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");

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

javascript regular expression test for 6 digit numbers only. comma seperated

and so this must pass:
454555, 939999 , 019999 ,727663
its for a user entering 6 digit invoice numbers. it should fail if a number is 5 or 7 digit and not 6. so 1234567, 123456 should fail, as one set is more than 6 numbers.
So far I have :
[0-9]{6}(\s*,*,\s*[0-9]{6})*
which only draw back is that it accepts 7 or more digit numbers. cant figure out if its even possible at this point to do both, test for 6 digits separated by a comma and one or more space, and all the digits have to be only 6 digits and fail if one is not.
any help appreciated. regular expressions are not my forte.
thanks
Norman

You can write it using regex like the function below.
const isPassword = (password: string) => /^\d{6}$/gm.test(password);
And here is an example test file below.
test('should recognize a valid password', () => {
expect(isPassword('123456')).toBe(true);
expect(isPassword('000000')).toBe(true);
});
test('should recognize an invalid password', () => {
expect(isPassword('asdasda1234')).toBe(false);
expect(isPassword('1234567')).toBe(false);
expect(isPassword('a123456a')).toBe(false);
expect(isPassword('11.11.11')).toBe(false);
expect(isPassword('aaaaaa')).toBe(false);
expect(isPassword('eeeeee')).toBe(false);
expect(isPassword('......')).toBe(false);
expect(isPassword('werwerwerwr')).toBe(false);
});

In order to validate the full string you can use this regex.
^(\s*\d{6}\s*)(,\s*\d{6}\s*)*,?\s*$
It works with six digits only, and you have to enter at least one 6 digit number.
It also works if you have a trailing comma with whitespaces.

It's accepting more than six digit numbers because you're not anchoring the text, and for some odd reason you're optionally repeating the comma. Try something like this:
^[0-9]{6}(?:\s*,\s*[0-9]{6})*$
Also note that [0-9] is equivalent to \d, so this can be rewritten more concisely as:
^\d{6}(?:\s*,\s*\d{6})*$

Your regex does not match 7 digits in a row, but it also doesn't enforce that it matches the whole string. It just has to match some substring in the string, so it would also match each of these:
"1234512345612345612345"
"NaNaNaN 123456, 123456 BOOO!"
"!##$%^&*({123456})*&^%$##!"
Just add the start of string (^) and end of string ($) anchors to enforce that the whole string matches and it will work correctly:
^[0-9]{6}(\s*,*,\s*[0-9]{6})*$
Also note that ,*, could be shortened to ,+, and if you only want one comma in a row, just use ,, not ,* or ,+.
You can also replace [0-9] with \d:
^\d{6}(\s*,\s*\d{6})*$

Using only regex:
var commaSeparatedSixDigits = /^(?:\d{6}\s*,\s*)*\d{6}$/;
if (myInput.test(commaSeparatedSixDigits)) console.log( "Is good!" );
This says:
^ - Starting at the beginning of the string
(?:…)* - Find zero or more of the following:
\d{6} - six digits
\s* - maybe some whitespace
, - a literal comma
\s* - maybe some whitespace
\d{6} - Followed by six digits
$ - Followed by the end of the string
Alternatively:
var commaSeparatedSixDigits = /^\s*\d{6}(?:\s*,\s*\d{6})*\s*$/;
I leave it as an exercise to you to decipher what's different about this.
Using JavaScript + regex:
function isOnlyCommaSeparatedSixDigitNumbers( str ){
var parts = srt.split(/\s*,\s*/);
for (var i=parts.length;i--;){
// Ensure that each part is exactly six digit characters
if (! /^\d{6}$/.test(parts[i])) return false;
}
return true;
}

I see a lot of complication here. Sounds to me like what you want is pretty simple:
/^(\d{6},)*\d{6}$/
Then we account for whitespace:
/^\s*(\d{6}\s*,\s*)*\d{6}\s*$/
But as others have noted, this is actually quite simple in JavaScript without using regex:
function check(input) {
var parts = input.split(',');
for (var i = 0, n = parts.length; i < n; i++) {
if (isNaN(+parts[i].trim())) {
return false;
}
}
return true;
}
Tested in the Chrome JavaScript console.

There isn;t any real need for a regexp. Limit the input to only 6 characters, only accept numbers and ensure that the input has 6 digits (not show here). So you would need:
HTML
<input type='text' name='invoice' size='10' maxlength='6' value='' onkeypress='evNumersOnly(event);'>
JavaScript
<script>
function evNumbersOnly( evt ) {
//--- only accepts numbers
//--- this handles incompatabilities between browsers
var theEvent = evt || window.event;
//--- this handles incompatabilities between browsers
var key = theEvent.keyCode || theEvent.which;
//--- convert key number to a letter
key = String.fromCharCode( key );
var regex = /[0-9]/; // Allowable characters 0-9.+-,
if( !regex.test(key) ) {
theEvent.returnValue = false;
//--- this prevents the character from being displayed
if(theEvent.preventDefault) theEvent.preventDefault();
}
}
</script>

