I have this text:
SYS.DBMS_SCHEDULER.RUN_JOB('CALCULO_AGIL_DT_POSTO');
I need this text:
CALCULO_AGIL_DT_POSTO
I have this regex in PHP:
'(?<=SYS\.DBMS_SCHEDULER\.RUN_JOB\()[^'].*(?<=\))
but it does not work properly.
I can't see the need for a look ahead here.
Just make a simple pattern like this:
$pattern = "/SYS\.DBMS_SCHEDULER\.RUN_JOB\([\'\"](.*?)[\'\"]\)\;/";
It will match what is in exactly this string between the ( and '.
I made it lazy just in case it finds another ') somewhere else.
https://regex101.com/r/c9Kl3E/2
Edit noticed now in your title you need to be able to match both ' and ".
Added that to the regex [\'\"]
this one should work too, get you gonna need group 2 :
^(.+?')([a-zA-Z_]+)?('.+)$
Related
Im trying to make a regex pattern for strings that contain [[Title#Night|Anchor]] or just [[Title|Anchor]] and extract Title and Anchor. Basically two variables, first part between [[ and | and second part between | and ]], no matter what type of characters are inside (excluding \n, \r).
I had tried writing different patterns and none worked like I wanted. The code can be seen here with a sample content that I need to apply to.
\[\[(.*?)|(.*?)\]\]
may this help you:
$str = ' [[Title#Night|Anchor]] ';
preg_match('/\[\[([\s\S]*?)\]\]/',$str,$match);
print_r(explode('|',$match[1]));
update:
preg_match('/\[\[([\s\S]*?)\|([\s\S]*?)\]\]/',$str,$match);
print_r($match);
update 2:
preg_match('/\[\[(.*?)\|(.*?)\]\]/',$str,$match);
print_r($match);
update 3:
preg_match('/\[\[([^|\n\r]*)\|([^\]\n\r]*)\]\]/',$str,$match);
print_r($match);
The following should work for alphanumeric entries, depends what you want Title or Anchor to contain.
\[\[([a-zA-Z0-9#]*)\|([a-zA-Z0-9]*)\]\]
I have to create translations for the project I work on. The simplest solution was to change all stings to a functioncall(string) so that I could get unique string hashes everytime.
My code has the following different t() function uses:
<label for="anon"><?php echo t('anonymously? (20% extra)')?></label>
exit(t("Success! You made {amount} oranges out of it!", array('amount' => $oranges)));
echo t('You failed.');
My current regexp is:
$transMatches = preg_match_all('/[^a-z]t\([^)(]+/m', $contents, $matches);
The problem is that it fails on #1 example, matchin "anonymously?".
What I really want to achieve is: "match t( then match either ' or " then match anything except what you matched for ' or " and )"
Idea: t\(['|"](.*?)[^'|"]\)?
I cannot make above regexp to work.
How could I do AND in regexp so that it matches "['|"] AND )" OR "['|"] AND, array"
Please help me on regexp and explain why it works.
Thank you!
Parsing function arguments may be quite complex, but you need to parse only the first argument which (for simplicity) can assume always to be string escaped either with ' or with ", thus those regexps may match"
"[^"\\\\]*(?:\\\\.[^"\\\\]*)*"
\'[^\'\\\\]*(?:\\\\.[^\'\\\\]*)*'
Therefore you just need to match:
'~[^\w\d]t\(\s*("[^"\\\\]*(?:\\\\.[^"\\\\]*)*"|\'[^\'\\\\]*(?:\\\\.[^\'\\\\]*)*\')~i'
[^\w\d] assumes that no test1t will match, \s* makes you space tolerant...
With this regexp you'll get results like:
'anonymously? (20% extra)'
"Success! You made {amount} oranges out of it!"
And I can't imagine situation where you would need to parse out array too, can you describe it in comment/question?
Is this what you need?
Using Backreferences in The Regular Expression - http://www.regular-expressions.info/brackets.html
But it looks strange what are you doing, and why?
Are you replacing the function call with some result? why dont you just let it call the function and return translation from it?
I want to find any pattern matching: ###-##-####
and replace the ###-##, with ***-**
but leave the -####
I tried this below, but nothing is being replaced at all.
preg_replace('/(^[\d]{3})(-)([\d]{2})(-[\d]{4}$)/','\2\4',$myText);
Any help is appreciated
Update, here is my entire code string as it currently stands, after trying a few of the suggestions below. I am comparing the second echo output to the first... and the social numbers all remain the same.
Also, as it was mentioned below, my string does contain more than just a social... it is thousands of characters long. which i think is my real issue. Sorry if i didnt clear that up in the beginning.
//Make the CSC credit report request.
