I have a text to validate using regex,
Text field is allowed to have all the characters a-z and 0-9 except for alphabets (i,o,q).
I tried something like this but cannot get it to work '/[^(oOiIqQ)]/'
A simple way for exclusions like this is to use negative lookahead. State what you want:
/^(?:[a-z0-9])+\z/i
Then exclude the items you don't want:
/^(?:(?![ioq])[a-z0-9])+\z/i
You cannot use parenthesis in [ ... ].
You have to use something like '/[0-9a-hj-npr-zA-HJ-NPR-Z]/'
If you want to be sure your text only has those characters, use:
'/^[0-9a-hj-npr-zA-HJ-NPR-Z]+$/'
So you can match a string containing any number of those characters, and only those.
Maybe someting like this /^[0-9a-hj-npr-z A-HJ-NPR-X]+$/
I would assume a little change to your's and would try:
^[^oOiIqQ]+$
This might works: [a-hj-npr-z] Maybe you can add the flag i at the end of your regexp for case insensibility.
(yours will allow EVERY characters except those you specified)
if (preg_match('#^[0-9a-hj-npr-z]+$#i', $string)) {
//string OK
}
There is a very simple solution for this which takes in consideration negative regexing (which makes the regex shorter and much readable)
[^ioqIOQ]+
Related
I want to find all strings looking like [!plugin=tesplugin arg=dfd arg=2!] and put them in array.
Important feature: the string could contain arg=uments or NOT(in some cases). and of course there could be any number of arg's. So the string could look like:
[!plugin=myname!] or [!plugin=whatever1 arg=22!] or even [!plugin=gal-one arg=1 arg=text arg=tx99!]. I need to put them all in $strarray items
Here is what i did...
$inp = "[!plugin=tesplugin arg=dfd!] sometxt [!plugin=second arg=1 arg=2!] 1sd";
preg_match_all('/\[!plugin=[a-z0-9 -_=]*!]/i', $inp, $str);
but $str[0][0] contains:
[!plugin=tesplugin arg=dfd!] sometxt [!plugin=second arg=1 arg=2!]
instead of putting each expression in a new array item..
I think my problem in regex.. but can't find one. Plz help...
The last ] needs to be escaped and the - in the character class needs to be at the start, end, or escaped. As is it is a range of ascii characters between a space and underscore.
\[!plugin=[a-z0-9 \-_=]*!\]
Regex101 Demo: https://regex101.com/r/zV4bO2/1
I can't understand a simple regex that I need for a control over a preg_match statment:
-I need that every words with even white space between, and accent and apostrophe are allowed
so something like: sjsjsjjsjsjs òòòò èèèèè ddddd ''' eerfk jefrkj sdc should be accepted
so i write something like: [a-zA-Z\xE0\xE8\xE9\xF9\xF2\xEC\x27\s]*
that take everything that is letters and some special HEX code for accent and apostrophe, but i can't understand how to concatenate the sentence:
[^\r\n]
I'd like to reject anything if there is an end of line or a return statement. The puntaction too but it seem to be allreafy solved with my regex
so something like:
adjnasdnjadsija adokasmdoasdmoa admoadsoasodoas END
sddaadsasd òòò
should be accepted until the words END
Is it the right code? i made several test but no result!
I test my regex over http://regex101.com/
Set the locale appropriately:
setlocale(LC_ALL,"it_IT");
Now you can use a much simpler regex:
/[\w\s]*/
This is because \w is locale-aware ^_^
This is the text sample:
$text = "asd dasjfd fdsfsd http://11111.com/asdasd/?s=423%423%2F gfsdf http://22222.com/asdasd/?s=423%423%2F
asdfggasd http://3333333.com/asdasd/?s=423%423%2F";
This is my regex pattern:
preg_match_all( "#http:\/\/(.*?)[\s|\n]#is", $text, $m );
That match the first two urls, but how do I match the last one? I tried adding [\s|\n|$] but that will also only match the first two urls.
