I want a regex solution to allow only
http://www.imdb.com/title/ttANYNumberWOrdetc/ links
Otherwise SHOW us error.. Incorrect link
I am not too good with regex
I just create this petren ..
preg_match('/http:\/\/www.imdb.com\/title\/(.*)\//is', 'http://www.imdb.com/title/tt0087469/', $result);
Its show me corect result but i think i missed some thing..
Thanks,
How about something like this: http://(?:www\.)?imdb.com/title/tt[^/]+/.
Example:
<?php
if ( preg_match('#^http://(?:www\.)?imdb\.com/title/tt[^/]+/$#', 'http://www.imdb.com/title/tt0448303/') )
echo 'Matches' . PHP_EOL;
Explanation:
The regular expression matches a string that starts with http:// followed either by imdb.com or www.imdb.com, then /title/tt followed by any character except for a / and that ends with a /.
The # is the delimiter, the ^ indicated the beginning of the string and the $ the end.
This should work:
if (preg_match("#^(http://www.|https://www.)imdb.com/title/tt([a-zA-Z0-9]+)(?:/)(?:^[a-zA-Z0-9]+)?$#s", 'http://www.imdb.com/title/tt0364845/', $matches)) {
echo 'yay';
} else {
echo 'nay';
}
Related
I want to check my link in a website, but I also want to check is it visible. I wrote this code:
$content = file_get_contents('tmp/test.html');
$pattern = '/<a\shref="http:\/\/mywebsite.com(.*)">(.*)<\/a>/siU';
$matches = [];
if(preg_match($pattern, $content, $matches)) {
$link = $matches[0];
$displayPattern = '/display(.?):(.?)none/si';
if(preg_match($displayPattern, $link)) {
echo 'not visible';
} else {
echo 'visible';
}
} else {
echo 'not found the link';
}
It works, but not perfect. If the link is like this:
<a class="sg" href="http://mywebsite.com">mywebsite.com</a>
the fist pattern won't work, but if I change the \s to (.*) it gives back string from the first a tag. The second problem is the two pattern. Is there any way to merge the first with negation of the second? The merged pattern has 2 results: visible or not found/invisible.
I'll try to guess.
You are having a problem if your code(one that you fetch with file_get_contents) looks like this
<a class="sg" href="http://mywebsite.com">mywebsite.com</a>
.
.
.
mywebsite.com
Your regex will return everything from first </a> tag because dot matches a new line(I guess you need it turned on, but if you dont, its 's' flag, so remove it)
Therefore
.*
will keep searching everything, so you need to make it greedy
(when its greedy it will stop searching once it finds what its looking for), like this
.*?
Your regex should look like this then
<a.*?href="http:\/\/mywebsite.com(.*?)">(.*?)<\/a>
A column in my spreadsheet contains data like this:
5020203010101/FIS/CASH FUND/SBG091241212
I need to extract the last part of string after forwward slash (/) i.e; SBG091241212
I tried the following regular expression but it does not seem to work:
\/.*$
Any Idea?
Try this:
'/(?<=\/)[^\/]*$/'
The reason your current REGEXP is failing is because your .* directive matches slashes too, so it anchors to the first slash and gives you everything after it (FIS/CASH FUND/SBG091241212).
You need to specify a matching group using brackets in order to extract content.
preg_match("/\/([^\/]+)$/", "5020203010101/FIS/CASH FUND/SBG091241212", $matches);
echo $matches[1];
You could do it like this without reg ex:
<?php
echo end(explode('/', '5020203010101/FIS/CASH FUND/SBG091241212'));
?>
this will do a positive lookbehind and match upto a value which does not contain a slash
like this
[^\/]*?(?<=[^\/])$
this will only highlight the match . i.e. the last part of the url
demo here : http://regex101.com/r/pF8pS2
Make use of substr() with strrpos() as a look behind.
