I need to validate range R1:C1-R2:C2 range
I tried with,
function isValid(rc) {
if (rc)
if (rc.trim() === "") return false;
var isPattern = /^[0-9]+:[0-9]+-[0-9]+:[0-9]+$/.test(rc);
if (!isPattern) return false;
return ((rc));
}
But the above code also takes 9:9-0:0 or 0:0-0:0
UPDATE:
How do I check whether the given range after a numeric validation at client side is correct in the server side. Because, at server side range may be invalid.
You need to check that there are some alphabetic characters, then numeric characters - this is how to match a field: [a-z]+\d+. It will match some characters, then some numbers. E.g. a1, ASD34. It won't match 12 or sab though.
If you put all together, you get this regex:
/^[a-z]+\d+:[a-z]+\d+\s*-\s*[a-z]+\d+:[a-z]+[0-9]+$/i
Explanation:
/../i matches case insensitively, so it's enough to specify [a-z], no need to type [a-zA-Z]
this matches a field: [a-z]+\d+, like R1 or B12. It won't match 12 or AB
\s*-\s* tolerates spaces around the dash
so we got: field:field-field:field and that's what you want.
You can play with this regex demo.
Checking if the values are actually okay, it takes more than a regex. You'll have to split the code by the dash and the colons and compare the values.
Related
I'm writing an application that requires color manipulation, and I want to know when the user has entered a valid hex value. This includes both '#ffffff' and '#fff', but not the ones in between, like 4 or 5 Fs. My question is, can I write a regex that determines if a character is present a set amount of times or another exact amount of times?
What I tried was mutating the:
/#(\d|\w){3}{6}/
Regular expression to this:
/#(\d|\w){3|6}/
Obviously this didn't work. I realize I could write:
/(#(\d|\w){3})|(#(\d|\w){6})/
However I'm hoping for something that looks better.
The shortest I could come up with:
/#([\da-f]{3}){1,2}/i
I.e. # followed by one or two groups of three hexadecimal digits.
You can use this regex:
/#[a-f\d]{3}(?:[a-f\d]{3})?\b/i
This will allow #<3 hex-digits> or #<6 hex-digits> inputs. \b in the end is for word boundary.
RegEx Demo
I had to find a pattern for this myself today but I also needed to include the extra flag for transparency (i.e. #FFF5 / #FFFFFF55). Which made things a little more complicated as the valid combinations goes up a little.
In case it's of any use, here's what I came up with:
var inputs = [
"#12", // Invalid
"#123", // Valid
"#1234", // Valid
"#12345", // Invalid
"#123456", // Valid
"#1234567", // Invalid
"#12345678", // Valid
"#123456789" // Invalid
];
var regex = /(^\#(([\da-f]){3}){1,2}$)|(^\#(([\da-f]){4}){1,2}$)/i;
inputs.forEach((itm, ind, arr) => console.log(itm, (regex.test(itm) ? "valid" : "-")));
Which should return:
#123 valid
#1234 valid
#12345 -
#123456 valid
#1234567 -
#12345678 valid
#123456789 -
I'm a bit stuck here.
With regards to the input of accounting data, the analyst requested a specific set of rules on the input of decimal data in text boxes.
Since I have not studied regular expressions at this point, and because I have very strict deadlines to attend with each a lot of work, I request your help.
The rules are (on blur):
IE8+ compatible;
Set default at 2 digits behind the comma;
Disallow other characters than ".", "," and numeric characters;
If there are multiple separators in the returned number, only keep the last one.
Explanation:
Employees may have different regional settings, and comma / dot may be used as either decimal or thousands separator.
In case an employee copy - pastes both the thousand and the decimal separator, it has to be ignored.
