I'm having a problem with the regex that work in Expresso but not in the javascript code. I know that there are also two other question with the same topic, but can't figure out how to implement an alternative to my regex so that it will work also in javascript.
So my expression is:
((?<=\bvar)\s\w*\s?)=\s?function(\s*\([^)]*\))
Used in javascript like that:
var functionsReg = /((?<=\bvar)\s\w*\s?)=\s?function(\s*\([^)]*\))/gm;
var match = functionsReg.exec(func);
and I'm expecting to be able to get for the values for each regex group.
like:
var name = match[0];
var params = match[1];
I found out that the problem is from the (?<=\bvar)
But I can not find the alternative for this syntax because in the end I want to be able to access the text like in the below image.
Just match the var in a non-capture group:
/(?:var)\s+(\w+)\s*=\s*function(\s*\([^)]*\))/g
Related
Hi,
I have this code:
var room = 'room2';
var exitroom = 'room1,room2,room3';
exitroom = exitroom.replace(/,${room},/,'');
console.log(exitroom);
you can try it here: https://jsfiddle.net/uq9w0Ls4/
my expected output is simply room1,room3 by taking room2 out but since it may change its position within the string I want to target the , no matter if it comes before or after the string but I cant figure out the regex logic here. I know I could just do simply:
var room = 'room2';
var exitroom = 'room1,room2,room3';
exitroom = exitroom.replace(room+',','').replace(','+room,'');
console.log(exitroom);
which works but I think regex would be a more direct approach.
Thank you.
First, by writing .replace(/,${room},/,'') you are not using the variable room.
To use a variable in a regex you should call new RegExp()
Second, if you want a regex that will match when the comma is before or after the word, you can use a group () with an Or | operator.
so it should look like this:
var reg = new RegExp(`(?:${room},|,${room})`, "g");
exitroom.replace(reg,'');
The ?: at the beginning of the group, is just so it should be a non-capturing group, it should work just fine also without it
Problem:
Extract image file name from CDN address similar to the following:
https://cdnstorage.api.com/v0/b/my-app.com/o/photo%2FB%_2.jpeg?alt=media&token=4e32-a1a2-c48e6c91a2ba
Two-stage Solution:
I am using two regular expressions to retrieve the file name:
var postLastSlashRegEx = /[^\/]+$/,
preQueryRegEx = /^([^?]+)/;
var fileFromURL = urlString.match(postLastSlashRegEx)[0].match(preQueryRegEx)[0];
// fileFromURL = "photo%2FB%_2.jpeg"
Question:
Is there a way I can combine both regular expressions?
I've tried using capture groups, but haven't been able to produce a working solution.
From my comment
You can use a lookahead to find the "?" and use [^/] to match any non-slash characters.
/[^/]+(?=\?)/
To remove the dependency on the URL needing a "?", you can make the lookahead match a question mark or the end of line indicator (represented by $), but make sure the first glob is non-greedy.
/[^/]+?(?=\?|$)/
You don't have to use regex, you can just use split and substr.
var str = "https://cdnstorage.api.com/v0/b/my-app.com/o/photo%2FB%_2.jpeg?alt=media&token=4e32-a1a2-c48e6c91a2ba".split("?")[0];
var fileName = temp.substr(temp.lastIndexOf('/')+1);
but if regex is important to you, then:
str.match(/[^?]*\/([^?]+)/)[1]
The code using the substring method would look like the following -
var fileFromURL = urlString.substring(urlString.lastIndexOf('/') + 1, urlString.lastIndexOf('?'))
I want to match the First url followed by a space using regex expression while typing in the input box.
For example :
if I type www.google.com it should be matched only after a space followed by the url
ie www.google.com<SPACE>
Code
$(".site").keyup(function()
{
var site=$(this).val();
var exp = /^http(s?):\/\/(\w+:{0,1}\w*)?(\S+)(:[0-9]+)?(\/|\/([\w#!:.?+=&%#!\-\/]))?/;
var find = site.match(exp);
var url = find? find[0] : null;
if (url === null){
var exp = /[-\w]+(\.[a-z]{2,})+(\S+)?(\/|\/[\w#!:.?+=&%#!\-\/])?/g;
var find = site.match(exp);
url = find? 'http://'+find[0] : null;
}
});
Fiddle
Please help, Thanks in advance
you should be using a better regex to correctly match the query & fragment parts of your url. Have a look here (What is the best regular expression to check if a string is a valid URL?) for a correct IRI/URI structured Regex test.
