Unexpected ] error in simple preg replace script [duplicate] - php

This question already has answers here:
preg_match() Unknown modifier '[' help
(2 answers)
Closed 8 years ago.
I have a script that downloads the latest newsletter from a group inbox on a spare touchscreen in our office. It works fine, but people keep accidentally unsubscribing us so I want to hide the unsubscribe link from the email.
$preg_replace seems like it would work because I can set up a pattern that simply removes any link withthe word "unsubscribe" in. I validated the pattern below using the tool at http://regex101.com/ , and it even picks up variations like "manage subscription" as well. It is ok if the odd legitimate link with the word subscribe also get removed - there won't be many and it's only for internal use.
However, when I execute I get an error.
Here's my code:
line 53: $pat='<\s*(a|A)\s+[^>]*>[^<>]*ubscri[^<>]*<\s*\/(a|A)\s*>';
line 54: $themail[bodycontent]= preg_replace($pat, ' ',$themail[bodycontent]);
and I get this error:
preg_replace() [function.preg-replace]: Unknown modifier ']' in /home/trev/public_html/bigscreen/screen-functions.php on line 54
It must be something really simple like an unescaped char but I have gone code blind and can't for the life of me see it.
How do I get this pattern:
<\s*(a|A)\s+[^>]*>[^<>]*ubscri[^<>]*<\s*\/(a|A)\s*>
to run in a simple php script?
Thanks

You haven't used any delimiters so it's treating the < character as the delimiter
Try something like this instead
$pat='#<\s*(a|A)\s+[^>]*>[^<>]*ubscri[^<>]*<\s*\/(a|A)\s*>#';

You have no delimiter. Or rather you do, but it's not the one you meant. PCRE is interpreting your first < as the opening delimiter (you can use matching brackets as delimiters - in fact, I use parentheses to help remind myself that the entire match is index 0). Then it sees the first > as the ending delimiter. Anything after that should be a modifier, but of course ] is not a modifier.
Wrap your regex with (...) to give it a proper set of delimiters.

$themail[bodycontent] should be either $themail['bodycontent'] or $themail[$bodycontent].
It's trying to parse bodycontent] ... as the array index.

Patterns used in preg_match need to be enclosed by a pair of delimiter characters.
For example, a / or a ~ at the start and end of the string.
Anything outside of these delimiters at the end of the string is considered to be a regex "modifier".
Your example doesn't have delimiters, so PHP is wrongly assuming that the < character is the delimiter. It therefore sees the next < character as the closing delimiter, and therefore, anything after that as a modifier. Obviously all that stuff is supposed to be inside the pattern and isn't valid as modifiers, which is why PHP is complaining.
Solution: Add a pair of modifier characters:
$pat='~<\s*(a|A)\s+[^>]*>[^<>]*ubscri[^<>]*<\s*\/(a|A)\s*>~';
^ ^
add this ...and this
(it doesn't have to be ~, you can choose your own modifier character to suit your needs. Best one to use is one that doesn't occur in your string (although you can escape it if it does)

Starting and ending of pattern with slash /
$pat='/<\s*(a|A)\s+[^>]*>[^<>]*ubscri[^<>]*<\s*\/(a|A)\s*>/';

Related

PHP - Comment System "Replace Http// urls" [duplicate]

