Develop Reference

Regex match cookie value and remove hyphens

I'm trying to extract out a group of words from a larger string/cookie that are separated by hyphens. I would like to replace the hyphens with a space and set to a variable. Javascript or jQuery.
As an example, the larger string has a name and value like this within it:
facility=34222%7CConner-Department-Store;
(notice the leading "C")
So first, I need to match()/find facility=34222%7CConner-Department-Store; with regex. Then break it down to "Conner Department Store"
var cookie = document.cookie;
var facilityValue = cookie.match( REGEX ); ??

var test = "store=874635%7Csomethingelse;facility=34222%7CConner-Department-Store;store=874635%7Csomethingelse;";
var test2 = test.replace(/^(.*)facility=([^;]+)(.*)$/, function(matchedString, match1, match2, match3){
return decodeURIComponent(match2);
});
console.log( test2 );
console.log( test2.split('|')[1].replace(/[-]/g, ' ') );

If I understood it correctly, you want to make a phrase by getting all the words between hyphens and disallowing two successive Uppercase letters in a word, so I'd prefer using Regex in that case.
This is a Regex solution, that works dynamically with any cookies in the same format and extract the wanted sentence from it:
var matches = str.match(/([A-Z][a-z]+)-?/g);
console.log(matches.map(function(m) {
return m.replace('-', '');
}).join(" "));
Demo:
var str = "facility=34222%7CConner-Department-Store;";
var matches = str.match(/([A-Z][a-z]+)-?/g);
console.log(matches.map(function(m) {
return m.replace('-', '');
}).join(" "));
Explanation:
Use this Regex (/([A-Z][a-z]+)-?/g to match the words between -.
Replace any - occurence in the matched words.
Then just join these matches array with white space.

Ok,
first, you should decode this string as follows:
var str = "facility=34222%7CConner-Department-Store;"
var decoded = decodeURIComponent(str);
// decoded = "facility=34222|Conner-Department-Store;"
Then you have multiple possibilities to split up this string.
The easiest way is to use substring()
var solution1 = decoded.substring(decoded.indexOf('|') + 1, decoded.length)
// solution1 = "Conner-Department-Store;"
solution1 = solution1.replace('-', ' ');
// solution1 = "Conner Department Store;"
As you can see, substring(arg1, arg2) returns the string, starting at index arg1 and ending at index arg2. See Full Documentation here
If you want to cut the last ; just set decoded.length - 1 as arg2 in the snippet above.
decoded.substring(decoded.indexOf('|') + 1, decoded.length - 1)
//returns "Conner-Department-Store"
or all above in just one line:
decoded.substring(decoded.indexOf('|') + 1, decoded.length - 1).replace('-', ' ')
If you want still to use a regular Expression to retrieve (perhaps more) data out of the string, you could use something similar to this snippet:
var solution2 = "";
var regEx= /([A-Za-z]*)=([0-9]*)\|(\S[^:\/?#\[\]\#\;\,']*)/;
if (regEx.test(decoded)) {
solution2 = decoded.match(regEx);
/* returns
[0:"facility=34222|Conner-Department-Store",
1:"facility",
2:"34222",
3:"Conner-Department-Store",
index:0,
input:"facility=34222|Conner-Department-Store;"
length:4] */
solution2 = solution2[3].replace('-', ' ');
// "Conner Department Store"
}
I have applied some rules for the regex to work, feel free to modify them according your needs.
facility can be any Word built with alphabetical characters lower and uppercase (no other chars) at any length
= needs to be the char =
34222 can be any number but no other characters
| needs to be the char |
Conner-Department-Store can be any characters except one of the following (reserved delimiters): :/?#[]#;,'
Hope this helps :)
edit: to find only the part
facility=34222%7CConner-Department-Store; just modify the regex to
match facility= instead of ([A-z]*)=:
/(facility)=([0-9]*)\|(\S[^:\/?#\[\]\#\;\,']*)/

You can use cookies.js, a mini framework from MDN (Mozilla Developer Network).
Simply include the cookies.js file in your application, and write:
docCookies.getItem("Connor Department Store");

Searching string from every position with regex so it finds overlapping matches

My regex doesn't find overlapping matches, is there a way to do find them?
Example regex that illustrates the problem: /(\d\*x\^(\d))\+(\d\*x\^(\d))/g
String to perform this regex on: 3*x^2+2*x^3+6*x^2
Matches: 3*x^2+2*x^3
Desired matches: 3*x^2+2*x^3, 2*x^3+6*x^2
I want to use this in a replace function, and check if the first digit is greater than the second, if that is the case, it should switch them around. Because it doesn't match overlapping matches, it doesn't work.
I checked another answer about this, but don't know how to implement it with the replace function in JavaScript.
I've got this so far:
var string1 = "3*x^2+2*x^3+6*x^2";
var newString1 = string1.replace(/(\d\*x\^(\d))\+(\d\*x\^(\d))/g, function (piece, $1, $2, $3, $4) {
if ($2 > $4) {
return $3 + "+" + $1;
} else {
return piece;
}
});
Or, if you know a better tool to solve this problem instead of regexp, like mentioned in the comments, please provide a solution using that tool.

