[ad_1] If you are trying to parse single or double quoted strings, it should be done in two steps. Validation, then parse for values. You could probably do both in a single regex with the use of a \G anchor, validating with \A\G and parsing with just the \G. If you are sure its valid, … Read more
[ad_1] The idea is to avoid the issue of new lines by not using the dot ($subject is your string): $pattern = ‘~ Project\hNo\.\h\d++\hDATE\h (?<date>\d{1,2}\/\d{1,2}\/\d{1,2}) \s++No\.\hQUESTION\hANSWER\s++ (?<No>\d++)\s++ # all characters but D or D not followed by “ate Required” (?<desc>(?>[^D]++|D(?!ate\hRequired))+) \D++ (?<date_required>\d{1,2}\/\d{1,2}\/\d{1,2}) ~x’; preg_match_all($pattern, $subject, $matches, PREG_SET_ORDER); print_r($matches); Note that i use possessive quantifiers and … Read more
[ad_1] Try this : put id for anchor tag <a id=”anchor1″ href=”https://stackoverflow.com/questions/25931155/site.com/register.php?mid=username&mode=bn&bid=1″> use below javascript <script> var href = document.getElementById(‘anchor1’).href; //get index of ? var indexStart = href.indexOf(‘?’); var indexLast = href.length; //get href from ? upto total length var params = href.substring(indexStart+1, indexLast); //get tokens with seperator as ‘&’ and iterate it var paramsArray … Read more
[ad_1] For this specific case, you don’t need a regular expression, you can just parse the url $url=”http://dev.time.com/wp-content/uploads/2014/08/sheep-shrek-300×199.jpg?w=360″; $parts = parse_url($url); echo $parts[‘host’] . $parts[‘path’]; Missed that you want to strip the size … $data = preg_replace(‘/(-\d+x\d+)\.jpg/’, ‘.jpg’, $data); 1 [ad_2] solved How to strip parts of the given url using preg_replace (php) [closed]
[ad_1] The regex would be this: /^[A-Z0-9]{5}-[A-Z0-9]{5}-[A-Z0-9]{5}-[A-Z0-9]{5}-[A-Z0-9]{5}$/ Regex explanation: Start regex expression /^ Capital A-Z and Numerical 0-9 [A-Z0-9] 5 long exactly {5} Dash between sets End expression $/ <?php $reg = ‘/^[A-Z0-9]{5}-[A-Z0-9]{5}-[A-Z0-9]{5}-[A-Z0-9]{5}-[A-Z0-9]{5}$/’; $string = ‘YTMG3-N6DKC-DKB77-7M9GH-8HVX7’; preg_match($reg, $string, $matches, PREG_OFFSET_CAPTURE); print_r($matches); ?> 7 [ad_2] solved Preg_match for a string key
[ad_1] You could use the below regex to extract all ipaddresses, phone number, dates. (?<ipaddress>\d{1,3}(?:\.\d{1,3}){3})|(?<Phonenumber>\d{10})|(?<Date>\d{1,2}\/\d{1,2}\/\d{4}) DEMO 2 [ad_2] solved Return Mobile Number, IP Address and Date Format in PHP String using regular expression [closed]
[ad_1] What this big regex commands mean? Pattern #1 Breakdown: / #start of pattern delimiter (@.*@) #Capture Group #1: match an @ sign, then zero or more (as many as possible) of any non-newline character, then another @ sign | #or (\.\.) #Capture Group #2: match a literal dot, then another literal dot | #or … Read more
[ad_1] Use /(?<=\[ID:)\d+/ pattern that match digits that is after [ID: string. $subject = “New tour subscription [ALKJ-102] [ID:12345]”; preg_match(“/(?<=\[ID:)\d+/”, $subject, $mat); echo $mat[0]; Check result in demo 1 [ad_2] solved Regular expression to extract id from email subject [closed]
[ad_1] if(preg_match($fullname_pattern,$fullname)) Remove Last “)” And add here (preg_match($password_pattern,$password))) Like this: if(preg_match($fullname_pattern,$fullname) && preg_match($email_pattern,$email) && preg_match($password_pattern,$password) ) { 1 [ad_2] solved More than one preg_match queries? [closed]
[ad_1] if (preg_match( ‘/(?<=\|\|) # Assert starting position directly after ||< [^,|]*, # Match any number of characters except , or |; then a , [^,|]*, # twice (so we have matched at least two commas) [^|]* # Then match anything except | characters (?=\|\|) # until we are right before || /x’, $subject, $regs)) … Read more
[ad_1] Use this regex :(?<=[()\/*+-])(?=[0-9()])|(?<=[0-9()])(?=[()\/*+-]) It will match every position between a digit or a parenthesis and a operator or a parenthesis.(?<=[()\/*+-])(?=[0-9()]) matches the position with a parenthesis or an operator at the left and a digit or parenthesis at the right(?<=[0-9()])(?=[()\/*+-]) is the same but with left and right reversed. Demo here 5 [ad_2] … Read more
[ad_1] I make a snippet for you, so you can start from here $str=”December 2012 Name: Jack Brown”; $ptr = “/^(?P<month>:Jan(?:uary)?|Feb(?:ruary)?|Dec(?:ember)?) (?P<year>:19[7-9]\d|2\d{3}) (Name:(?P<name>(.*)))/”; preg_match($ptr, $str, $data); echo ‘For ‘.trim($data[‘name’]).’ – ‘.$data[‘month’].’ ‘.$data[‘year’]; the result will be ‘For Jack Brown – December 2012’ this is a array Array ( [0] => December 2012 Name: Jack Brown … Read more
[ad_1] If you want to test if a string contains at least one alphanumeric and at least one non-alphanumeric string, use the following regex: /^(?=.*[a-z0-9])(?=.*[^a-z0-9])/i Breakup: / start of regex ^ match the start of the string (?= if the following is present there: .* anything at all (except newlines), then [a-z0-9] an alphanumeric character … Read more
[ad_1] The solution is the data cutting between {}. Try this: $result = preg_replace(“#\{img:(.*)\}#U”,'<img src=\’$1\’ style=\’width:100%\’ />’,$content); 0 [ad_2] solved Match content between parenthesis and get the url and change it to img tag using RegExp PHP [duplicate]
[ad_1] You can try this regex pattern: (?<=mailfrom=)[^;]+(?=;) (?<=mailfrom=) // assert ‘mailfrom=’ on the left [^;]+ // 1 or more characters not ‘;’ (?=;) // assert ‘;’ on the right Demo 0 [ad_2] solved Find email address between specific substring and semicolon