I am trying to verify str with the code below. My final goal is to allow this style of input:
18.30 Saturday_lastMatch 3/10
However, the code I have can't even work for the basic usage (98.5% str will be of this format):
19.30 Friday 15/5
var regex= /[0-9]{2}[\.:][0-9]{2} [A-Z][a-z]{4,7} [0-9]\/[0-9]{2}/;
if(!str.match(regex)) {
//"Bad format, match creation failed!");
}
What am I missing?
There are a number of problems with your regex.
The date & time matching portions at the beginning and end don't allow for 1 or 2 digit numbers as they should.
You may want to consider anchoring the regex at the beginning and end with ^ and $, respectively.
The literal dot in the character class doesn't need to be escaped.
Try this:
var regex= /^[0-9]{1,2}[.:][0-9]{1,2} [A-Z][a-z]{5,8} [0-9]{1,2}\/[0-9]{1,2}$/;
The final part of your regular expression that checks day/month needs to be expanded. It currently only matches #/##, but it should allow ##/# as well. The simplest fix would be to allow either one or two digits on either side (e.g. 12/31)
var regex= /[0-9]{2}[\.:][0-9]{2} [A-Z][a-z]{4,7} [0-9]{1,2}\/[0-9]{1,2}/;
Related
I'm trying to match my timestamp format, but I need to detect whether it is invalid format or not (I need to do something about the invalid format)
Currently, I need to match a space character inside my timestamp:
examples:
[02:21.10,E] or [02:21.10,C#] //correct format
[02:21.10, E] or [10.21.10,E ] //incorrect format, there is a space, but i need to match it with my incorrectRegex
for my correct regex, I use this regex and it is valid:
/\[(\d\d:\d\d\.\d\d),(.*?)\]/g
and for my incorrect regex, I use
/\[\d\d:\d\d\.\d\d,\s*?\]/g
but it didn't find any match for my incorrect format
nb: I'm using a javascript
Try this:
\[\d\d[.:]\d\d[.:]\d\d,(?:[ ][^\]]+?|[^\]]+?[ ])\]
Live Demo.
var re = /\[\d\d[.:]\d\d[.:]\d\d,(?:[ ][^\]]+?|[^\]]+?[ ])\]/;
console.log(re.test('[02:21.10,E]'));;
console.log(re.test('[02:21.10,C#]'));
console.log(re.test('[02:21.10, E]'));
console.log(re.test('[10.21.10,E ]'));
console.log(re.test('[10.21.10,C# ]'));
console.log(re.test('[02:54.97,C#]single[02:55.61,A ]'));
PS: I assume it is not an error that your second invalid sample [10.21.10,E ] does not contain a : after the first two digits. I have applied the same for the second dot/colon.
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 -
My question is simple but takes work. I tried lots of regex expressions to check my datetime is ok or not, but though I am sure my regex exprerssion is correct it always return to me isnotok with ALERT. Can you check my code?
validateForLongDateTime('22-03-1981')
function validateForLongDateTime(date){
var regex=new RegExp("/^\d{2}[.-/]\d{2}[.-/]\d{4}$/");
var dateOk=regex.test(date);
if(dateOk){
alert('ok');
}else{
alert('notok');
}
}
There are at least 2 issues with the regex:
It has unescaped forward slashes
The hyphen in the character classes is unescaped and forms a range (matching only . and /) that is not what is necessary here.
The "fixed" regex will look like:
/^\d{2}[.\/-]\d{2}[.\/-]\d{4}$/
See demo
However, you cannot validate dates with it since it will also match 37-67-5734.
Here is an SO post with a comprehensive regex approach that looks viable
Here is my enahanced version with a character class for the delimiter:
^(?:(?:31([\/.-])(?:0?[13578]|1[02]))\1|(?:(?:29|30)([\/.-])(?:0?[1,3-9]|1[0-2])\2))(?:(?:1[6-9]|[2-9]\d)?\d{2})$|^(?:29([\/.-])0?2\3(?:(?:(?:1[6-9]|[2-9]\d)?(?:0[48]|[2468][048]|[13579][26])|(?:(?:16|[2468][048]|[3579][26])00))))$|^(?:0?[1-9]|1\d|2[0-8])([\/.-])(?:(?:0?[1-9])|(?:1[0-2]))\4(?:(?:1[6-9]|[2-9]\d)?\d{2})$
Here is an SO post showing another approach using Date.parse
this way you can validate date between 1 to 31 and month 1 to 12
var regex = /^(0[1-9]|[12][0-9]|3[01])[- \/.](0[1-9]|1[012])[- \/.](19|20)\d\d$/
see this demo here https://regex101.com/r/xP1bD2/1
I have a regex which allows only to enter integers and floats in a text box.
Regex Code:-
("^[0-9]*(?:[.][0-9]*|)$");
But it gives an error when the user enters whitespace at the beginning and end of the entered values. I want the user to allow spaces at the beginning and at the end as optional, so I changed the regex as below but it didn't work.
Note: Spaces may be spaces or tabs.
Test Case: User might enter:
"10","10.23"," 10","10 "," 10.23","10.23 "
Any number of spaces are allowed.
("^(?:\s)*[0-9]*(?:[.][0-9]*|)$")
I am newbie with regex, so any help will be highly appreciated.
Thank you.
Try this:
/^\s*[0-9]*(?:[.][0-9]*|)\s*$/;
You don't have to wrap a single entity in a group to repeat it, and I have added a second zero-or-more-spaces at the end which is what you are missing to make it work.