$strCscResponse = $Csc->makeRequest($strFixedFormatRecord);
echo "<br/><br/><pre>" . $strCscResponse . "</pre><br/><br/>";
$strCscResponse = str_replace("!", " ", $strCscResponse);
$strCscResponse = preg_replace('/^\d{3}-\d{2}(-\d{4})$/','***-**$1',$strCscResponse);
echo "<br/><br/><pre>" . $strCscResponse . "</pre><br/><br/>";
update
I'd like to mark all the answers and "the answer" just because i didnt clarify the string has more than just a social in it. thank you for the help with this issue, embarrisingly enough it has been driving me wild for a couple days now.
There is one possible problem: you might not be matching the right string (if you are trying to find SSNs buried in a large block of text) - the ^ and $ anchors will only match beginning of string (or sometimes beginning of line) - if this is not what you want, but instead you want to find SSNs in a long string, you need to get rid of those anchors.
The other problem, potentially, is that you seem to want to replace things with asterisks, but you do not include asterisks in your replacement expression. you need to use a replacement expression like
`***-**\4`
Try this regex:
(\d{3})(-)(\d{2})(-\d{4})
Try this:
preg_replace('/^\d{3}-\d{2}(-\d{4})$/','***-**$1',$myText);
you have ^ and $ in your pattern, but I see no m modifier, so this
will only match if ###-##-#### is the entire string.
[\d] can be
shortened to \d
your \2\4 will leave --####, if you wanted *-####
you can simply have *\4
I know I've seen this done a lot in places, but I need something a little more different than the norm. Sadly When I search this anywhere it gets buried in posts about just making the link into an html tag link. I want the PHP function to strip out the "http://" and "https://" from the link as well as anything after the .* so basically what I am looking for is to turn A into B.
A: http://www.youtube.com/watch?v=spsnQWtsUFM
B: www.youtube.com
If it helps, here is my current PHP regex replace function.
ereg_replace("[[:alpha:]]+://[^<>[:space:]]+[[:alnum:]/]", "\\0", htmlspecialchars($body, ENT_QUOTES)));
It would probably also be helpful to say that I have absolutely no understanding in regular expressions. Thanks!
EDIT: When I entered a comment like this blahblah https://www.facebook.com/?sk=ff&ap=1 blah I get html like this<a class="bwl" href="blahblah https://www.facebook.com/?sk=ff&ap=1 blah">www.facebook.com</a> which doesn't work at all as it is taking the text around the link with it. It works great if someone only comments a link however. This is when I changed the function to this
preg_replace("#^(.*)//(.*)/(.*)$#",'<a class="bwl" href="\0">\2</a>', htmlspecialchars($body, ENT_QUOTES));
This is the simples and cleanest way:
$str = 'http://www.youtube.com/watch?v=spsnQWtsUFM';
preg_match("#//(.+?)/#", $str, $matches);
$site_url = $matches[1];
EDIT: I assume that the $str had been checked to be a URL in the first place, so I left that out. Also, I assume that all the URLs will contain either 'http://' or 'https://'. In case the url is formatted like this www.youtube.com/watch?v=spsnQWtsUFM or even youtube.com/watch?v=spsnQWtsUFM, the above regexp won't work!
EDIT2: I'm sorry, I didn't realize that you were trying to replace all strings in a whole test. In that case, this should work the way you want it:
$str = preg_replace('#(\A|[^=\]\'"a-zA-Z0-9])(http[s]?://(.+?)/[^()<>\s]+)#i', '\\1\\3', $str);
I am not a regex whizz either,
^(.*)//(.*)/(.*)$
\2
was what worked for me when I tried to use as find and replace in programmer's notepad.
^(.)// should extract the protocol - referred as \1 in the second line.
(.)/ should extract everything till the first / - referred as \2 in the second line.
(.*)$ captures everything till the end of the string. - referred as \3 in the second line.
Added later
^(.*)( )(.*)//(.*)/(.*)( )(.*)$
\1\2\4 \7
This should be a bit better, but will only replace just 1 URL
The \0 is replaced by the entire matched string, whereas \x (where x is a number other than 0 starting at 1) will be replaced by each subpart of your matched string based on what you wrap in parentheses and the order those groups appear. Your solution is as follows:
ereg_replace("[[:alpha:]]+://([^<>[:space:]]+[:alnum:]*)[[:alnum:]/]", "\\1
I haven't been able to test this though so let me know if it works.
I think this should do it (I haven't tested it):
preg_match('/^http[s]?:\/\/(.+?)\/.*/i', $main_url, $matches);
$final_url = ''.$matches[1].'';
I'm surprised no one remembers PHP's parse_url function:
$url = 'http://www.youtube.com/watch?v=spsnQWtsUFM';
echo parse_url($url, PHP_URL_HOST); // displays "www.youtube.com"
I think you know what to do from there.
