RegEx to get ALL Strings between two Strings - javascript

I seem to have a love/hate relationship with RegEx in that I love how incredibly powerful it is, but at the same time, I don't quite understand all of the nuances of it yet.
I've got rather lengthy JSON feed that I need to parse and capture ALL of the matches between two specific strings. I've included a link to the regex101.com example with a few of the JSON results.
regex101.com Example
I'm trying to match every string between each /content/usergenerated and /jcr:content
...
I guess what I should really be trying to match is a string that starts with /content/webAppName/en/home and ends before /jcr:content
The path that I care about will always start with /content/webAppName/en/home

you have to use "positive look-ahead" that match a sequence of digits if they are followed by something
https://regex101.com/r/fU1iD1/4

Just wrap the two things you're looking to remove in parenthesis, and then remove them from the output. So...
(\/content\/usergenerated)(.*)(\/jcr\:content)
replaced by
/2
Which is everything in the middle of those two.
edit: Sorry, didn't look at your example :) - there was a deleted answer that said to add the g modifier, which looks like it works.

/content/usergenerated/content/webAppName/en/home([a-zA-Z/-]+)/jcr:content
This should work. It matches 3 out of 4 don't know why it doesn't match one of em. You could use exec() in a loop till it returns null and get hold of the object[1] which contains data for the first and only capture group.
all the best.
PS: I used gmi in options for the regex.

Related

regex pattern a then b then a, one or more times

I want to use regex to find a repeating pattern but I can't work out how to do it. I want to get all matches where there is a number then a plus symbol then another number, including repeats.
So for example, if the string is "5+2-6+10+3", I want to end up with ["5+2", "6+10+3"]
So far I've got
\d*\+\d*
But that doesn't capture the final "+3" in the example above. I had a few attempts using brackets for capturing groups but I couldn't get the output I wanted.
Any help would be much appreciated!
Solved using the following (thanks The fourth bird):
\d+(?:\+\d+)+

Javascript substring check using indexOf or search on a date string with forward slash /

I am surprised to not to find any post regarding this, I must be missing something very trivial. I have a small JavaScript function to check if a string matches an object's properties. Simple stuff right? It works easily with all strings except those which contain a forward slash.
"‎04‎/‎08‎/‎2015‎".indexOf('4') // returns 2 :good
"‎04‎/‎08‎/‎2015‎".indexOf('4/') // returns -1 :why?
The same issue appears to be with .search() function as well. I encountered this issue while working on date strings.
Please note that I don't want to use regex based solution for performance reasons. Thanks for your help in advance!
Your string has invisible Unicode characters in it. The "left-to-right mark" (hex 200E) appears around the two slash characters as well as at the beginning and the end of the string.
If you type the code in on your browser console instead of cutting and pasting, you'll see that it works as expected.

Capturing optional part of URL with RegExp

While writing an API service for my site, I realized that String.split() won't do it much longer, and decided to try my luck with regular expressions. I have almost done it but I can't find the last bit. Here is what I want to do:
The URL represents a function call:
/api/SECTION/FUNCTION/[PARAMS]
This last part, including the slash, is optional. Some functions display a JSON reply without having to receive any arguments. Example: /api/sounds/getAllSoundpacks prints a list of available sound packs. Though, /api/sounds/getPack/8Bit prints the detailed information.
Here is the expression I have tried:
req.url.match(/\/(.*)\/(.*)\/?(.*)/);
What am I missing to make the last part optional - or capture it in whole?
This will capture everything after FUNCTION/ in your URL, independent of the appearance of any further / after FUNCTION/:
FUNCTION\/(.+)$
The RegExp will not match if there is no part after FUNCTION.
This regex should work by making last slash and part after optional:
/^\/[^/]*\/[^/]*(?:\/.*)?$/
This matches all of these strings:
/api/SECTION/FUNCTION/abc
/api/SECTION
/api/SECTION/
/api/SECTION/FUNCTION
Your pattern /(.*)/(.*)/?(.*) was almost correct, it's just a bit too short - it allows 2 or 3 slashes, but you want to accept anything with 3 or 4 slashes. And if you want to capture the last (optional) slash AND any text behind it as a whole, you simply need to create a group around that section and make it optional:
/.*/.*/.*(?:/.+)?
should do the trick.
Demo. (The pattern looks different because multiline mode is enabled, but it still works. It's also a little "better" because it won't match garbage like "///".)

What's wrong with this regular expression to find URLs?

I'm working on a JavaScript to extract a URL from a Google search URL, like so:
http://www.google.com/search?client=safari&rls=en&q=thisisthepartiwanttofind.org&ie=UTF-8&oe=UTF-8
Right now, my code looks like this:
var checkForURL = /[\w\d](.org)/i;
var findTheURL = checkForURL.exec(theURL);
I've ran this through a couple regex testers and it seems to work, but in practice the string I get returned looks like this:
thisisthepartiwanttofind.org,.org
So where's that trailing ,.org coming from?
I know my pattern isn't super robust but please don't suggest better patterns to use. I'd really just like advice on what in particular I did wrong with this one. Thanks!
Remove the parentheses in the regex if you do not process the .org (unlikely since it is a literal). As per #Mark comment, add a + to match one or more characters of the class [\w\d]. Also, I would escape the dot:
var checkForURL = /[\w\d]+\.org/i;
What you're actually getting is an array of 2 results, the first being the whole match, the second - the group you defined by using parens (.org).
Compare with:
/([\w\d]+)\.org/.exec('thisistheurl.org')
→ ["thisistheurl.org", "thisistheurl"]
/[\w\d]+\.org/.exec('thisistheurl.org')
→ ["thisistheurl.org"]
/([\w\d]+)(\.org)/.exec('thisistheurl.org')
→ ["thisistheurl.org", "thisistheurl", ".org"]
The result of an .exec of a JS regex is an Array of strings, the first being the whole match and the subsequent representing groups that you defined by using parens. If there are no parens in the regex, there will only be one element in this array - the whole match.
You should escape .(DOT) in (.org) regex group or it matches any character. So your regex would become:
/[\w\d]+(\.org)/
To match the url in your example you can use something like this:
https?://([0-9a-zA-Z_.?=&\-]+/?)+
or something more accurate like this (you should choose the right regex according to your needs):
^https?://([0-9a-zA-Z_\-]+\.)+(com|org|net|WhatEverYouWant)(/[0-9a-zA-Z_\-?=&.]+)$

How to match between characters but not include them in the result

Say I have a string "&something=variable&something_else=var2"
I want to match between &something= and &, so I'll write a regular expression that looks like:
/(&something=).*?(&)/
And the result of .match() will be an array:
["&something=variable&", "&something=", "&"]
I've always solved this by just replacing the start and end elements manually but is there a way to not include them in the match results at all?
You're using the wrong capturing groups. You should be using this:
/&something=(.*?)&/
This means that instead of capturing the stuff you don't want (the delimiters), you capture what you do want (the data).
You can't avoid them showing up in your match results at all, but you can change how they show up and make it more useful for you.
If you change your match pattern to /&something=(.+?)&/ then using your test string of "&something=variable&something_else=var2" the match result array is ["&something=variable&", "variable"]
The first element is always the entire match, but the second one, will be the captured portion from the parentheses, which is much more useful, generally.
I hope this helps.
If you are trying to get variable out of the string, using replace with backreferences will get you what you want:
"&something=variable&something_else=var2".replace(/^.*&something=(.*?)&.*$/, '$1')
gives you
"variable"

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