I have this Regex expression
UriPatternToMatch= new Regex(#"(href|src)=""[\d\w\/:##%;$\(\)~_\?\+\-=\\\.&]*",
RegexOptions.Compiled | RegexOptions.IgnoreCase)
This is working fine to pickup all URLS including http,ftp and others , but it picks up text within "<" special characters as URL too
for example it will wrongly pick up the text below as a URL too ( adding a photo instead of text below)
I believe something like ^< is what is needed , but where do I add it ?
Thanks
You need to use negative lookahead like this:
(?!.*?<)
Related
What would be a quick way to extract the value of the title attributes for an HTML table:
...
<li>Proclo</li>
<li>Proclus</li>
<li>Ptolemy</li>
<li>Pythagoras</li></ul><h3>S</h3>
...
so it would return Proclo, Proclus, Ptolemy, Pythagoras,.... in strings for each line. I'm reading the file using a StreamReader. I'm using C#.
Thank you.
This C# regex will find all title values:
(?<=\btitle=")[^"]*
The C# code is like this:
Regex regex = new Regex(#"(?<=\btitle="")[^""]*");
Match match = regex.Match(input);
string title = match.Value;
The regex uses positive lookbehind to find the position where the title value starts. It then matches everything up to the ending double quote.
Use the regexp below
title="([^"]+)"
and then use Groups to browse through matched elements.
EDIT: I have modified the regexp to cover the examples provided in comment by #Staffan Nöteberg
I'm trying to add something along the lines of this regex logic.
For Input:
reading/
reading/123
reading/456
reading/789
I want the regex to match only
reading/123
reading/456
reading/789
Excluding reading/.
I've tried reading\/* but that doesn't work because it includes reading/
You must escape your backslashes in Hugo, \\/\\d+.
I'm trying to create a custom filter in Google Analytic to remove the query parts of the url which I don't want to see. The url has the following structure
[domain]/?p=899:2000:15018702722302::NO:::
I would like to create a regex which skips the first 12 characters (that is until:/?p=899:2000), and what ever is going to be after that replace it with nothing.
So I made this one: https://regex101.com/r/Xgbfqz/1 (which could be simplified to .{0,12}) , but I actually would like to skip those and only let the regex match whatever is going to be after that, so that I'll be able to tell in Google Analytics to replace it with "".
The part in the url that is always the same is
?p=[3numbers]:[0-4numbers]
Thank you
Your regular expression:
\/\?p=\d{3}\:\d{0,4}(.*)
Tested in Golang RegEx 2 and RegEx101
It search for /p=###:[optional:####] and capture the rest of the right side string.
(extra) JavaScript:
paragraf='[domain]/?p=899:2000:15018702722302::NO:::'
var regex= /\/\?p=\d{3}\:\d{0,4}(.*)/;
var match = regex.exec(paragraf);
alert('The rest of the right side of the string: ' + match[1]);
Easily use "[domain]/?p=899:2000:15018702722302::NO:::".substr(12)
You can try this:
/\?p\=\d{3}:\d{0,4}
Which matches just this: ?p=[3numbers]:[0-4numbers]
Not sure about replacing though.
https://regex101.com/r/Xgbfqz/1
I have an url like below and wanted to use RegEx to extract segments like: Id:Reference, Title:dfgdfg, Status.Title:Current Status, CreationDate:Logged...
This is the closest pattern I got [=,][^,]*:[^,]*[,&] but obviously the result is not as expected, any better ideas?
P.S. I'm using [^,] to matach any characters except , because , will not exist the segment.
This is the site using for regex pattern matching.
http://regexpal.com/
The URL:
http://localhost/site/=powerManagement.power&query=_Allpowers&attributes=Id:Reference,Title:dfgdfg,Status.Title:Current Status,CreationDate:Logged,RaiseUser.Title:标题,_MinutesToBreach&sort_by=CreationDate"
Thanks,
You haven't specified what programming language you use. But almost all with support this:
([\p{L}\.]+):([\p{L}\.]+)
\p{L} matches a Unicode character in any language, provided that your regex engine support Unicode. RegEx 101.
You can extract the matches via capturing groups if you want.
In python:
import re
matchobj = re.match("^.*Id:(.*?),Title:(.*?),.*$", url, )
Id = matchobj.group(1)
Title = matchobj.group(2)
I am trying to write a pattern for extracting the path for files found in img tags in HTML.
String string = "<img src=\"file:/C:/Documents and Settings/elundqvist/My Documents/My Pictures/import dialog step 1.JPG\" border=\"0\" />";
My Pattern:
src\\s*=\\s*\"(.+)\"
Problem is that my pattern will also include the 'border="0" part of the img tag.
What pattern would match the URI path for this file without including the 'border="0"?
Your pattern should be (unescaped):
src\s*=\s*"(.+?)"
The important part is the added question mark that matches the group as few times as possible
This one only grabs the src only if it's inside of an tag and not when it is written anywhere else as plain text. It also checks if you've added other attributes before or after the src attribute.
Also, it determines whether you're using single (') or double (") quotes.
\<img.+src\=(?:\"|\')(.+?)(?:\"|\')(?:.+?)\>
So for PHP you would do:
preg_match("/\<img.+src\=(?:\"|\')(.+?)(?:\"|\')(?:.+?)\>/", $string, $matches);
echo "$matches[1]";
for JavaScript you would do:
var match = text.match(/\<img.+src\=(?:\"|\')(.+?)(?:\"|\')(?:.+?)\>/)
alert(match[1]);
Hopefully that helps.
Try this expression:
src\s*=\s*"([^"]+)"
I solved it by using this regex.
/<img.*?src="(.*?)"/g
Validated in https://regex101.com/r/aVBUOo/1
You want to play with the greedy form of group-capture. Something like
src\\s*=\\s*\"(.+)?\"
By default the regex will try and match as much as possible
I am trying to write a pattern for extracting the path for files found in img tags in HTML.
Can we have an autoresponder for "Don't use regex to parse [X]HTML"?
Problem is that my pattern will also include the 'border="0" part of the img tag.
Not to mention any time 'src="' appears in plain text!
If you know in advance the exact format of the HTML you're going to be parsing (eg. because you generated it yourself), you can get away with it. But otherwise, regex is entirely the wrong tool for the job.
I'd like to expand on this topic as usually the src attribute comes unquoted so the regex to take the quoted and unquoted src attribute is:
src\s*=\s*"?(.+?)["|\s]