This question already has answers here:
How do I replace certain parts of my string?
(5 answers)
Closed 2 years ago.
I'm creating a simple comment system connected by Steam API. Every Steam user connected in my website can automatically post things. But i'm changing some functions to replace things like the URLs.
My question is: When a user post something like,
"Hello I'm nice, have a look at http://www.cute.com"
Automatically replaces the http:// for the link without changing the http:// in the string.
Maybe something like this?
<?php
$str = "helloo im nice, have a look http://www.cute.com";
echo preg_replace("/http:\/\/(.+)\.(.+)\.(.+)/", "<a href='http://$1.$2.$3'>$1.$2.$3</a>", $str);
?>
This will convert any link into an anchor (or an a tag).
Alternative added
Alternatively, it might be a good idea to add support for https as well. In which case the following might be useful.
<?php
$str = "helloo im nice, have a look http://www.cute.com";
echo preg_replace("/http(s?):\/\/(.+)\.(.+)\.(.+)/", "<a href='http$1://$2.$3.$4'>http$1://$2.$3.$4</a>", $str);
?>
This takes advantage of the ? modifier which means "one or more of the preceding character". In this case it is the "s" character since it is "http" and "https" both match.
Explanation
This uses RegEx (or Regular Expressions) to create this.
The first parameter of the preg_replace function takes the RegEx (I like to test mine here: http://regexr.com/).
All RegExs must start and end with a forward slash. The bits inbetween are as follows.
http: is simply selecting a string that starts with "http:"
\/\/ is called "escaping" and that will select two forward slashes. Since forward slashes are special characters used in RegEx (start and end of a statement) they need to be escaped so that PHP doesn't think the RegEx has ended sooner.
(.+) The brackets are also special characters (though not escaped) and they are known as "capture groups". What this is used for is so that I can see what is between the "http://" and the ".com" (or whatever extension is used). The full stop (or period or ".") character selects anything.
\. Further on the escaping. Since full stop is used as a special character, we have to escape this one. What that means so far is that we are selecting "http://" then anything and then stopping at a full stop.
(.+) Last but not least is the final capture group. This, again selects anything from the string so that have our final capture group and RegEx complete.
Modifiers:
? means "one or more of the preceding character". This means that /tests?/ would match test and tests since s is the preceding character and in the first example we have 0 and in the second there is 1
+ means "one of more of the preceding character". In this case we are saying one of more of anything which means we expect at least one character to be provided.
The second parameter is our replace part.
In short, the $1 and $2 sections are to reference the two brackets from the above RegEx.
Some further reading
The PHP function I used
More information on Regular Expressions
RegEx capture groups
$string = 'helloo im nice, have a look http://www.cute.com';
$string = str_replace('http://', '', $string);
echo $string;

Trying To Find Forward Slash In Preg_Match

I've been searching for hours trying to find a solution to this. I am trying to determing if the REQUEST URI is legit and break it down from there.
$samplerequesturi = "/variable/12345678910";
To determine if it is legit, the first section variable is only letters and is variable in length. The second section is numbers, which should have 11 total. My problem is escaping the forward slash so it is matched in the uri. I've tried:
preg_match("/^[\/]{1}[a-z][\/]{1}[0-9]{11}+$/", $samplerequesturi)
preg_match("/^[\\/]{1}[a-z][\\/]{1}[0-9]{11}+$/", $samplerequesturi)
preg_match("/^#/#{1}[a-z]#/#{1}[0-9]{11}+$/", $samplerequesturi)
preg_match("/^|/|{1}[a-z]|/|{1}[0-9]{11}+$/", $samplerequesturi)
Among others which I can't remember now.
The request usually errors out:
preg_match(): Unknown modifier '|'
preg_match(): Unknown modifier '#'
preg_match(): Unknown modifier '['
Edit:
I guess I should state that the REQUEST URI is already known. I'm trying to prove the whole string to make sure it isn't a bogus string ie to make sure there the 1st set is only lower case letters, and the 2nd set is only 11 numbers.
/ is not the only thing you can use as a delimiter. In fact, you can use almost any non-slphanumeric character. Personally I like to use () because it reminds me that the first item of the result array is the entire match and it also never needs escaping in the pattern.
preg_match("(^/([a-z]+)/(\d+)$)i",$samplerequesturi,$out);
var_dump($out);
That should do it.
If you want to use regex (which I don't think is necessary in this case, simply splitting on "/" should be fine:
$samplerequesturi = "/variable/12345678910";
preg_match("#^/([A-Za-z]+)/(\d+)$#", $samplerequesturi, $out);
echo $out[1];
echo $out[2];
should get you going
Your problem may be that you are using the / forward-slash as a regex delimiter (at the start and end of the regex expression). Switch to using a character other than the forward-slash, such as a # hash symbol or any other symbol which will never need to appear in this particular expression. Then you won't need to escape the forward-slash character at all in the expression.