Don't try to match \n (there's no line break after all!) and instead use $ (which will match to the end of the string).
Edit:
I'd love to hear why my initial idea doesn't work, so in case you know it, let me know. I'd guess because [] tries to match one character, while end of line isn't one? :)
This one will work:
preg_match_all('#http://(\S+)#is', $text, $m);
Note that you don't have to escape the / due to them not being the delimiting character, but you'd have to escape the \ as you're using double quotes (so the string is parsed). Instead I used single quotes for this.
I'm not familar with PHP, so I don't have the exact syntax, but maybe this will give you something to try. the [] means a character class so |$ will literally look for a $. I think what you'll need is another look ahead so something like this:
#http:\/\/(.*)(?=(\s|$))
I apologize if this is way off, but maybe it will give you another angle to try.
See What is the best regular expression to check if a string is a valid URL?
It has some very long regular expressions that will match all urls.
I need a regular expression for php that outputs everything between <!--:en--> and <!--:-->.
So for <!--:en-->STRING<!--:--> it would output just STRING.
EDIT: oh and the following <!--:--> nedds to be the first one after <!--:en--> becouse there are more in the text..
The one you want is actually not too complicated:
/<!--:en-->(.*?)<!--:-->/gi
Your matches will be in capture group 1.
Explanation:
The .*? is a lazy quantifier. Basically, it means "keep matching until you find the shortest string that will still fit this pattern." This is what will cause the matching to stop at the first instance of <!--:-->, rather than sucking up everything until the last <!--:--> in the document.
Usage is something like preg_match("/<!--:en-->(.*?)<!--:-->/gi", $input) if I recall my PHP correctly.
If you have just that input
$input = '<!--:en-->STRING<!--:-->';
You can try with
$output = strip_tags($input);
Try:
^< !--:en-- >(.*)< !--:-- >$
I don't think any of the other characters need to be escaped.
<!--:en--\b[^>]*>(.*?)<!--:-->
This will match the things between your tags. This will break if you nest your tags, but you didnt say you were doing that :)
Here is the subject:
http://www.mysite.com/files/get/937IPiztQG/the-blah-blah-text-i-dont-need.mov
What I need using regex is only the bit before the last / (including that last / too)
The 937IPiztQG string may change; it will contain a-z A-Z 0-9 - _
Here's what I tried:
$code = strstr($url, '/http:\/\/www\.mysite\.com\/files\/get\/([A-Za-z0-9]+)./');
EDIT: I need to use regex because I don't actually know the URL. I have string like this...
a song
more text
oh and here goes some more blah blah
I need it to read that string and cut off filename part of the URLs.
You really don't need a regexp here. Here is a simple solution:
echo basename(dirname('http://www.mysite.com/files/get/937IPiztQG/the-blah-blah-text-i-dont-need.mov'));
// echoes "937IPiztQG"
Also, I'd like to quote Jamie Zawinski:
"Some people, when confronted with a problem, think 'I know, I'll use regular expressions.' Now they have two problems."
This seems far too simple to use regex. Use something similar to strrpos to look for the last occurrence of the '/' character, and then use substr to trim the string.
/http:\/\/www.mysite.com\/files\/get\/([^/]+)\/
How about something like this? Which should capture anything that's not a /, 1 or more times before a /.
The greediness of regexp will assure this works fine ^.*/
The strstr() function does not use a regular expression for any of its arguments it's the wrong function for regex replacement.
Are you thinking of preg_replace()?
But a function like basename() would be more appropriate.
Try this
$ok=preg_match('#mysite\.com/files/get/([^/]*)#i',$url,$m);
if($ok) $code=$m[1];
Then give a good read to these pages
http://www.php.net/preg_match
preg_replace
Note
the use of "#" as a delimiter to avoid getting trapped into escaping too many "/"
the "i" flag making match insensitive
(allowing more liberal spellings of the MySite.com domain name)
the $m array of captured results