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
Related
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 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]+
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 :)
I'm having trouble using preg_match to find and replace a string. The string of interest is:
<span style="font-size:0.6em">EXPIRATION DATE: 04/30/2011</span>
I need to target and replace the date, "04/30/2011" with a different date. Can someone throw me a bone a give me the regular expression to match this pattern using preg_match in PHP? I also need it to match in such a way that it only replaces up to the first closing span and not closing span tags later in the code, e.g.:
<span style="font-size:0.6em">EXPIRATION DATE: 04/30/2011</span><span class="hello"></span>
I'm not versed in regex, and although I've spent the last hour trying to learn enough to make this work, I'm utterly failing. Thanks so much!
EDIT: As you can see this has gotten me exhausted. I did mean preg_replace, not preg_match.
If you're after a replacement, consider using preg_replace(), something like
preg_replace('#(\d{2})/(\d{2})/(\d{4})#', '<new date>', $string);
How about this:
$toBeFoundPattern = '/([0-9][0-9])\/([0-9][0-9])\/([0-9][0-9][0-9][0-9])/';
$toBeReplacedPattern = '$2.$1.$3';
$inString = '<span style="font-size:0.6em">EXPIRATION DATE: 04/30/2011</span>';
// Will convert from US date format 04/30/2011 to european format 30.04.2011
echo preg_replace( $toBeFoundPattern, $toBeReplacedPattern, $inString );
and prints
EXPIRATION DATE: 30.04.2011
Patterns always begin and end with identical so called delimiter characters. Often the character / is used.
$1 references the string, which matched the first string matched by ([0-9][0-9]), $2 references be (...) and $3 the four letters matched by the last (...).
[...] matched a single character, which is one of those listed inside the brackets. E.g. [a-z] matches all lower case letters.
To use the special meaning character / inside of a pattern, you need to escape it by \ to make it be the literal slash character.
Update: Using {..} as pointed out below is shorthand for repeated patterns.
Regex should be:
(0[1-9]|1[012])[- /.](0[1-9]|[12][0-9]|3[01])[- /.](19|20)\d\d
If you want to only match one instance, this is OK. For multiple instances, use preg_match_all instead. Taken from http://www.regular-expressions.info/regexbuddy/datemmddyyyy.html.
Edit: are you looking to just search and replace inside a PHP script or do you want to do some javascript live replacement?
I am having problems with RegEx in PHP and can't seem to find the answer.
I have a string, which is 3 letters, all caps ie COS.
the letters will change but always be 3 chars long and in caps, it will also be in the center of another string, surrounded by commas.
I need a regEx to find 3 caps letter inside a string and cahnge them from COS to 'COS'
(im doing this to amend a sql insert string)
I can't seem to find the regEx unless i use spercifit letter but the letters will change.
I need something along the lines of
[A-z]{3} then replace with '[A-Z]' (I know this isnt anywere near correct, just shorthand)
Anyone any suggestions?
Cheers
EDIT:
Just wanted to add incase anyone comes accross this question at a later date:
the sql insert string (provided from an external source and ftp's to my server daily)
contained the 3 capital string twice, once with commas and once with out
so I had to also remove the double commas added from the first regEx
$sqlString = preg_replace('/([A-Z]{3})/', "'$1'", $isqlString);
$sqlString = preg_replace('/\'\'([A-Z]{3})\'\'/', "'$1'", $sqlStringt);
Thanks everyone
You were actually very close. You could use:
echo preg_replace('/([A-Z]{3})/', "'$1'", 'COS'); //will output 'COS'
For MySQL statements I would advise to use the function mysql_real_escape_string() though.
$string = preg_replace('/([A-Z]{3})/', "'$1'", $string);
http://php.net/manual/en/function.preg-replace.php
Assuming it's like you said, "three capital letters surrounded by commas, e.g.
Foo bar,COS,Foo Bar
You can use look-ahead and look-behinds and find the letters:
(?<=,)([A-Z]{3})(?=,)
Then a simple replace to surround with single quotes will be adequate:
'$1'
All together, Here's it working.
preg_replace('/(^|\b)([A-Z]{3})(\b|$)/', "'${2}'", $string);