I am looking to find a random alpha-numeric ID from a URL returned in some JSON. Using the example:
"responseURL" : "http://sutureself.com/userid/123abc"
... I want to just match on the 123abc. So far I have:
http://sutureself.com/userid/([a-z0-9]+)
... but this matches the whole URL. What do I need to add to only match this ID? Note, the length of the ID can differ.
How about this:
[^/]+(?="})
Working regex example:
http://regex101.com/r/pT3fG8
Looks like you need to set the template to "$1". See comments in this related post.
Related
For Matomo outgoing link tracking I need the regex pattern, which matched the following URLs:
https://www.example.com/product/?sku=12345&utm_source=123456789
and
https://www.example.com/product/?utm_source=123456789
"https://www.example.com/" and "utm_source=123456789" are always fixed in the URL, just "product/" or "category/product/" change and must replaced by regex pattern.
Thanks
Maybe this example can help you reach your goal:
(?<=https:\/\/www\.example\.com\/).+(?=utm_source=123456789)
It looks for any characters between these two groups:
https://www.example.com/
utm_source=123456789
Given the examples:
https://www.example.com/product/?sku=12345&utm_source=123456789
https://www.example.com/product/?utm_source=123456789
Your matches would be:
product/?sku=12345&
product/?
I have data from two URLS that I need to combine using REGEX
/online-teaching
/online-teaching?fbclid
I have /(online-teaching)|(online teaching)
I can't figure out how to include the url with the ? and the one without.
Thanks!
How about something as simple as:
online-teaching(?:.+)?
Regex demo
Match online-teaching and anything that follows, if it exists (might need to constraint for specific characters instead of matching all with . to have a valid URL, but I'll leave that up to you).
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 a response like below
{"id":9,"announcementName":"Test","announcementText":"<p>TestAssertion</p>\n","effectiveStartDate":"03/01/2016","effectiveEndDate":"03/02/2016","updatedDate":"02/29/2016","status":"Active","moduleName":"Individual Portal"}
{"id":103,"announcementName":"d3mgcwtqhdu8003","announcementText":"<p>This announcement is a test announcement”,"effectiveStartDate":"03/01/2016","effectiveEndDate":"03/02/2016","updatedDate":"02/29/2016","status":"Active","moduleName":"Individual Portal"}
{"id":113,"announcementName":"asdfrtwju3f5gh7f21","announcementText":"<p>This announcement is a test announcement”,"effectiveStartDate":"03/02/2016","effectiveEndDate":"03/03/2016","updatedDate":"02/29/2016","status":"InActive","moduleName":"Individual Portal"}
I am trying get the value of id (103) of announcementName d3mgcwtqhdu8003.
I am using below regEx pattern to get the id
"id":(.*?),"announcementName":"${announcementName}","announcementText":"
But it is matching everything from the first id to the announcementName. and returning
9,"announcementName":"Test","announcementText":"<p>TestAssertion</p>\n","effectiveStartDate":"03/01/2016","effectiveEndDate":"03/02/2016","updatedDate":"02/29/2016","status":"Active","moduleName":"Individual Portal"}
{"id":103,"announcementName":"d3mgcwtqhdu8003","announcementText":
But I want to match only from the id just before the required announcementName.
How can I do this in RegEx . Can someone please help me on this ?
As an answer here as well. Either use appropriate JSON functions, if not, a simple regex like:
"id":(\d+)
will probably do as the IDs are numeric.
Basically, I have list of URLs that look like this:
http://auctionnetwork.com.my/auctiondate_img.php?id=244003
and I want to extract auctiondate_244003, how would I do that with regex?
I want the output to be "auctiondate_244003".
You didn't define exactly what you are looking for.
How about something like: \/([^\/]*)img\.php\?id=([0-9]+)?
You'll have to concatenate the 1st and second match groups to get what you need
See https://regex101.com/r/dD2hW6/1