But here's a rudimentary version:
var regex = /[-\w]+(\.[a-z]{2,})+(\/?)([^\s]+)/g;
var text = 'test google.com/?q=foo basdasd www.url.com/test?q=asdasd#cheese something else';
console.log(text.match(regex));
Expected Result:
["google.com/?q=foo", "www.url.com/test?q=asdasd#cheese"]
If you really want to check for URLs, make sure you include scheme, port, username & password checks just to be safe.
In the context of what you're trying to achieve, you should really put in some delay so that you don't impact browser performance. Regex tests can be expensive when you use complex rules especially so when running the same rule every time a new character is entered. Just think about what you're trying to achieve and whether or not there's a better solution to get there.
With a lookahead:
var exp = /[-\w]+(\.[a-z]{2,})+(\S+)?(\/|\/[\w#!:.?+=&%#!\-\/])?(?= )/g;
I only added this "(?= )" to your regex.
Fiddle
I've got some string that contain invisible characters, but they are in somewhat predictable places. Typically the surround the piece of text I want to extract, and then after the 2nd occurrence I want to keep the rest of the text.
I can't seem to figure out how to both key off of the invisible characters, and exclude them from my result. To match invisibles I've been using this regex: /\xA0\x00-\x09\x0B\x0C\x0E-\x1F\x7F/ which does seem to work.
Here's an example: [invisibles]Keep as match 1[invisibles]Keep as match 2
Here's what I've been using so far without success:
/([\xA0\x00-\x09\x0B\x0C\x0E-\x1F\x7F]+)(.+)([\xA0\x00-\x09\x0B\x0C\x0E-\x1F\x7F]+)/(.+)
I've got the capture groups in there, but it's bee a while since I've had to use regex's in this way, so I know I'm missing something important. I was hoping to just make the invisible matches non-capturing groups, but it seems that JavaScript does not support this.
Something like this seems like what you want. The second regex you have pretty much works, but the / is in totally the wrong place. Perhaps you weren't properly reading out the group data.
var s = "\x0EKeep as match 1\x0EKeep as match 2";
var r = /[\xA0\x00-\x09\x0B\x0C\x0E-\x1F\x7F]+(.+)[\xA0\x00-\x09\x0B\x0C\x0E-\x1F\x7F]+(.+)/;
var match = s.match(r);
var part1 = match[1];
var part2 = match[2];
I am complete novice at regex and Javascript. I have the following problem: need to check into a textfield the existence of one (1) or many (n) consecutive * (asterisk) character/characters eg. * or ** or *** or infinite (n) *. Strings allowed eg. *tomato or tomato* or **tomato or tomato** or as many(n)*tomato many(n)*. So, far I had tried the following:
var str = 'a string'
var value = encodeURIComponent(str);
var reg = /([^\s]\*)|(\*[^\s])/;
if (reg.test(value) == true ) {
alert ('Watch out your asterisks!!!')
}
By your question it's hard to decipher what you're after... But let me try:
Only allow asterisks at beginning or at end
If you only allow an arbitrary number (at least one) of asterisks either at the beginning or at the end (but not on both sides) like:
*****tomato
tomato******
but not **tomato*****
Then use this regular expression:
reg = /^(?:\*+[^*]+|[^*]+\*+)$/;
Match front and back number of asterisks
If you require that the number of asterisks at the biginning matches number of asterisks at the end like
*****tomato*****
*tomato*
but not **tomato*****
then use this regular expression:
reg = /^(\*+)[^*]+\1$/;
Results?
It's unclear from your question what the results should be when each of these regular expressions match? Are strings that test positive to above regular expressions fine or wrong is on you and your requirements. As long as you have correct regular expressions you're good to go and provide the functionality you require.
I've also written my regular expressions to just exclude asterisks within the string. If you also need to reject spaces or anything else simply adjust the [^...] parts of above expressions.
Note: both regular expressions are untested but should get you started to build the one you actually need and require in your code.
If I understand correctly you're looking for a pattern like this:
var pattern = /\**[^\s*]+\**/;
this won't match strings like ***** or ** ***, but will match ***d*** *d or all of your examples that you say are valid (***tomatos etc).If I misunderstood, let me know and I'll see what I can do to help. PS: we all started out as newbies at some point, nothing to be ashamed of, let alone apologize for :)
After the edit to your question I gather the use of an asterisk is required, either at the beginning or end of the input, but the string must also contain at least 1 other character, so I propose the following solution:
var pattern = /^\*+[^\s*]+|[^\s*]+\*+$/;
'****'.match(pattern);//false
' ***tomato**'.match(pattern);//true
If, however *tomato* is not allowed, you'll have to change the regex to:
var pattern = /^\*+[^\s*]+$|^[^\s*]+\*+$/;
Here's a handy site to help you find your way in the magical world of regular expressions.