The snippet below just uses split, reverse, join.
It orders the pieces between the '+' based on the reversed pieces.
Thus sorts them by the exponent.
For example 3*x^2+2*x^3+6*x^2 becomes 3*x^2+6*x^2+2*x^3
var str = "3*x^2+2*x^3+6*x^2";
var result = str
.split('').reverse().join('') // reverse the string
.split('+') // split into an array by the +
.sort().reverse() // sort the array descending
.join('+') // join the array by + to a string
.split('').reverse().join(''); // reverse the string
console.log(result);
As for how to use regex to get each piece and the piece after this?
That can be done via a positive lookahead that has a capture group in it.
var str = "3*x^2+2*x^3+6*x^2";
var re = /(\d\*x\^(\d))(?=\+?(\d\*x\^(\d))?)/g;
var m;
while (m = re.exec(str)) {
console.log(m[1], "exponent:"+m[2], m[3], "exponent:"+m[4]);
}

What's the JS RegExp for this specific string?

I have a rather isolated situation in an inventory management program where our shelf locations have a specific format, which is always Letter: Number-Letter-Number, such as Y: 1-E-4. Most of us coworkers just type in "y1e4" and are done with it, but that obviously creates issues with inconsistent formats in a database. Are JS RegExp's the ideal way to automatically detect and format these alphanumeric strings? I'm slowly wrapping my head around JavaScript's Perl syntax, but what's a simple example of formatting one of these strings?

spec: detect string format of either "W: D-W-D" or "WDWD" and return "W: D-W-D"
This function will accept any format and return undefined if it doesnt match, returns the formatted string if a match does occur.
function validateInventoryCode(input) {
var regexp = /^([a-zA-Z]+)(?:\:\s*)?(\d+)-?(\w+)-?(\d+)$/
var r = regexp.exec(input);
if(r != null) {
return `${r[1]}: ${r[2]}-${r[3]}-${r[4]}`;
}
}
var possibles = ["y1e1", "y:1e1", "Y: 1r3", "y: 32e4", "1:e3e"];
possibles.forEach(function(posssiblity) {
console.log(`input(${posssiblity}), result(${validateInventoryCode(posssiblity)})`);
})
function validateInventoryCode(input) {
var regexp = /^([a-zA-Z]+)(?:\:\s*)?(\d+)-?(\w+)-?(\d+)$/
var r = regexp.exec(input);
if (r != null) {
return `${r[1]}: ${r[2]}-${r[3]}-${r[4]}`;
}
}

I understand the question as "convert LetterNumberLetterNumber to Letter: Number-Letter-Number.
You may use
/^([a-z])(\d+)([a-z])(\d+)$/i
and replace with $1: $2-$3-$4
Details:
^ - start of string
([a-z]) - Group 1 (referenced with $1 from the replacement pattern) capturing any ASCII letter (as /i makes the pattern case-insensitive)
(\d+) - Group 2 capturing 1 or more digits
([a-z]) - Group 3, a letter
(\d+) - Group 4, a number (1 or more digits)
$ - end of string.
See the regex demo.
var re = /^([a-z])(\d+)([a-z])(\d+)$/i;
var s = 'y1e2';
var result = s.replace(re, '$1: $2-$3-$4');
console.log(result);
OR - if the letters must be turned to upper case:
var re = /^([a-z])(\d+)([a-z])(\d+)$/i;
var s = 'y1e2';
var result = s.replace(re,
(m,g1,g2,g3,g4)=>`${g1.toUpperCase()}: ${g2}-${g3.toUpperCase()}-${g4}`
);
console.log(result);

this is the function to match and replace the pattern: DEMO
function findAndFormat(text){
var splittedText=text.split(' ');
for(var i=0, textLength=splittedText.length; i<textLength; i++){
var analyzed=splittedText[i].match(/[A-z]{1}\d{1}[A-z]{1}\d{1}$/);
if(analyzed){
var formattedString=analyzed[0][0].toUpperCase()+': '+analyzed[0][1]+'-'+analyzed[0][2].toUpperCase()+'-'+analyzed[0][3];
text=text.replace(splittedText[i],formattedString);
}
}
return text;
}

i think it's just as it reads:
y1e4
Letter, number, letter, number:
/([A-z][0-9][A-z][0-9])/g
And yes, it's ok to use regex in this case, like form validations and stuff like that. it's just there are some cases on which abusing of regular expressions gives you a bad performance (into intensive data processing and the like)
Example
"HelloY1E4world".replace(/([A-z][0-9][A-z][0-9])/g, ' ');
should return: "Hello world"
regxr.com always comes in handy

How can I split this string in JavaScript?

