String.contains() accepts CharSequence as argument and doesn’t “trigger” regexes that it or processing string may contain. Returns true if and only if this string contains the specified sequence of char values. From here. solved If I use string.contains() on a string that may have regular expressions [closed]
You could try: r’\?id=([a-zA-Z\.]+)’ For your regex, like so: def get_id(toParse) regex = r’\?id=([a-zA-Z\.]+)’ x = re.findall(regex, toParse)[0] return x Regex – By adding r before the actual regex code, we specify that it is a raw string, so we don’t have to add multiple backslashes before every command, which is better explained here. ? … Read more
This will return the elements you want: import re s=””‘journey (a,b) from station south chennai to station punjab chandigarh journey (c,d) from station jammu katra to city punjab chandigarh journey (e) from station journey (c,d) from station ANYSTRING jammu katra to ANYSTRING city punjab chandigarh ”’ matches_single = re.findall(‘journey (\([^,]+,[^,]+\)) from (\S+ \S+\s{0,1}\S*) to (\S+ … Read more
s/(\d{1,3}\.\d{1,3})\.233(\.\d{1,3})/$1.234$2/ Will do what you want in PCRE 5 solved Swap multiple parts of an IP using a Perl REGEX [duplicate]
Pretty simple: In [8]: course_name Out[8]: ‘Post Graduate Certificate Programme in Retail Management (PGCPRM) (Online)’ In [9]: print re.sub(‘\([A-Z]+\)\s*’, ”, course_name) Post Graduate Certificate Programme in Retail Management (Online) In [17]: print re.search(‘\(([A-Z]+)\)\s*’, course_name).groups()[0] PGCPRM 0 solved Extract only first match using python regular expression [duplicate]
Have managed to work on a solution that suits my needs: .*item-list[\s\S]*?<\/ul> This grabs everything between item-list and the </ul>. May of use to something solved Regular Expression match term within ul class
You can use a regex like this: -\d+x\d+(\.\w+)$ Working demo The code you can use is: $re = “/-\\d+x\\d+(\\.\\w+)$/”; $str = “http://website.dev/2014/05/my-file-name-here-710×557.png”; $subst=”\1″; $result = preg_replace($re, $subst, $str, 1); The idea is to match the resolution -NumbersXNumbers using -\d+x\d+ (that we’ll get rid of it) and then capture the file extension by using (\.\w+)$ using … Read more
From comment: but if I do not know the text between the delimiters? So it sounds like the replacement values are positional. Best way to do this is a regular expression appendReplacement loop: public static String replace(String input, String… values) { StringBuffer buf = new StringBuffer(); Matcher m = Pattern.compile(“\\|{2}(.*?)\\|{2}”).matcher(input); for (int i = 0; … Read more
If the rest of the url never changes, then you don’t need a regular expression. Just store the first part of the url in a string: $url_prefix = “http:// www.badoobook.com/clips/index.php?page=videos§ion=view&vid_id=”; then append your identifier to it: $id = 100162; $full_url = $url_prefix + $id; solved regular expression [closed]
Whenever you think “I need to match a pattern”, you should think “Regular Expressions” as a good starting point. See doco. It is a little trickier since the input file is unicode. import re import codecs with codecs.open(“test.unicode.txt”,”rb”, “utf-8″) as f: words = [] for line in f.readlines(): matches = re.match(b”solution:.+\[(?P<word>\w+).*\]”, line, flags=re.U) if matches: … Read more
This should match any of the specified characters and no others. /^[$0-9a-z+\-\/*]+$/ Note: the ‘-‘ character needs to be escaped in selection groups like this since it generally signifies a range of possible characters (i.e. a-z or 0-9). 3 solved Want regex for accepting $,0-9,a-z,+,-,/,*? [closed]
It is not entirely clear what the conditions are. Based on your comments I think you look for the following regex: r”Профессиональная ГИС \”Панорама\” \(версия 12(\.\d+){2,}, для платформы \”x(32|64)\”\)” This will match any sequence of dots and numbers after the 12. (so for instance 12.4.51.3.002 is allowed), and furthermore both x64 and x32 are allowed … Read more
As the perlvar man-page explains: $+[0] is the offset into the string of the end of the entire [last successful] match. This is the same value as what the pos function returns when called on the variable that was matched against. $-[0] is the offset of the start of the last successful match. solved What … Read more
I assume X and Y can’t begin with 0. [1-9]\d{0,2} matches a number from 1 to 3 digits that doesn’t begin with 0. So the regexp to extract X and Y should be: ^([1-9]\d{0,2})000([1-9]\d{0,2})000$ Then you can use re.sub() to remove the zeroes between X and Y. regex = re.compile(r’^([1-9]\d{0,2})000([1-9]\d{0,2})000$’); i = 14000010000 istr = … Read more
You can achieve that by either using an alternation or an optional group: With alternation (https://regex101.com/r/qIjNWF/1): 400 people drive dark blue cars|400 people drive cars The pipe symbol is working like a logical OR resp. alternation. So, it matches either the left side or the right side. With an optional group (https://regex101.com/r/p5Z7eU/1/): 400 people drive … Read more