Develop Reference

Show all regex matches from String

I have the following string: aaabaaa
I want to match the following pattern: aba, aabaa, aaabaaa
Right now I have the following regex pattern: /(([a-z])\2*)((?!\2)[a-z])\1/g
It only matches aaabaaa but not the other ones. The pattern has to work for all characters and the characters can be surrounded by other characters.
Works for:
'mnonnopoo' -> ['non', 'opo']
'asasd' -> ['asa']
Doesn't work for:
'aaaasaaaaa' -> ['asa', 'aasaa', 'aaasaaa', 'aaaasaaaa']
'aaabaazsaaasbbabba' -> ['aba', 'aabaa', 'bab', 'bbabb']
regexr.com/63igv

As people indicated in the comment, regex don't match multiple times for the same string, not in javascript at least.
If you used the on-greedy regex,
/([a-z])((?!\1)[a-z])\1/g
it would find
'aaaasaaaaa' -> ['asa']
'aaabaazsaaasbbabba' -> ['aba', 'bab']
So you could use greedy regex
/([a-z]*?)((?!\1)[a-z])\1/g
Which would leave you with
'aaaasaaaaa' -> ['aaaasaaaa']
'aaabaazsaaasbbabba' -> ['aabaa', 'bbabb']
Then sub the results recursively to make the following regex:
/(?!.*bbabb.*)(?!.*aabaa.*)(?!.*aaaasaaaa.*)([a-z]*?)((?!\1)[a-z])\1/g
It then matches 'aaasaaa', 'aba', and 'bab' and you simply keep looping till there is no match.
Though I'm sure there are probably better ways to this without regex.

javascript regex insert new element into expression

I am passing a URL to a block of code in which I need to insert a new element into the regex. Pretty sure the regex is valid and the code seems right but no matter what I can't seem to execute the match for regex!
//** Incoming url's
//** url e.g. api/223344
//** api/11aa/page/2017
//** Need to match to the following
//** dir/api/12ab/page/1999
//** Hence the need to add dir at the front
var url = req.url;
//** pass in: /^\/api\/([a-zA-Z0-9-_~ %]+)(?:\/page\/([a-zA-Z0-9-_~ %]+))?$/
var re = myregex.toString();
//** Insert dir into regex: /^dir\/api\/([a-zA-Z0-9-_~ %]+)(?:\/page\/([a-zA-Z0-9-_~ %]+))?$/
var regVar = re.substr(0, 2) + 'dir' + re.substr(2);
var matchedData = url.match(regVar);
matchedData === null ? console.log('NO') : console.log('Yay');
I hope I am just missing the obvious but can anyone see why I can't match and always returns NO?
Thanks

Let's break down your regex
^\/api\/ this matches the beginning of a string, and it looks to match exactly the string "/api"
([a-zA-Z0-9-_~ %]+) this is a capturing group: this one specifically will capture anything inside those brackets, with the + indicating to capture 1 or more, so for example, this section will match abAB25-_ %
(?:\/page\/([a-zA-Z0-9-_~ %]+)) this groups multiple tokens together as well, but does not create a capturing group like above (the ?: makes it non-captuing). You are first matching a string exactly like "/page/" followed by a group exactly like mentioned in the paragraph above (that matches a-z, A-Z, 0-9, etc.
?$ is at the end, and the ? means capture 0 or more of the precending group, and the $ matches the end of the string
This regex will match this string, for example: /api/abAB25-_ %/page/abAB25-_ %
You may be able to take advantage of capturing groups, however, and use something like this instead to get similar results: ^\/api\/([a-zA-Z0-9-_~ %]+)\/page\/\1?$. Here, we are using \1 to reference that first capturing group and match exactly the same tokens it is matching. EDIT: actually, this probably won't work, since the text after /api/ and the text after /page/ will most likely be different, carrying on...
Afterwards, you are are adding "dir" to the beginning of your search, so you can now match someting like this: dir/api/abAB25-_ %/page/abAB25-_ %
You have also now converted the regex to a string, so like Crayon Violent pointed out in their comment, this will break your expected funtionality. You can fix this by using .source on your regex: var matchedData = url.match(regVar.source); https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp/source
Now you can properly match a string like this: dir/api/11aa/page/2017 see this example: https://repl.it/Mj8h

As mentioned by Crayon Violent in the comments, it seems you're passing a String rather than a regular expression in the .match() function. maybe try the following:
url.match(new RegExp(regVar, "i"));
to convert the string to a regular expression. The "i" is for ignore case; don't know that's what you want. Learn more here:
https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/RegExp

Test if a sentence is matching a text declaration using regex

I want to test if a sentence like type var1,var2,var3 is matching a text declaration or not.
So, I used the following code :
var text = "int a1,a2,a3",
reg = /int ((([a-z_A-Z]+[0-9]*),)+)$/g;
if (reg.test(text)) console.log(true);
else console.log(false)
The problem is that this regular expression returns false on text that is supposed to be true.
Could someone help me find a good regular expression matching expressions as in the example above?