What I've done so far, but doesn't fulfill the requirements:
http://jsfiddle.net/NxFHL/1/
$('#test_rules.numeric').on('blur', function(){
var myString = $('#test_rules.numeric').val();
myString = parseFloat(myString.replace(/[^\d.-]/g, ''));
myString = toFixed(myString, 2);
console.log(myString);
});
function toFixed(value, precision) {
var power = Math.pow(10, precision || 0);
return
String(Math.round(value * power) / power);
}
The regular expression used doesn't work correctly as it only accepts dot, not comma.
Also I am not sure about how I should make it so that only the last separator stays (so that the thousands separator gets ignored).
Try this function:
function parseInputNum(val) {
var sep = val.lastIndexOf('.') > val.lastIndexOf(',')? '.' : ',';
var arr = val.replace(new RegExp('[^\\d'+sep+']+', 'g'), '')
.match(new RegExp('(\\d+(?:['+sep+']\\d+|))$'));
return arr? arr[1].replace(/[,]/g, '.') : false;
}
You can use this pattern:
^(?:[1-9](?:[0-9]{0,2}(?:([.,])[0-9]{3})?(?:\1[0-9]{3})*|[0-9]*)|0)(?!\1)[.,][0-9]{2}$
This pattern will check numbers like:
123,45
12,45
0.45
123,456,789.12
1234.56
123.456.789,12
but not numbers like:
12.23.45,12
012,345,678,91
1234,567.89
123,456,78
To convert the string into a number you must remove the thousand delimiter before. This can easily be done since the delimiter (if present) is in the capturing group 1. You must probably too replace the , by the . if it is used as decimal separator.
I am trying to validate a string, that should contain letters numbers and special characters &-._ only. For that I tried with a regular expression.
var pattern = /[a-zA-Z0-9&_\.-]/
var qry = 'abc&*';
if(qry.match(pattern)) {
alert('valid');
}
else{
alert('invalid');
}
While using the above code, the string abc&* is valid. But my requirement is to show this invalid. ie Whenever a character other than a letter, a number or special characters &-._ comes, the string should evaluate as invalid. How can I do that with a regex?
Add them to the allowed characters, but you'll need to escape some of them, such as -]/\
var pattern = /^[a-zA-Z0-9!##$%^&*()_+\-=\[\]{};':"\\|,.<>\/?]*$/
That way you can remove any individual character you want to disallow.
Also, you want to include the start and end of string placemarkers ^ and $
Update:
As elclanrs understood (and the rest of us didn't, initially), the only special characters needing to be allowed in the pattern are &-._
/^[\w&.\-]+$/
[\w] is the same as [a-zA-Z0-9_]
Though the dash doesn't need escaping when it's at the start or end of the list, I prefer to do it in case other characters are added. Additionally, the + means you need at least one of the listed characters. If zero is ok (ie an empty value), then replace it with a * instead:
/^[\w&.\-]*$/
Well, why not just add them to your existing character class?
var pattern = /[a-zA-Z0-9&._-]/
If you need to check whether a string consists of nothing but those characters you have to anchor the expression as well:
var pattern = /^[a-zA-Z0-9&._-]+$/
The added ^ and $ match the beginning and end of the string respectively.
Testing for letters, numbers or underscore can be done with \w which shortens your expression:
var pattern = /^[\w&.-]+$/
As mentioned in the comment from Nathan, if you're not using the results from .match() (it returns an array with what has been matched), it's better to use RegExp.test() which returns a simple boolean:
if (pattern.test(qry)) {
// qry is non-empty and only contains letters, numbers or special characters.
}
Update 2
In case I have misread the question, the below will check if all three separate conditions are met.
if (/[a-zA-Z]/.test(qry) && /[0-9]/.test(qry) && /[&._-]/.test(qry)) {
// qry contains at least one letter, one number and one special character
}
Try this regex:
/^[\w&.-]+$/
Also you can use test.