regular expression to validate URL not working correctly in PHP

I am using a regular expression to validate URL. This expression works very well in JavaScript, But in PHP it gives me this error
A PHP Error was encountered
Severity: Warning
Message: preg_match() [function.preg-match]: Unknown modifier '('
Filename: home/auth.php
Line Number: 1596
A PHP Error was encountered
Severity: Warning
Message: preg_match() [function.preg-match]: Unknown modifier '('
Filename: home/auth.php
Line Number: 1601
This is my expression
$pattern ="/^(http|https|ftp)\:\/\/www\.([a-zA-Z0-9\.\-]+(\:[a-zA-Z0-9\.&%\$\-]+)*#)*(\.){1}((25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9])\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9]|0)\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9]|0)\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[0-9])|([a-zA-Z0-9\-]+\.)*[a-zA-Z0-9\-]+\.(com|edu|gov|int|mil|net|org|biz|arpa|info|name|pro|aero|coop|museum|[a-zA-Z]{2}))(\:[0-9]+)*(/($|[a-zA-Z0-9\.\,\?\'\\\+&%\$#\=~_\-]+))*$/";
This is the php function
public function valid_url($data)
{
$data = trim($data);
if(!$data)
{
return TRUE;
}
$pattern ="/^(http|https|ftp)\:\/\/www\.([a-zA-Z0-9\.\-]+(\:[a-zA-Z0-9\.&%\$\-]+)*#)*(\.){1}((25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9])\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9]|0)\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9]|0)\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[0-9])|([a-zA-Z0-9\-]+\.)*[a-zA-Z0-9\-]+\.(com|edu|gov|int|mil|net|org|biz|arpa|info|name|pro|aero|coop|museum|[a-zA-Z]{2}))(\:[0-9]+)*(/($|[a-zA-Z0-9\.\,\?\'\\\+&%\$#\=~_\-]+))*$/";
$valid = preg_match($pattern,$data);
if(!$valid)
{
$data = "http://".$data;
$valid = preg_match($pattern,$data);
}
if(!$valid)
{
$this->form_validation->set_message('valid_url', 'Please enter a valid URL.');
return FALSE;
}
else
{
return TRUE;
}
}
I am not very good at regular expressions so I could not figure out the issue, please help me correct the regular expression.
Wow, that is a big expression. I found several faults in it, and I shall hopefully explain them to you. Let's break it apart:
$pattern ="/
Here was your first mistake. As a forward slash is used in multiple sections of a url, you should use a different delimiter. I would suggest a tilde ~, as this is not used in a url very often. This would mean you don't have to keep escaping the forward slash every where with \/.
^(http|https|ftp)\:\/\/www\.([a-zA-Z0-9\.\-]+
This character class contains the next error. Within a character class, a dot just means a dot. There is no need to escape it. Furthermore, with placing the dash at the end, it also does not need escaping as it cannot possibly mean a range. The character class can be shortened to become [a-zA-Z0-9.-]+.
(\:[a-zA-Z0-9\.&%\$\-]+
Here we have the next error, & within the character class. This will match an & or an a or an m or a ;, not just an &. You don't need to convert it to the html code as doing so will mean to match any of the characters that the code contains. And using the previous knowledge, you don't need to escape the dot, or the dash if it is at the end. You also don't need to escape the dollar sign, as in a character class it just means a dollar. Remember, within a character class, all meta characters are just standard characters except the caret ^, the backslash \, the closing square bracket ], the dash - (but this can be left if it's at the end), and whatever you choose as your delimiter, e.g. tilde ~. This character class can then become, [a-zA-Z0-9.&%$-]+.
)*#)*(\.){1}