I have strings like this:
ab
rx'
wq''
pok'''
oyu,
mi,,,,
Basically, I want to split the string into two parts. The first part should have the alphabetical characters intact, the second part should have the non-alphabetical characters.
The alphabetical part is guaranteed to be 2-3 lowercase characters between a and z; the non-alphabetical part can be any length, and is gauranteed to only be the characters , or ', but not both in the one string (e.g. eex,', will never occur).
So the result should be:
[ab][]
[rx][']
[wq]['']
[pok][''']
[oyu][,]
[mi][,,,,]
How can I do this? I'm guessing a regular expression but I'm not particularly adept at coming up with them.

Regular expressions have is a nice special called "word boundary" (\b). You can use it, well, to detect the boundary of a word, which is a sequence of alpha-numerical characters.
So all you have to do is
foo.split(/\b/)
For example,
"pok'''".split(/\b/) // ["pok", "'''"]

If you can 100% guarantee that:
Letter-strings are 2 or 3 characters
There are always one or more primes/commas
There is never any empty space before, after or in-between the letters and the marks
(aside from line-break)
You can use:
/^([a-zA-Z]{2,3})('+|,+)$/gm
var arr = /^([a-zA-Z]{2,3})('+|,+)$/gm.exec("pok'''");
arr === ["pok'''", "pok", "'''"];
var arr = /^([a-zA-Z]{2,3})('+|,+)$/gm.exec("baf,,,");
arr === ["baf,,,", "baf", ",,,"];
Of course, save yourself some sanity, and save that RegEx as a var.
And as a warning, if you haven't dealt with RegEx like this:
If a match isn't found -- if you try to match foo','' by mixing marks, or you have 0-1 or 4+ letters, or 0 marks... ...then instead of getting an array back, you'll get null.
So you can do this:
var reg = /^([a-zA-Z]{2,3})('+|,+)$/gm,
string = "foobar'',,''",
result_array = reg.exec(string) || [string];
In this case, the result of the exec is null; by putting the || (or) there, we can return an array that has the original string in it, as index-0.
Why?
Because the result of a successful exec will have 3 slots; [*string*, *letters*, *marks*].
You might be tempted to just read the letters like result_array[1].
But if the match failed and result_array === null, then JavaScript will scream at you for trying null[1].
So returning the array at the end of a failed exec will allow you to get result_array[1] === undefined (ie: there was no match to the pattern, so there are no letters in index-1), rather than a JS error.

You could try something like that:
function splitString(string){
var match1 = null;
var match2 = null;
var stringArray = new Array();
match1 = string.indexOf(',');
match2 = string.indexOf('`');
if(match1 != 0){
stringArray = [string.slice(0,match1-1),string.slice(match1,string.length-1];
}
else if(match2 != 0){
stringArray = [string.slice(0,match2-1),string.slice(match2,string.length-1];
}
else{
stringArray = [string];
}
}

var str = "mi,,,,";
var idx = str.search(/\W/);
if(idx) {
var list = [str.slice(0, idx), str.slice(idx)]
}
You'll have the parts in list[0] and list[1].
P.S. There might be some better ways than this.

yourStr.match(/(\w{2,3})([,']*)/)

if (match = string.match(/^([a-z]{2,3})(,+?$|'+?$)/)) {
match = match.slice(1);
}

Regex using javascript to return just numbers

If I have a string like "something12" or "something102", how would I use a regex in javascript to return just the number parts?

Regular expressions:
var numberPattern = /\d+/g;
'something102asdfkj1948948'.match( numberPattern )
This would return an Array with two elements inside, '102' and '1948948'. Operate as you wish. If it doesn't match any it will return null.
To concatenate them:
'something102asdfkj1948948'.match( numberPattern ).join('')
Assuming you're not dealing with complex decimals, this should suffice I suppose.

You could also strip all the non-digit characters (\D or [^0-9]):
let word_With_Numbers = 'abc123c def4567hij89'
let word_Without_Numbers = word_With_Numbers.replace(/\D/g, '');
console.log(word_Without_Numbers)

For number with decimal fraction and minus sign, I use this snippet:
const NUMERIC_REGEXP = /[-]{0,1}[\d]*[.]{0,1}[\d]+/g;
const numbers = '2.2px 3.1px 4px -7.6px obj.key'.match(NUMERIC_REGEXP)
console.log(numbers); // ["2.2", "3.1", "4", "-7.6"]
Update: - 7/9/2018
Found a tool which allows you to edit regular expression visually: JavaScript Regular Expression Parser & Visualizer.
Update:
Here's another one with which you can even debugger regexp: Online regex tester and debugger.
Update:
Another one: RegExr.
Update:
Regexper and Regex Pal.