You have a couple of mistekes.
As you wrote, the last coma is required at the end of the line.
I suppose you also want to match int abc123 as correct string, so you need to include letter to other characters
Avoid using capturing groups for just testing strings.
const str = 'int a1,a2,a3';
const regex = /int (?:[a-zA-Z_](?:[a-zA-Z0-9_])*(?:\,|$))+/g
console.log(regex.test(str));

You will need to add ? after the comma ,.
This token ? matches between zero and one.
Notice that the last number in your text a3 does not have , afterward.
int ((([a-z_A-Z]+[0-9]*),?)+)$

Extract specific chars from a string using a regex

I need to split an email address and take out the first character and the first character after the '#'
I can do this as follows:
'bar#foo'.split('#').map(function(a){ return a.charAt(0); }).join('')
--> bf
Now I was wondering if it can be done using a regex match, something like this
'bar#foo'.match(/^(\w).*?#(\w)/).join('')
--> bar#fbf
Not really what I want, but I'm sure I miss something here! Any suggestions ?

Why use a regex for this? just use indexOf to get the char at any given position:
var addr = 'foo#bar';
console.log(addr[0], addr[addr.indexOf('#')+1])
To ensure your code works on all browsers, you might want to use charAt instead of []:
console.log(addr.charAt(0), addr.charAt(addr.indexOf('#')+1));
Either way, It'll work just fine, and This is undeniably the fastest approach
If you are going to persist, and choose a regex, then you should realize that the match method returns an array containing 3 strings, in your case:
/^(\w).*?#(\w)/
["the whole match",//start of string + first char + .*?# + first string after #
"groupw 1 \w",//first char
"group 2 \w"//first char after #
]
So addr.match(/^(\w).*?#(\w)/).slice(1).join('') is probably what you want.

If I understand correctly, you are quite close. Just don't join everything returned by match because the first element is the entire matched string.
'bar#foo'.match(/^(\w).*?#(\w)/).splice(1).join('')
--> bf

Using regex:
matched="",
'abc#xyz'.replace(/(?:^|#)(\w)/g, function($0, $1) { matched += $1; return $0; });
console.log(matched);
// ax

The regex match function returns an array of all matches, where the first one is the 'full text' of the match, followed by every sub-group. In your case, it returns this:
bar#f
b
f
To get rid of the first item (the full match), use slice:
'bar#foo'.match(/^(\w).*?#(\w)/).slice(1).join('\r')

Use String.prototype.replace with regular expression:
'bar#foo'.replace(/^(\w).*#(\w).*$/, '$1$2'); // "bf"

Or using RegEx
^([a-zA-Z0-9])[a-zA-Z0-9.!#$%&'*+\/=?^_`{|}~-]+#([a-zA-Z0-9-])[a-zA-Z0-9-]+(?:\.[a-zA-Z0-9-]+)*$
Fiddle

JS XRegExp Replace all non characters

My objective is to replace all characters which are not dash (-) or not number or not letters in any language in a string.All of the #!()[], and all other signs to be replaced with empty string. All occurences of - should not be replaced also.
I have used for this the XRegExp plugin but it seems I cannot find the magic solution :)
I have tryed like this :
var txt = "Ad СТИНГ (ALI) - Englishmen In New York";
var regex = new XRegExp('\\p{^N}\\p{^L}',"g");
var b = XRegExp.replace(txt, regex, "")
but the result is : AСТИН(AL EnglishmeINeYork ... which is kind of weird
If I try to add also the condition for not removing the '-' character leads to make the RegEx invalid.

\\p{^N}\\p{^L} means a non-number followed by a non-letter.
Try [^\\p{N}\\p{L}-] that means a non-number, non-letter, non-dash.
A jsfiddle where to do some tests... The third XRegExp is the one you asked.

\p{^N}\p{^L}
is a non-number followed by a non-letter. You probably meant to say a character that is neither a letter nor a number:
[^\p{N}\p{L}]

// all non letters/numbers in a string => /[^a-zA-z0-9]/g
I dont know XRegExp.
but in js Regexp you can replace it by
b.replace(/[^a-zA-z0-9]/g,'')