$result = preg_replace('%(http[s]?://)(\S+)%', '\2', $subject);
The code with regex does not work completely.
I made this code. It is much more comprehensive, but it works:
See the result here: http://cht.dk/data/php-scripts/inc_functions_links.php
See the source code here: http://cht.dk/data/php-scripts/inc_functions_links.txt
I am trying to pull the anchor text from a link that is formatted this way:
<h3><b>File</b> : i_want_this</h3>
I want only the anchor text for the link : "i_want_this"
"variable_text" varies according to the filename so I need to ignore that.
I am using this regex:
<a href=\"\/en\/browse\/file\/variable_text\">(.*?)<\/a>
This is matching of course the complete link.
PHP uses a pretty close version to PCRE (PERL Regex). If you want to know a lot about regex, visit perlretut.org. Also, look into Regex generators like exspresso.
For your use, know that regex is greedy. That means that when you specify that you want something, follwed by anything (any repetitions) followed by something, it will keep on going until that second something is reached.
to be more clear, what you want is this:
<a href="
any character, any number of times (regex = .* )
">
any character, any number of times (regex = .* )
</a>
beyond that, you want to capture the second group of "any character, any number of times". You can do that using what are called capture groups (capture anything inside of parenthesis as a group for reference later, also called back references).
I would also look into named subpatterns, too - with those, you can reference your choice with a human readable string rather than an array index. Syntax for those in PHP are (?P<name>pattern) where name is the name you want and pattern is the actual regex. I'll use that below.
So all that being said, here's the "lazy web" for your regex:
<?php
$str = '<h3><b>File</b> : i_want_this</h3>';
$regex = '/(<a href\=".*">)(?P<target>.*)(<\/a>)/';
preg_match($regex, $str, $matches);
print $matches['target'];
?>
//This should output "i_want_this"
Oh, and one final thought. Depending on what you are doing exactly, you may want to look into SimpleXML instead of using regex for this. This would probably require that the tags that we see are just snippits of a larger whole as SimpleXML requires well-formed XML (or XHTML).
I'm sure someone will probably have a more elegant solution, but I think this will do what you want to done.
Where:
$subject = "<h3><b>File</b> : i_want_this</h3>";
Option 1:
$pattern1 = '/(<a href=")(.*)(">)(.*)(<\/a>)/i';
preg_match($pattern1, $subject, $matches1);
print($matches1[4]);
Option 2:
$pattern2 = '()(.*)()';
ereg($pattern2, $subject, $matches2);
print($matches2[4]);
Do not use regex to parse HTML. Use a DOM parser. Specify the language you're using, too.
Since it's in a captured group and since you claim it's matching, you should be able to reference it through $1 or \1 depending on the language.
$blah = preg_match( $pattern, $subject, $matches );
print_r($matches);
The thing to remember is that regex's return everything you searched for if it matches. You need to specify that only care about the part you've surrounded in parenthesis (the anchor text). I'm not sure what language you're using the regex in, but here's an example in Ruby:
string = 'i_want_this'
data = string.match(/<a href=\"\/en\/browse\/file\/variable_text\">(.*?)<\/a>/)
puts data # => outputs 'i_want_this'
If you specify what you want in parenthesis, you can reference it:
string = 'i_want_this'
data = string.match(/<a href=\"\/en\/browse\/file\/variable_text\">(.*?)<\/a>/)[1]
puts data # => outputs 'i_want_this'
Perl will have you use $1 instead of [1] like this:
$string = 'i_want_this';
$string =~ m/<a href=\"\/en\/browse\/file\/variable_text\">(.*?)<\/a>/;
$data = $1;
print $data . "\n";
Hope that helps.
I'm not 100% sure if I understand what you want. This will match the content between the anchor tags. The URL must start with /en/browse/file/, but may end with anything.
#(.*?)#
I used # as a delimiter as it made it clearer. It'll also help if you put them in single quotes instead of double quotes so you don't have to escape anything at all.
If you want to limit to numbers instead, you can use:
#(.*?)#
If it should have just 5 numbers:
#(.*?)#
If it should have between 3 and 6 numbers:
#(.*?)#
If it should have more than 2 numbers:
#(.*?)#
This should work:
<a href="[^"]*">([^<]*)
this says that take EVERYTHING you find until you meet "
[^"]*
same! take everything with you till you meet <
[^<]*
The paratese around [^<]*
([^<]*)
group it! so you can collect that data in PHP! If you look in the PHP manual om preg_match you will se many fine examples there!
Good luck!
And for your concrete example:
<a href="/en/browse/file/variable_text">([^<]*)
I use
[^<]*
because in some examples...
.*?
can be extremely slow! Shoudln't use that if you can use
[^<]*
You should use the tool Expresso for creating regular expression... Pretty handy..
http://www.ultrapico.com/Expresso.htm