beginning of a string ^\d+
Then looks for a patch of whitespace after it \s+
When you use a backslash before certain letters it represents something...
\d represents a digit 0,1,2,3,4,5,6,7,8,9.
\s represents a space .
Add a plus sign (+) to the end and you can have...
\d+ a series of digits (number)
\s+ multiple spaces (typos etc.)
The same regex I gave you on your other question still applies. You just have to use preg_replace() instead.
Search for /^[\s\d]+/ and replace with the empty string. Eg:
$str = preg_replace(/^[\s\d]+/, '', $str);
This will remove digits and spaces in any order from the beginning of the string. For something that removes only a number followed by spaces, see BoltClock's answer.
If the input strings all have the same ecpected format and you will receive the same result from left trimming all numbers and spaces (no matter the order of their occurrence at the front of the string), then you don't actually need to fire up the regex engine.
I love regex, but know not to use it unless it provides a valuable advantage over a non-regex technique. Regex is often slower than non-regex techniques.
Use ltrim() with a character mask that includes spaces and digits.
Code: (Demo)
var_export(
ltrim('420 911 90210 666 keep this part', ' 0..9')
);
Output:
'keep this part'
It wouldn't matter if the string started with a space either. ltrim() will greedily remove all instances of spaces or numbers from the start of the string intil it can't anymore.
Got some trouble with my preg_match.
The code.
$text = "tel: 012 213 123. mobil: 0303 11234 \n address: street 14";
$regex_string = '/(tel|Tel|TEL)[\s|:]+(.+)[\.|\n]/';
preg_match($regex_string , $text, $match);
And I get this result in $match[2]
"012 213 123. mobil: 023 123 123"
First question.
I want the regex to stop at the .(dot) but it doesent.
Can someone explain to why it isnt?
Second question.
preg_match uses () to get their match.
Is it possible to skip the parentheses surrounding the different "Tel" and still get the same functionality?
Thnx all stackoverflow is great :D
This should do:
/tel(?:\s|:)+([^.]+)(?:\.|$)/i
+ is a greedy quantifier, which means it'll match as many characters as possible.
To your second question: in this particular case you just need to use case-insensitive match (i flag). Generally, you could use (?:...) syntax, example of which you could see in the end match. Square brackets are used for character classes.
If you're simply trying to extract a phone number out of that line, and it's guaranteed to be 11 numbers, you could simply use this:
$text = 'tel: 012 213 123. mobil: 0303 11234';
$phone_number = substr(preg_replace('/[^\d]/', '', $text), 0, 11);`
With your example, $phone_number would be 0122131230.
How this works is any non-digit is replaced with an empty string, removing it, and then the first 11 numbers are returned.
No idea - your regex works for me (I get "012 213 123" in $match[2] with your code). The fact that the mobile phone differs between the two might indicate that it's not really the output of your code; check again.
Some other things - if you happen to have more dots in the line ("tel: xxx. phone: xxx. fax: xxx" for example), you will get bad results - use non-greedy operators ("get least chunk that matches" .*? instead of "get biggest chunk that matches" .*) or limit the repeated characters ("any number of non-periods" [^.]*). Also, you could spare yourself the trouble by making the regex case-insensitive (unless you really hate people typing "tEl").
Your other question: (?:stuff) will match "stuff" just like (stuff), but will not capture it.
Useful link: http://www.regular-expressions.info/
Why do you have pipes in your character classes [\.|\n] and [\s|:]? Character classes (stuff in square brackets []) are by definition like an OR relationship, so you don't need the pipe... unless you really are trying to match pipe |.
As for question #1, I'm not sure what's cusiong your problem, but usually this has to do with greedy quantifiers. The (.+) quantifier is greedy, so it matches as much as it can while still matching the entire pattern. Greedy quantifiers don't care what comes after them in the pattern. Since a period . matches any character other than new line characters, it can match a period, and so it does match a period. To make a quantifier non-greedy you can use a question mark ?.
For your second question In RegEx uses parenthesis to group things and to store them. If you want to group (tel|Tel|TEL) but not store it in $match you can put a ?: at after the open parenthesis:
(?:tel|Tel|TEL)
Do you mean you want to match only the number, so you don't have to strip off the tel: and the dot? Try this:
/tel[:\s]+\K[^.]+/i
The i makes it case-insensitive.
[:\s] matches a colon or whitespace (the | doesn't mean "or" in a character class, it just matches a |).
[^.]+ matches one or more non-dots; it stops matching when it sees a dot or the end of the line, so you don't have to match the dot if you don't want it in the result.
Finally, \K means "forget about whatever you've matched so far and pretend the match really started here"--a little gem of a feature that's only available in Perl and PHP (that I know of).