echo substr($str,strrpos($str,'/')+1); //"prints" SBG091241212
Demo
You can 'explode' the string:
$temp = explode('/',$input);
if (!empty($temp)){
$myString = $temp[count($temp)-1];
}
You can also use:
$string = '5020203010101/FIS/CASH FUND/SBG091241212';
echo basename($string);
http://www.php.net/manual/en/function.basename.php
I don't know almost anything about regex, but I need to allow only one space between letters, this is what I have done with another questiona and answer, plus random tries with regex101:
/^(\d){1,}(\:[A-Za-z0-9-]+([ a-zA-Z0-9-]+)?)?\:(\d){1,}(?:\.(\d){1,2})?$/m
The fomat should be:
[integer]:[optional label :][integer/decimal]
Example:
12:aaa:12.31
56:a s f:15
34:45.8
I have done some random tries without any success,I'm only able to allow infinite space, could someone help me? I have also looked at others answers, but I couldn't implement in my regex.
Check if error:
preg_match_all('/^(\d+)(:[A-Z0-9-]+(?: [A-Z0-9-]+)*)?:(\d+(?:\.\d{1,2})?)$/mi', $_POST['ratetable'], $out);
if($out[0]!=explode("\n",$_POST['ratetable'])){
header('Content-Type: application/json; charset=utf-8');
echo json_encode(array(0=>'Invalid price table at line: '.implode(", ", array_diff_key(array_flip(explode("\n",$_POST['ratetable'])),array_flip($out[0])))));
exit();
}
You could make the regex a bit shorter and allow a single space in between each letter like this:
/^(\d+)(:[A-Z0-9-]+(?: [A-Z0-9-]+)*)?:(\d+(?:\.\d{1,2})?)$/gmi
regex101 demo
If you want to capture only the label (and exclude the :) in the capture, you can use this one.
I think
^\d+:([^:]+:)?\d+(\.\d+)?$
might do it...
It means: <int> + ":" + [label + ":"] + <float> where label can be anything but :.
You're nearly there:
(?: [a-zA-Z0-9-]+)*
the above would mean: space followed by at least one in the character group; the latter group any number of times, without capturing
You can use preg_match
$str = 'Another Example String 1_2-3';
echo $str . "<br />\n";
if (preg_match('#^(?:[\w-]+ ?)+$#', $str)){
echo 'Legal format!';
} else {
echo 'Illegal format!';
}
^\d+:([\w\d]+ ?)*:?\d+(\.\d+)?$
Live demo
I have a regex that validate a specific url but it not really working. I want to validate urls like this -----> https : // example.co.nz/#![RANDOM_KEYS_HERE].
I want to do it only with https. Most importantly, the input of the user need to match https : // example.co.nz/#! but after the #!, the user can put anything he like.
Here is the code:
I know that the code is fked up xD I have a basic knowledge in that lol
#^https://example+\.[co\.nz][a-z0-9-_.]+\.[a-z]{2,4}#i
If anyone could help me to do it, it would be great! thanks!
Erm... not even close. Your regex reads as follows:
Starting from the beginning of the string...
Match literally https://exampl
Match one or more e
Match a literal .
Match one of any of these: cnoz.
Match one or more of these: a-z0-9-_.
Match a literal .
Match between 2 and 4 letters
This is nothing like what you're looking for. After all, I don't think you want this to pass:
https://exampleeeeeeeeeeee.complete.and.total.failure.-_-.lol
Instead, try this:
(^https://example\.co\.nz/#!(.*))
This regex reads as follows:
Starting from the beginning of the string...
Match literally https://example.co.nz/#!
Capture everything thereafter
Try this out:
^https:\/\/example\.co\.nz\/\#\!(.*)$
The parentheses at the end will do a sub-expression match which should allow you to pull out the ID.
if (preg_match('/^https:\/\/example\.co\.nz\/\#\!(.*)$/', $searchString, $matches)) {
$id = $matches[1];
}
if (preg_match('%^https://example\.co\.nz/#!(.+)$%i', $subject)) {
# Successful match
} else {
# Match attempt failed
}
Or you can get your [RANDOM_KEYS_HERE] part with this one
if (preg_match('%^https://example\.co\.nz/#!(.+)$%i', $subject, $regs)) {
$result = $regs[0];
} else {
$result = "";
}
You don't need regexp there. You just need to find out if string starts with some substring. Check this:
if(strpos($url, 'https://example.co.nz/#!')===0)
{
echo 'url is OK';
}
else
{
echo 'url is wrong';
}
http://www.php.net/manual/en/function.strpos.php
I hope this helps.
