If I have string like this "location":"Denton TX" (including the double quotes)
I'd like to get just Denton TX. How can I write regular expression for that thing?
I tried
function getInfo($info,$content){
echo 'getting INFO';
preg_match_all("\.".$info."\.:\"[^\"]*(.*?)\.,",$content,$matches,PREG_PATTERN_ORDER);
print_r($matches);
return $matches[0];
}
and I put 'location' into $info but it doesn't work.
Simple get every character between quotes. You was too close.
"\.".$info."\.:\"([^\"]*)\"
I'd go with this:
"\"".$info."\":\"([^\"]+)\""
I'm not sure why you wrote this particular bit, but the main problem in your regex is the [^\"]*, which is too greedy and will consume everything, and leave (.*?) empty (well, unless is starts with a double quote, I guess). I also don't really understand why you what a comma in the end, but maybe your example wasn't descrptive... In any case, the regex I provided should match your example, provided $info contains the string location.
Try this: (?<=:")(.*)(?=")
I use a similar one to remove element tags from an XML response in an Android app. Good luck
Related
I have the following content in a string (query from the DB), example:
$fulltext = "Thank you so much, {gallery}art-by-stephen{/gallery}. As you know I fell in love with it from the moment I saw it and I couldn’t wait to have it in my home!"
So I only want to extract what it is between the {gallery} tags, I'm doing the following but it does not work:
$regexPatternGallery= '{gallery}([^"]*){/gallery}';
preg_match($regexPatternGallery, $fulltext, $matchesGallery);
if (!empty($matchesGallery[1])) {
echo ('<p>matchesGallery: '.$matchesGallery[1].'</p>');
}
Any suggestions?
Try this:
$regexPatternGallery= '/\{gallery\}(.*)\{\/gallery\}/';
You need to escape / and { with a \ before it. And you where missing start and end / of the pattern.
http://www.phpliveregex.com/p/fn1
Similar to Andreas answer but differ in ([^"]*?)
$regexPatternGallery= '/\{gallery\}([^"]*?)\{\/gallery\}/';
Don't forget to put / at the beginning and the end of the Regex string. That's a must in PHP, different from other programming languages.
{,},/ are characters that can be confused as a Regex logic, so you have to escape it using \ like \{.
Use ? to make the string to non-greedy, thus saves memory. It avoids error when facing this kind of string "blabla {galery}you should only get this{/gallery} but you also got this instead.{/gallery} Rarely happens but be careful anyway".
Try this RegEx:
\{gallery\}(.*?)\{\/gallery\}
The problem with your RegEx was that you did not escape the / in the closing {gallery}. You also need to escape { and }.
You should use .*? for a lazy match, otherwise if there are 2 tags in one string, it will combine them. I.e. {gallery}by-joe{/gallery} and {gallery}by-tim{/gallery} would end up as:
by-joe{/gallery} and {gallery}by-tim
However, using a lazy match, you would get 2 results:
by-joe
by-tim
Live Demo on Regex101
all,
I am using preg_match to filter some data, and it is strange that, it dose not work correctly. I am new to regex, and I used a php live regex website to check my regex, which works correctly. So I have no idea what is wrong here.
I would like to have preg_match to find something like "a\_b" in the $string:
$string="aaa\_bbb:ccc"
if(preg_match("/[a-zA-Z]\\_[a-zA-Z]/", $string)){
$snew = str_replace('\_', "_", $string);
}
But it is strange that even I have a $string like in this example above, the result of preg_match is 0. But when I change it to
preg_match("/\\_[a-zA-Z]/", $string)
It works fine and return 1. But of course that is not what I want. Any idea?
Thanks very much~
You don't really need the preg_match at all, from what I can see.
However the problem you're having with it is to do with escaping.
You have this: "/[a-zA-Z]\\_[a-zA-Z]/"
You've correctly identified that the backslash needs to be escaped, however, you've missed a subtle issue:
Regular expressions in PHP are strings. This means that you need to escape it as a string as well as a regular expression. In effect, this means that to correctly escape a backslash so it is matched as an actual backslash character in your pattern, you actually need to have four backslashes.
"/[a-zA-Z]\\\\_[a-zA-Z]/"
It's not pretty, but that's how it is.
Hope that helps.
use:
if(preg_match("/[a-zA-Z]\\\\_[a-zA-Z]/", $string))
instead
You don't need the preg_match altogether, instead just do a replace using this regex:
/([a-zA-Z])\\\\_([a-zA-Z])/
and then replace with $1_$2, like this:
$result = preg_replace("/([a-zA-Z])\\\\_([a-zA-Z])/", "$1_$2$, $string);
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.
In PHP, I need to be able to figure out if a string contains a URL. If there is a URL, I need to isolate it as another separate string.
For example: "SESAC showin the Love! http://twitpic.com/1uk7fi"
I need to be able to isolate the URL in that string into a new string. At the same time the URL needs to be kept intact in the original string. Follow?
I know this is probably really simple but it's killing me.
Something like
preg_match('/[a-zA-Z]+:\/\/[0-9a-zA-Z;.\/?:#=_#&%~,+$]+/', $string, $matches);
$matches[0] will hold the result.
(Note: this regex is certainly not RFC compliant; it may fetch malformed (per the spec) URLs. See http://www.faqs.org/rfcs/rfc1738.html).
this doesn't account for dashes -. needed to add -
preg_match('/[a-zA-Z]+:\/\/[0-9a-zA-Z;.\/\-?:#=_#&%~,+$]+/', $_POST['string'], $matches);
URLs can't contain spaces, so...
\b(?:https?|ftp)://\S+
Should match any URL-like thing in a string.
The above is the pure regex. PHP preg_* and string escaping rules apply before you can use it.
$test = "SESAC showin the Love! http://twitpic.com/1uk7fi";
$myURL= strstr ($test, "http");
echo $myURL; // prints http://twitpic.com/1uk7fi
I'm new to preg_replace() and I've been trying to get this to work, I couldn't so StackOverflow is my last chance.
I have a string with a few of these:
('pm_IDHERE', 'NameHere');">
I want it to be replaced with nothing, so it would require 2 wildcards for NameHere and pm_IDHERE.
But I've tried it and failed myself, so could someone give me the right code please, and thanks :)
Update:
You are almost there, you just have to make the replacement an empty string and escape the parenthesis properly, otherwise they will be treated as capture group (which you don't need btw):
$str = preg_replace("#\('pm_.+?', '.*?'\);#si", "", $str);
You probably also don't need the modifiers s and i but that is up to you.
Old answer:
Probably str_replace() is sufficient:
$str = "Some string that contains pm_IDHERE and NameHere";
$str = str_replace(array('pm_IDHERE', 'NameHere'), '', $str);
If this is not what you mean and pm_IDHERE is actually something like pm_1564 then yes, you probably need regular expressions for that. But if NameHere has no actual pattern or structure, you cannot replace it with regular expression.
And you definitely have to explain better what kind of string you have and what kind of string you have want to replace.