If you want only digits:
var value = '675-805-714';
var numberPattern = /\d+/g;
value = value.match( numberPattern ).join([]);
alert(value);
//Show: 675805714
Now you get the digits joined

I guess you want to get number(s) from the string. In which case, you can use the following:
// Returns an array of numbers located in the string
function get_numbers(input) {
return input.match(/[0-9]+/g);
}
var first_test = get_numbers('something102');
var second_test = get_numbers('something102or12');
var third_test = get_numbers('no numbers here!');
alert(first_test); // [102]
alert(second_test); // [102,12]
alert(third_test); // null

IMO the #3 answer at this time by Chen Dachao is the right way to go if you want to capture any kind of number, but the regular expression can be shortened from:
/[-]{0,1}[\d]*[\.]{0,1}[\d]+/g
to:
/-?\d*\.?\d+/g
For example, this code:
"lin-grad.ient(217deg,rgba(255, 0, 0, -0.8), rgba(-255,0,0,0) 70.71%)".match(/-?\d*\.?\d+/g)
generates this array:
["217","255","0","0","-0.8","-255","0","0","0","70.71"]
I've butchered an MDN linear gradient example so that it fully tests the regexp and doesn't need to scroll here. I think I've included all the possibilities in terms of negative numbers, decimals, unit suffixes like deg and %, inconsistent comma and space usage, and the extra dot/period and hyphen/dash characters within the text "lin-grad.ient". Please let me know if I'm missing something. The only thing I can see that it does not handle is a badly formed decimal number like "0..8".
If you really want an array of numbers, you can convert the entire array in the same line of code:
array = whatever.match(/-?\d*\.?\d+/g).map(Number);
My particular code, which is parsing CSS functions, doesn't need to worry about the non-numeric use of the dot/period character, so the regular expression can be even simpler:
/-?[\d\.]+/g

var result = input.match(/\d+/g).join([])

Using split and regex :
var str = "fooBar0123".split(/(\d+)/);
console.log(str[0]); // fooBar
console.log(str[1]); // 0123

The answers given don't actually match your question, which implied a trailing number. Also, remember that you're getting a string back; if you actually need a number, cast the result:
item=item.replace('^.*\D(\d*)$', '$1');
if (!/^\d+$/.test(item)) throw 'parse error: number not found';
item=Number(item);
If you're dealing with numeric item ids on a web page, your code could also usefully accept an Element, extracting the number from its id (or its first parent with an id); if you've an Event handy, you can likely get the Element from that, too.

As per #Syntle's answer, if you have only non numeric characters you'll get an Uncaught TypeError: Cannot read property 'join' of null.
This will prevent errors if no matches are found and return an empty string:
('something'.match( /\d+/g )||[]).join('')

Here is the solution to convert the string to valid plain or decimal numbers using Regex:
//something123.777.321something to 123.777321
const str = 'something123.777.321something';
let initialValue = str.replace(/[^0-9.]+/, '');
//initialValue = '123.777.321';
//characterCount just count the characters in a given string
if (characterCount(intitialValue, '.') > 1) {
const splitedValue = intitialValue.split('.');
//splittedValue = ['123','777','321'];
intitialValue = splitedValue.shift() + '.' + splitedValue.join('');
//result i.e. initialValue = '123.777321'
}

If you want dot/comma separated numbers also, then:
\d*\.?\d*
or
[0-9]*\.?[0-9]*
You can use https://regex101.com/ to test your regexes.

Everything that other solutions have, but with a little validation
// value = '675-805-714'
const validateNumberInput = (value) => {
let numberPattern = /\d+/g
let numbers = value.match(numberPattern)
if (numbers === null) {
return 0
}
return parseInt(numbers.join([]))
}
// 675805714

One liner
I you do not care about decimal numbers and only need the digits, I think this one liner is rather elegant:
/**
* #param {String} str
* #returns {String} - All digits from the given `str`
*/
const getDigitsInString = (str) => str.replace(/[^\d]*/g, '');
console.log([
'?,!_:/42\`"^',
'A 0 B 1 C 2 D 3 E',
' 4 twenty 20 ',
'1413/12/11',
'16:20:42:01'
].map((str) => getDigitsInString(str)));
Simple explanation:
\d matches any digit from 0 to 9
[^n] matches anything that is not n
* matches 0 times or more the predecessor
( It is an attempt to match a whole block of non-digits all at once )
g at the end, indicates that the regex is global to the entire string and that we will not stop at the first occurrence but match every occurrence within it
Together those rules match anything but digits, which we replace by an empty strings. Thus, resulting in a string containing digits only.