Note: You have not posted the code you use to create the RegExp object, but if it is new RegExp(string), remember to escape your backslashes (by doubling them):
var r = new RegExp("^\\s*[0-9]*(?:[.][0-9]*|)\\s*$");
Also, as #Blender suggests, this can be simplified to:
/^\s*[0-9]*(?:\.[0-9]*)?\s*$/;
Or, using \d instead of [0-9]:
/^\s*\d*(?:\.\d*)?\s*$/;
You don't necessarily need a Regular Expression: !isNaN(Number(textboxvalue.trim())) would be sufficient.
Otherwise, try /^\s{0,}\d+\.{0,1}\d+\s{0,}$/. Test:
var testvalues = ["10","10.23"," 10","10 "," 10.23","10.23 ","10.24.25"];
for (var i=0;i<testvalues.length;i+=1){
console.log(/^\s{0,}\d+\.{0,1}\d+\s{0,}$/.test(testvalues[i]));
}
//=> 6 x true, 1 x false
I have strings which contains thousand separators, however no string-to-number function wants to consume it correctly (using JavaScript). I'm thinking about "preparing" the string by stripping all thousand separators, leaving anything else untoched and letting Number/parseInt/parseFloat functions (I'm satisfied with their behavious otherwise) to decide the rest. But it seems what i have no idea which RegExp can do that!
Better ideas are welcome too!
UPDATE:
Sorry, answers enlightened me how badly formulated question it is. What i'm triyng to achieve is: 1) to strip thousand separators only if any, but 2) to not disturb original string much so i will get NaNs in the cases of invalid numerals.
MORE UPDATE:
JavaScript is limited to English locale for parsing, so lets assume thousand separator is ',' for simplicity (naturally, it never matches decimal separator in any locale, so changing to any other locale should not pose a problem)
Now, on parsing functions:
parseFloat('1023.95BARGAIN BYTES!') // parseXXX functions just "gives up" on invalid chars and returns 1023.95
Number('1023.95BARGAIN BYTES!') // while Number constructor behaves "strictly" and will return NaN
Sometimes I use rhw loose one, sometimes strict. I want to figure out the best approach for preparing string for both functions.
On validity of numerals:
'1,023.99' is perfectly well-formed English number, and stripping all commas will lead to correct result.
'1,0,2,3.99' is broken, however generic comma stripping will give '1023.99' which is unlikely to be a correct result.
welp, I'll venture to throw my suggestion into the pot:
Note: Revised
stringWithNumbers = stringwithNumbers.replace(/(\d+),(?=\d{3}(\D|$))/g, "$1");
should turn
1,234,567.12
1,023.99
1,0,2,3.99
the dang thing costs $1,205!!
95,5,0,432
12345,0000
1,2345
into:
1234567.12
1023.99
1,0,2,3.99
the dang thing costs $1205!!
95,5,0432
12345,0000
1,2345
I hope that's useful!
EDIT:
There is an additional alteration that may be necessary, but is not without side effects:
(\b\d{1,3}),(?=\d{3}(\D|$))
This changes the "one or more" quantifier (+) for the first set of digits into a "one to three" quantifier ({1,3}) and adds a "word-boundary" assertion before it. It will prevent replacements like 1234,123 ==> 1234123. However, it will also prevent a replacement that might be desired (if it is preceded by a letter or underscore), such as A123,789 or _1,555 (which will remain unchanged).
A simple num.replace(/,/g, '') should be sufficient I think.
Depends on what your thousand separator is
myString = myString.replace(/[ ,]/g, "");
would remove spaces and commas.
This should work for you
var decimalCharacter = ".",
regex = new RegExp("[\\d" + decimalCharacter + "]+", "g"),
num = "10,0000,000,000.999";
+num.match(regex).join("");
To confirm that a numeral-string is well-formed, use:
/^(\d*|\d{1,3}(,\d{3})+)($|[^\d])/.test(numeral_string)
which will return true if the numeral-string is either (1) just a sequence of zero or more digits, or (2) a sequence of digits with a comma before each set of three digits, or (3) either of the above followed by a non-digit character and who knows what else. (Case #3 is for floats, as well as your "BARGAIN BYTES!" examples.)
Once you've confirmed that, use:
numeral_string.replace(/,/g, '')
which will return a copy of the numeral-string with all commas excised.
You can use s.replaceAll("(\\W)(?=\\d{3})","");
This regex gets all alpha-numeric character with 3 characters after it.
Strings like 4.444.444.444,00 € will be 4444444444,00 €
I have used the following in a commercial setting, and it has worked often:
numberStr = numberStr.replace(/[. ,](\d\d\d\D|\d\d\d$)/g,'$1');
In the above example, thousands can be marked with a decimal, a comma, or a space.
In some cases ( like a price of 1000,5 Euros) the above doesn't work. If you need something more robust, this should work 100% of the time:
//convert a comma or space used as the cent placeholder to a decimal
$priceStr = $priceStr.replace(/[, ](\d\d$)/,'.$1');
$priceStr = $priceStr.replace(/[, ](\d$)/,'.$1');
//capture cents
var $hasCentsRegex = /[.]\d\d?$/;
if($hasCentsRegex.test($priceStr)) {
var $matchArray = $priceStr.match(/(.*)([.]\d\d?$)/);
var $priceBeforeCents = $matchArray[1];
var $cents = $matchArray[2];
} else{
var $priceBeforeCents = $priceStr;
var $cents = "";
}
//remove decimals, commas and whitespace from the pre-cent portion
$priceBeforeCents = $priceBeforeCents.replace(/[.\s,]/g,'');
//re-create the price by adding back the cents
$priceStr = $priceBeforeCents + $cents;