if ( pattern.test( qry ) ) {
// valid
}
let pattern = /^(?=.*[0-9])(?=.*[!##$%^&*])(?=.*[a-z])(?=.*[A-Z])[a-zA-Z0-9!##$%^&*]{6,16}$/;
//following will give you the result as true(if the password contains Capital, small letter, number and special character) or false based on the string format
let reee =pattern .test("helLo123#"); //true as it contains all the above
I tried a bunch of these but none of them worked for all of my tests. So I found this:
^(?=.*\d)(?=.*[a-z])(?=.*[A-Z])(?=.*[^a-zA-Z0-9])(?!.*\s).{8,15}$
from this source: https://www.w3resource.com/javascript/form/password-validation.php
Try this RegEx: Matching special charecters which we use in paragraphs and alphabets
Javascript : /^[a-zA-Z]+(([\'\,\.\-_ \/)(:][a-zA-Z_ ])?[a-zA-Z_ .]*)*$/.test(str)
.test(str) returns boolean value if matched true and not matched false
c# : ^[a-zA-Z]+(([\'\,\.\-_ \/)(:][a-zA-Z_ ])?[a-zA-Z_ .]*)*$
Here you can match with special char:
function containsSpecialChars(str) {
const specialChars = /[`!##$%^&*()_+\-=\[\]{};':"\\|,.<>\/?~]/;
return specialChars.test(str);
}
console.log(containsSpecialChars('hello!')); // 👉️ true
console.log(containsSpecialChars('abc')); // 👉️ false
console.log(containsSpecialChars('one two')); // 👉️ false
I am writing some javascript to validate user input in a text area. Here are the requirements for input format:
Input must be in the format of two hex values separated by a space (i.e. A7 B3 9D).
Input must be valid hex values.
I have requirement #2 sorted out with a regExpression valueOne.match("[0-9A-Fa-f]{1}") (or at least I hope that is the recommended way to do it). So I am just looking for some input on how to go about handling requirement number one in a simple and efficient way.
Thanks!
This regex will do it:
/^[0-9A-F]{2}(\s[0-9A-F]{2})*$/i
That is:
^ // beginning of string
[0-9A-F]{2} // two characters of 0-9 or A-F
(\s[0-9A-F]{2})* // zero or more instances of a space followed by
// two characters of 0-9 or A-F
$ // end of string
Where the i flag at the end makes it case insensitive.
To use it:
var valueOne = // set to your textarea's value here
if (/^[0-9A-F]{2}(\s[0-9A-F]{2})*$/i.test(valueOne)) {
// is OK, do something
} else {
// is not OK, do something
}
I have a text box and it says "Phone:" as the standard here for phone number is (XXX)-XXX-XXXX
I'd like to have a javascript that automatically puts my numbers into that format if it's not in that format, so if you typed 9993334444 then it would change it automatically on the fly as I'm typing to (999)-333-4444 I have looked around Google for Javascript Phone Regex to no success, maybe a Regex isn't what I'm looking for?
you want to add an onkeyup event with a regex like
this.value = this.value.replace(/^\(?([0-9][0-9][0-9]){1}\)?-?([0-9][0-9][0-9][0-9]){1}-?([0-9][0-9][0-9]){1}$/, '($1)-$2-$3');
Check out http://jsfiddle.net/R8enX/
/ means start/end a regex string
^ means start of matching string
$ means end of matching string
? means 0 or 1 instances (make braces and dashes optional)
[0-9] means any single digit
(x){1} tags x as an expression that we can reference in the replacement with a $ sign
EDIT: realized I missed a digit on the last group of numbers, the jsfiddle will only work (properly) with 3 digits in the last group
To build somewhat on #Andrews answer you can check for a valid (local)phone number via this method. If the number is shorter or larger than 10 digits, it collapses back into an invalid number
-
<input type="text" onBlur="localNumber(this.value)"/>
<div id="output"></div>
-
<script>
var localNumber = function(str){
repl = str.replace(/^([0-9]{3})([0-9]{3})([0-9]{4})$/, "($1)-$2-$3");
outp = document.getElementById('output');
if( repl.match(/\W/) )
{
outp.innerHTML = repl;
}
else
{
outp.innerHTML = 'Invalid number for this region';
}
}
</script>