My current regular expression should be correct, though I wouldn't expect so, it doesn't work properly. It won't return "Got Match"
My currrent code is as follows:
$id = "http://steamcommunity.com/id/TestID";
if (preg_match("^http://steamcommunity\.com/id/.*?\n$", $id)) {
print "Got match!\n";
}
You're missing delimiters on your regex:
if (preg_match("#^http://steamcommunity\.com/id/.*?\n$#", $id)) {
^--here ^--here
Note that I've used # as the delimiter here, since that saves you having to escape all of the internal / charrs, if you'd used the traditional / as the delimiter.
You need a delimiter, like this:
if (preg_match("#^http://steamcommunity\.com/id/.*?$#", $id)) {
^ ^
And what's with the newline at the end? Surely you don't need that.
You're missing delimiters. For example:
"#^http://steamcommunity\.com/id/.*?\n$#"
Also, you're trying to match a newline (\n) that isn't in your string.
You need to add the pattern delimiter:
$id = "http://steamcommunity.com/id/TestID";
if (preg_match("#^http://steamcommunity\.com/id/.*?(\n|$)#", $id)) {
print "Got match!\n";
}
There are a couple of things that are wrong with it. First of all, you need to delimit the start and end of your regex with a character. I used #. You're also matching for a new line at the end of your regex, which you don't have and likely won't ever have in your string.
<?php
$id = "http://steamcommunity.com/id/TestID";
if (preg_match("#^http://steamcommunity\.com/id/.*?$#", $id)) {
print "Got match!\n";
}
?>
http://codepad.viper-7.com/L7XctT
First of all, your regex shouldn't even compile because it's missing delimiters.
if (preg_match("~^http://steamcommunity\.com/id/.*?\n$~", $id)) {
^---- these guys here -----^
Second of all, why do you have a \n if your string doesn't contain a new line?
And finally, why are you using regex at all? Effectively, you are just trying to match a constant string. This should be equivalent to what you are trying to match:
if (strpos($id, 'http://steamcommunity.com/id/') === 0) {
You need to have starting and ending delimiter in your pattern like /pattern/ or #pattern# or with brackets (pattern). Why is that? To have some pattern modifiers after ending delimiter like #pattern#i (ignore case)
preg_match('(^http://steamcommunity\.com/id/.*?\n$)', $id)
As the say your patten is start and end wrong. (Delimiter)
But this will be a better match of a 64-bit Steam ID. (Minimum 17 and Maximum 25 numbers)
if( preg_match("#^http://steamcommunity\.com/id/([0-9]{17,25})#i", $id, $matches) )
{
echo "Got match! - ".$matches;
}
I believe that there is no need for you to require that the string must end with a line break.
Explanation.
http://steamcommunity\.com/id/([0-9]{17,25})
^--- string ---^^-- Regexp --^
[0-9] - Match a number between 0 to 9
{17,25} - Make 17 to 25 matches
() - Returns match
Or use pattern as those (It is the same):
/^http:\/\/steamcommunity\.com\/id\/([0-9]{17,25})/i
(^http://steamcommunity\.com/id/([0-9]{17,25}))i
Regular Expressions PHP Tutorial
Online regular expression testing <- Dont use delimiter.
<?php
# URL that generated this code:
# http://txt2re.com/index-php.php3?s=http://steamcommunity.com/id&-1
$txt='http://steamcommunity.com/id';
$re1='(http:\\/\\/steamcommunity\\.com\\/id)'; # HTTP URL 1
if ($c=preg_match_all ("/".$re1."/is", $txt, $matches))
{
$httpurl1=$matches[1][0];
print "($httpurl1) \n";
}
#-----
# Paste the code into a new php file. Then in Unix:
# $ php x.php
#-----
?>
Resorces:
http://txt2re.com/index.php3?s=http://steamcommunity.com/id&-1