How to check for uppercase alphabets in an input string, using jQuery

I am using following code snippet, but its not working :-(
//First four characters of input Text should be ALPHABATES (Letters)
if (($("#txtId").val()).length >= 4) {
var firstFourChars = $("#txtId").val().substring(0, 4);
var pattern = new RegExp('[^A-Z]');
if (firstFourChars.match(pattern))
isValid = true;
else
isValid = false;
}

change /[^A-Z]/ to /^[A-Z]/
example :
var a = "ABCJabcd";
console.log(a.match(/^[A-Z]{4}/));

you don't need to use substring(). Your regexp can do all the work for you. The RegExp you are using matches against characters that are NOT between A and Z. As Avinash said, ^[A-Z]{4} will match if your first 4 characters are uppercase. "^" at the beginning of your regexp tells that the following should be the beginning of the string. When placed inside square brackets, it reverts the range of characters you want to match.

The regex should be /[^A-Z]{4}/ if you want to match the 4 lowercase characters.

To detect in the middle of the big papers change /^[A-Z]/ to /[A-Z]/
Example text: " asşldla ABCJ abcd AÇALASD"
$('.Order input').change(function (){ucheck($(this).val())});
$('.Order input').keyup(function (){ucheck($(this).val())});
function ucheck(a) {
if(a.match(/[A-ZĞÜŞİÖÇ]{4}/)){
$('.Order #Error').html(' UPPERCASE');
}else{$('.Order #Error').html('Capitalize');}
}

If they need to be capital:
const startsWithCapitals = /^[A-Z]{4}/.test(string);
Or if they just need to be letters, add an i for ignore case:
const startsWithLetters = /^[a-z]{4}/i.test(string);
^ means start of the string and {number} means x copies

How to match with javascript and regex?

I have the following HTML:
<span id="UnitCost5">$3,079.95 to $3,479.95</span>
And i want to use Javascript and Regex to get all number matches.
So i want my script function to return: 3,079.95 AND 3,479.95
Note the text may be different so i need the solution as generic as posible, may be it will be like this:
<span id="UnitCost5">$3,079.95 And Price $3,479.95</span>

All the numbers would be matched by:
\.?\d[\d.,]*
This assumes the numbers you look for can start with a decimal dot. If they cannot, this would work (and maybe produce less false positives):
\d[\d.,]*
Be aware that different local customs exist in number formatting.
I assume that you use appropriate means to get hold of the text value of the HTML nodes you wish to process, and that HTML parsing is not part of the excercise.

You don't want to capture all numbers, otherwise you would get the 5 in the id, too. I would guess, what you're looking for is numbers looking like this: $#,###.##
Here goes the expression for that:
/\$[0-9]{1,3}(,[0-9]{3})*(\.[0-9]+)?/
\$ The dollar sign
[0-9]{1,3} One to three digits
(,[0-9]{3})* [Optional]: Digit triplets, preceded by a comma
(\.[0-9]+)? [Optional]: Even more digits, preceded by a period

/(?:\d{1,3},)*\d{1,3}(?:\.\d+)?/g;
Let's break that into parts for explanations:
(?:\d{1,3},)* - Match any numbers separated by a thousand-divider
\d{1,3} - Match the numbers before the decimal point
(?:.\d+) - Match an arbitrary number of decimals
Flag 'g' - Make a global search to find all matches in the string
You can use it like this:
var regex = /(?:\d{1,3},)*\d{1,3}(?:\.\d+)?/g;
var numbers = "$3,079.95 And Price $3,479.95".match(regex);
// numbers[0] = 3,079.95
// numbers[1] = 3,479.95

A very simple solution is the following one. Note that it will also match some invalid number strings like $..44,.777.
\$[0-9,.]+

(function () {
var reg = /\$([\d\.,]+)\s[\w\W]+\s\$([\d\.,]+)$/;
// this function used to clean inner html
function trim(str) {
var str = str.replace(/^\s\s*/, ''),
ws = /\s/,
i = str.length;
while (ws.test(str.charAt(--i)));
return str.slice(0, i + 1);
}
function getNumbersFromElement(elementId) {
var el = document.getElementById(elementId),
text = trim(el.innerHTML),
firstPrice,
secondPrice,
result;
result = reg.exec(text);
if (result[1] && result[2]) {
// inside this block we have valid prices
firstPrice = result[1];
secondPrice = result[2];
// do whatever you need
return firstPrice + ' AND ' + secondPrice;
} else {
return null; // otherwise
}
}
// usage:
getNumbersFromElement('UnitCost5');
})();

The following will return an array of all prices found in the string
function getPrices(str) {
var reg = /\$([\d,.]+)/g;
var prices =[];
var price;
while((price = reg.exec(str))!=null) {
prices.push(price);
}
return prices;
}
edit: note that the regex itself may return some false positives