Part of this might be an error, it might not be. Basically, is there any need to capture the dot here? If there is not a need to capture it, leave the brackets alone. However, there is a definite error in the repetition. {1} is completely and utterly superfluous. Everything in there has to be repeated at least once. This is just making the code messy. The above can shortened into, )*#)*\..
((25[0-5]|2[0-4][0-9]|[0-1]{1}
Again, the {1} is not needed. Remove it, ((25[0-5]|2[0-4][0-9]|[0-1].
[0-9]{2}|[1-9]{1}[0-9]{1}
And again twice, this becomes [0-9]{2}|[1-9][0-9].
You keep doing this, the next block of code you have can be shortened:
|[1-9])\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9]|0)\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[1-9]|0)\.(25[0-5]|2[0-4][0-9]|[0-1]{1}[0-9]{2}|[1-9]{1}[0-9]{1}|[0-9])
Into
|[1-9])\.(25[0-5]|2[0-4][0-9]|[0-1][0-9]{2}|[1-9][0-9]|[1-9]|0)\.(25[0-5]|2[0-4][0-9]|[0-1][0-9]{2}|[1-9][0-9]|[1-9]|0)\.(25[0-5]|2[0-4][0-9]|[0-1][0-9]{2}|[1-9][0-9]|[0-9])
It's not amazingly better, but every little helps. Next:
|([a-zA-Z0-9\-]+\.)*[a-zA-Z0-9\-]+
The two character classes can be optimized, |([a-zA-Z0-9-]+\.)*[a-zA-Z0-9-]+.
\.(com|edu|gov|int|mil|net|org|biz|arpa|info|name|pro|aero|coop|museum|[a-zA-Z]{2})
This is very restrictive, but I assume you have it like this for a reason so I'll leave it.
)(\:[0-9]+)*(/
And here is the cause of your error. You did not escape the forward slash. However, I am going to leave it as using a different delimiter would avoid this and also tidy up your pattern.
($|[a-zA-Z0-9\.\,\?\'\\\+&%\$#\=~_\-]+))*$/";
That character class can be greatly shortened now knowing that we don't need to escape everything within them. It can become, ($|[a-zA-Z0-9.,?'\\+&%$#=~_-]+))*$/";.
Using everything we now know your pattern can be made much prettier and easier to handle.
It can become instead:
$pattern = "~^(http|https|ftp)://www\.([a-zA-Z0-9.-]+(:[a-zA-Z0-9.&%$-]+)*#)*((25[0-5]|2[0-4][0-9]|[0-1][0-9]{2}|[1-9][0-9]|[1-9])\.(25[0-5]|2[0-4][0-9]|[0-1][0-9]{2}|[1-9][0-9]|[1-9]|0)\.(25[0-5]|2[0-4][0-9]|[0-1][0-9]{2}|[1-9][0-9]|[1-9]|0)\.(25[0-5]|2[0-4][0-9]|[0-1][0-9]{2}|[1-9][0-9]|[0-9])|([a-zA-Z0-9-]+\.)+(com|edu|gov|int|mil|net|org|biz|arpa|info|name|pro|aero|coop|museum|[a-zA-Z]{2}))(:[0-9]+)*(/($|[a-zA-Z0-9.,?'\\+&%$#=\~_-]+))*$~";
Now that you have a smaller expression, finding faults and more customization should be a little easier.
Just a quick note
I keep noticing that you have used the following syntax at the beginning of some groupings, (\:. I have removed the backslash as it is not needed for a colon. However, were you trying to make it so the group was not captured? If so, the syntax for that is, (?:.
Edit:: You can also optimize the pattern further by utilizing character classes
\d = [0-9]
\w = [a-zA-Z0-9_]
Adding i to the end of the last pattern delimiter turns case insensitivity on too. Which means, instead of writing [a-zA-Z] you can just write [a-z] instead.
Also, the http|https can just become https?
So you pattern could be shortened further too:
$pattern = "~^(https?|ftp)://www\.([a-z\d.-]+(:[a-z\d.&%$-]+)*#)*((25[0-5]|2[0-4]\d|[0-1]\d{2}|[1-9]\d|[1-9])\.(25[0-5]|2[0-4]\d|[0-1]\d{2}|[1-9]\d|[1-9]|0)\.(25[0-5]|2[0-4]\d|[0-1]\d{2}|[1-9]\d|[1-9]|0)\.(25[0-5]|2[0-4]\d|[0-1]\d{2}|[1-9]\d|\d)|([a-z\d-]+\.)+(com|edu|gov|int|mil|net|org|biz|arpa|info|name|pro|aero|coop|museum|[a-z]{2}))(:\d+)*(/($|[\w.,?'\\+&%$#=\~-]+))*$~i";
I see one error:
[0-9]+)*(/($
to
[0-9]+)*(\/($
or to
[0-9]+)*(($
if the / is supposed to be an ender, which it's not supposed to be.
But seriously, is there no other way you can achieve this? This string is really hard to troubleshoot.
Why don't use standard php function filter_var?
http://lv.php.net/manual/ru/function.filter-var.php

