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 try to get an URL from a String of the following format:
RANDOMRUBBISHhttps://www.my-url.com/randomfirstname_randomlastnameRANDOMRUBBISH
I already tried some things, especially the the look before/after, which I used before successfully on another url format (starts https... ends .html, this was working).
But seems I'm too stupid to figure out the regex for the kind of string mentioned above. I just want the URL part from https.... to the end of the random last name. Is this even possible?
Any Ideas?
If you can guarantee that randomfirstname_randomlastname is all lowercase and RANDOMRUBBISH is all uppercase, you can use character classes [a-z] and [A-Z]. The language the regex is for will determine how to use these.
This is example works in javascript:
var str = "RANDOMRUBBISHhttps://www.my-url.com/randomfirstname_randomlastnameRANDOMRUBBISH";
var match = /https:\/\/www\.my-url\.com\/[a-z]*/.exec(str);
I'm trying to write a regex which will parse the hash portion of a URL, removing whichever conventionally-formatted hashbang may be present.
For example, I wish to remove any of the following:
#
#/
#!
#!/
This is what I currently have:
/[(?:#|#\/|#!|#!\/)]+/
However, this is capturing an empty group at the start, and splitting the remaining strings. For example,
"#!/E/F".split(/[(?:#|#\/|#!|#!\/)]/); // ["", "", "", "E", "F"]
Whereas the desirable outcome is simply a single group
["E/F"]
Could someone please point out the error in my regex?
[If it makes a difference, I produced the above output using the JavaScript console in Firebug.]
Use string.replace instead of string.split.
#!?\/?
Use the above regex and then replace the match with empty string.
> '#!/E/F'.replace(/#!?\/?/g, '');
'E/F'
DEMO
Your regex seems awfully complicated. Maybe this is more what you're looking for:
"#!/E/F".split(/(#!/|#/|#!|#)/);
Did you checkout the Javascript regex documentation?
It might be different from what you imagined, since I don't understand why you're using the : and ? in your regex.
If you're using Javascript then you can just use:
location.assign(location.href.replace(/#.*$/, ""));
However if you only want to remove above listed hashtags then use:
var repl = location.href.replace(/#(!\/?|\/)?$/, '');
I am trying to write a regex to get the filename from a url if it exists.
This is what I have so far:
(?:[^/][\d\w\.]+)+$
So from the url http://www.foo.com/bar/baz/filename.jpg, I should match filename.jpg
Unfortunately, I match anything after the last /.
How can I tighten it up so it only grabs it if it looks like a filename?
The examples above fails to get file name "file-1.name.zip" from this URL:
"http://sub.domain.com/sub/sub/handler?file=data/file-1.name.zip&v=1"
So I created my REGEX version:
[^/\\&\?]+\.\w{3,4}(?=([\?&].*$|$))
Explanation:
[^/\\&\?]+ # file name - group of chars without URL delimiters
\.\w{3,4} # file extension - 3 or 4 word chars
(?=([\?&].*$|$)) # positive lookahead to ensure that file name is at the end of string or there is some QueryString parameters, that needs to be ignored
This one works well for me.
(\w+)(\.\w+)+(?!.*(\w+)(\.\w+)+)
(?:.+\/)(.+)
Select all up to the last forward slash (/), capture everything after this forward slash. Use subpattern $1.
Non Pcre
(?:[^/][\d\w\.]+)$(?<=\.\w{3,4})
Pcre
(?:[^/][\d\w\.]+)$(?<=(?:.jpg)|(?:.pdf)|(?:.gif)|(?:.jpeg)|(more_extension))
Demo
Since you test using regexpal.com that is based on javascript(doesnt support lookbehind), try this instead
(?=\w+\.\w{3,4}$).+
I'm using this:
(?<=\/)[^\/\?#]+(?=[^\/]*$)
Explanation:
(?<=): positive look behind, asserting that a string has this expression, but not matching it.
(?<=/): positive look behind for the literal forward slash "/", meaning I'm looking for an expression which is preceded, but does not match a forward slash.
[^/\?#]+: one or more characters which are not either "/", "?" or "#", stripping search params and hash.
(?=[^/]*$): positive look ahead for anything not matching a slash, then matching the line ending. This is to ensure that the last forward slash segment is selected.
Example usage:
const urlFileNameRegEx = /(?<=\/)[^\/\?#]+(?=[^\/]*$)/;
const testCases = [
"https://developer.mozilla.org/en-US/docs/Web/API/MutationObserverInit#yo",
"https://developer.mozilla.org/static/fonts/locales/ZillaSlab-Regular.subset.bbc33fb47cf6.woff2",
"https://developer.mozilla.org/static/build/styles/locale-en-US.520ecdcaef8c.css?is-nice=true"
];
testCases.forEach(testStr => console.log(`The file of ${testStr} is ${urlFileNameRegEx.exec(testStr)[0]}`))
It might work as well:
(\w+\.)+\w+$
You know what your delimiters look like, so you don't need a regex. Just split the string. Since you didn't mention a language, here's an implementation in Perl:
use strict;
use warnings;
my $url = "http://www.foo.com/bar/baz/filename.jpg";
my #url_parts = split/\//,$url;
my $filename = $url_parts[-1];
if(index($filename,".") > 0 )
{
print "It appears as though we have a filename of $filename.\n";
}
else
{
print "It seems as though the end of the URL ($filename) is not a filename.\n";
}
Of course, if you need to worry about specific filename extensions (png,jpg,html,etc), then adjust appropriately.
> echo "http://www.foo.com/bar/baz/filename.jpg" | sed 's/.*\/\([^\/]*\..*\)$/\1/g'
filename.jpg
Assuming that you will be using javascript:
var fn=window.location.href.match(/([^/])+/g);
fn = fn[fn.length-1]; // get the last element of the array
alert(fn.substring(0,fn.indexOf('.')));//alerts the filename
Here is the code you may use:
\/([\w.][\w.-]*)(?<!\/\.)(?<!\/\.\.)(?:\?.*)?$
names "." and ".." are not considered as normal.
you can play with this regexp here https://regex101.com/r/QaAK06/1/:
In case you are using the JavaScript URL object, you can use the pathname combined with the following RegExp:
.*\/(.[^(\/)]+)
Benefit:
It matches anything at the end of the path, but excludes a possible trailing slash (as long as there aren't two trailing slashes)!
Try this one instead:
(?:[^/]*+)$(?<=\..*)
This is worked for me, no matter if you have '.' or without '.' it take the sufix of url
\/(\w+)[\.|\w]+$
Can someone assist in creating a Regex for the following situation:
I have about 2000 records for which I need to do a search/repleace where I need to make a replacement for a known item in each record that looks like this:
<li>View Product Information</li>
The FILEPATH and FILE are variable, but the surrounding HTML is always the same. Can someone assist with what kind of Regex I would substitute for the "FILEPATH/FILE" part of the search?
you may match the constant part and use grouping to put it back
(<li>View Product Information</li>)
then you should replace the string with $1your_replacement$2, where $1 is the first matching group and $2 the second (if using python for instance you should call Match.group(1) and Match.group(2))
You would have to escape \ chars if you're using Java instead.