What does the Unknown modifier 'c' mean in Regex? [duplicate]

This question already has answers here:
Warning: preg_replace(): Unknown modifier
(3 answers)
Closed 3 years ago.
I'm a newbie with regular expressions and i need some help :).
I have this:
$url = '<img src="http://mi.url.com/iconos/oks/milan.gif" alt="Milan">';
$pattern = '/<img src="http:\/\/mi.url.com/iconos/oks/(.*)" alt="(.*)"\>/i';
preg_match_all($pattern, $url, $matches);
print_r($matches);
And I get this error:
Warning: preg_match_all() [function.preg-match-all]: Unknown modifier 'c'
I want to select that 'milan.gif'.
How can I do that?
If you’re using / as delimiter, you need to escape every occurrence of that character inside the regular expression. You didn’t:
/<img src="http:\/\/mi.url.com/iconos/oks/(.*)" alt="(.*)"\>/i
^
Here the marked / is treated as end delimiter of the regular expression and everything after is is treated as modifier. i is a valid modifier but c isn’t (see your error message).
So:
/<img src="http:\/\/mi\.url\.com\/iconos\/oks\/(.*)" alt="(.*)"\>/i
But as Pekka already noted in the comments, you shouldn’t try to use regular expressions on a non-regular language like HTML. Use an HTML parser instead. Take a look at Best methods to parse HTML.
The problem is that you haven't escaped the forward slashes in the url string (you have escaped the ones in the http:// part, but not the url path).
Therefore the first one it comes across it (which is after .com), it thinks is the end of the regex, so it treats everything after that slash as the 'modifier' codes.
The next character ('i') is a valid modifier (as you know, since you're actually using it in your example), so that passes the test. However the next character ('c') is not, so it throws an error, which is what you're seeing.
To fix it, simply escape the slashes. So your example would look like this:
$pattern = '/<img src="http:\/\/mi.url.com\/iconos\/oks\/(.*)" alt="(.*)"\\>/i';
Hope that helps.
Note, as someone has already said, it's generally not advisable to use regex to match HTML, since HTML can be too complex to match accurately. It's generally preferrable to use a DOM parser. In your example, the regex could fail if the alt attribute or the end of the image URL contains unexpected characters, or if the quoting in the HTML code isn't as you expect.

Weird error using preg_match and unicode

if (preg_match('(\p{Nd}{4}/\p{Nd}{2}/\p{Nd}{2}/\p{L}+)', '2010/02/14/this-is-something'))
{
// do stuff
}
The above code works. However this one doesn't.
if (preg_match('/\p{Nd}{4}/\p{Nd}{2}/\p{Nd}{2}/\p{L}+/u', '2010/02/14/this-is-something'))
{
// do stuff
}
Maybe someone could shed some light as to why the one below doesn't work. This is the error that is being produced:
A PHP Error was encountered
Severity: Warning
Message: preg_match()
[function.preg-match]: Unknown
modifier '\'
Try this: (delimit the regex with ())
if (preg_match('#\p{Nd}{4}/\p{Nd}{2}/\p{Nd}{2}/\p{L}+#', '2010/02/14/this-is-something'))
{
// do stuff
}
Edited
The modifier u is available from PHP 4.1.0 or greater on Unix and from PHP 4.2.3 on win32.
Also as nvl observed, you are using / as the delimiter and you are not escaping the / present in the regex. So you'lll have to use:
/\p{Nd}{4}\/\p{Nd}{2}\/\p{Nd}{2}\/\p{L}+/u
To avoid this escaping you can use a different set of delimiters like:
#\p{Nd}{4}/\p{Nd}{2}/\p{Nd}{2}/\p{L}+#
or
#\p{Nd}{4}/\p{Nd}{2}/\p{Nd}{2}/\p{L}+#
As a tip, if your delimiter is present in your regex, its better to choose a different delimiter not found in the regex. This keeps the regex clean and short.
In the second regex you're using / as the regex delimiter, but you're also using it in the regex. The compiler is trying to interpret this part as a complete regex:
/\p{Nd}{4}/
It thinks the next character after the second / should be a modifier like 'u' or 'm', but it sees a backslash instead, so it throws that cryptic exception.
In the first regex you're using parentheses as regex delimiters; if you wanted to add the u modifier, you would put it after the closing paren:
'(\p{Nd}{4}/\p{Nd}{2}/\p{Nd}{2}/\p{L}+)u'
Although it's legal to use parentheses or other bracketing characters ({}, [], <>) as regex delimiters, it's not a good idea IMO. Most people prefer to use one of the less common punctuation characters. For example:
'~\p{Nd}{4}/\p{Nd}{2}/\p{Nd}{2}/\p{L}+~u'
'%\p{Nd}{4}/\p{Nd}{2}/\p{Nd}{2}/\p{L}+%u'
Of course, you could also escape the slashes in the regex with backslashes, but why bother?

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