This question already has answers here:
Escaping a forward slash in a regular expression
(6 answers)
Closed 8 years ago.
In my RadGrid I am using Filter for DATE Column. Date format is like this 16/12/1990. Filter textbox should allow only Numbers and /. How to write JavaScript function to do this?
function CharacterCheckDate(text, e)
{
var regx, flg;
regx = /[^0-9/'' ]/
flg = regx.test(text.value);
if (flg)
{
var val = text.value;
val = val.substr(0, (val.length) - 1)
text.value = val;
}
}
You don't have to worry about / in character class, (thanks Robin for pointing that out). For example,
console.log(/[^\d/]/.test("/"));
# false
console.log(/[^\d/]/.test("a"));
# true
If you are really in doubt, simply escape it with backslash, like this
regx = /[^0-9\/'' ]/
Also, you don't need to specify ' twice, once is enough.
regx = /[^0-9\/' ]/
Instead of using numbers explicitly, you can use \d character class, like this
regx = /[^\d\/' ]/
So, you could have written your RegEx, like this
regx = /[^\d/' ]/
Related
This question already has an answer here:
Reference - What does this regex mean?
(1 answer)
Closed 4 years ago.
I'm looking for some assistance with JavaScript/Regex when trying to format a string of text.
I have the following IDs:
00A1234/A12
0A1234/A12
A1234/A12
000A1234/A12
I'm looking for a way that I can trim all of these down to 1234/A12. In essence, it should find the first letter from the left, and remove it and any preceding numbers so the final format should be 0000/A00 or 0000/AA00.
Is there an efficient way this can be acheived by Javascript? I'm looking at Regex at the moment.
Instead of focussing on what you want to strip, look at what you want to get:
/\d{4}\/[A-Z]{1,2}\d{2}/
var str = 'fdfhfjkqhfjAZEA0123/A45GHJqffhdlh';
match = str.match(/\d{4}\/[A-Z]{1,2}\d{2}/);
if (match) console.log(match[0]);
You could seach for leading digits and a following letter.
var data = ['00A1234/A12', '0A1234/A12', 'A1234/A12', '000A1234/A12'],
regex = /^\d*[a-z]/gi;
data.forEach(s => console.log(s.replace(regex, '')));
Or you could use String#slice for the last 8 characters.
var data = ['00A1234/A12', '0A1234/A12', 'A1234/A12', '000A1234/A12'];
data.forEach(s => console.log(s.slice(-8)));
You could use this function. Using regex find the first letter, then make a substring starting after that index.
function getCode(s){
var firstChar = s.match('[a-zA-Z]');
return s.substr(s.indexOf(firstChar)+1)
}
getCode("00A1234/A12");
getCode("0A1234/A12");
getCode("A1234/A12");
getCode("000A1234/A12");
A regex such as this will capture all of your examples, with a numbered capture group for the bit you're interested in
[0-9]*[A-Z]([0-9]{4}/[A-Z]{1,2}[0-9]{2})
var input = ["00A1234/A12","0A1234/A12","A1234/A12","000A1234/A12"];
var re = new RegExp("[0-9]*[A-Z]([0-9]{4}/[A-Z]{1,2}[0-9]{2})");
input.forEach(function(x){
console.log(re.exec(x)[1])
});
This question already has an answer here:
Why this javascript regex doesn't work?
(1 answer)
Closed 2 years ago.
I was very surprised that I didn't find this already on the internet.
is there's a regular expression that validates only digits in a string including those starting with 0 and not white spaces
here's the example I'm using
function ValidateNumber() {
var regExp = new RegExp("/^\d+$/");
var strNumber = "010099914934";
var isValid = regExp.test(strNumber);
return isValid;
}
but still the isValid value is set to false
You could use /^\d+$/.
That means:
^ string start
\d+ a digit, once or more times
$ string end
This way you force the match to only numbers from start to end of that string.
Example here: https://regex101.com/r/jP4sN1/1
jsFiddle here: https://jsfiddle.net/gvqzknwk/
Note:
If you are using the RegExp constructor you need to double escape the \ in the \d selector, so your string passed to the RegExp constructor must be "^\\d+$".
So your function could be:
function ValidateNumber(strNumber) {
var regExp = new RegExp("^\\d+$");
var isValid = regExp.test(strNumber); // or just: /^\d+$/.test(strNumber);
return isValid;
}
This question already has answers here:
How do I replace all occurrences of a string in JavaScript?
(78 answers)
Closed 3 years ago.
I have this
var date = $('#Date').val();
this get the value in the textbox what would look like this
12/31/2009
Now I do this on it
var id = 'c_' + date.replace("/", '');
and the result is
c_1231/2009
It misses the last '/' I don't understand why though.
You need to set the g flag to replace globally:
date.replace(new RegExp("/", "g"), '')
// or
date.replace(/\//g, '')
Otherwise only the first occurrence will be replaced.
Unlike the C#/.NET class library (and most other sensible languages), when you pass a String in as the string-to-match argument to the string.replace method, it doesn't do a string replace. It converts the string to a RegExp and does a regex substitution. As Gumbo explains, a regex substitution requires the global flag, which is not on by default, to replace all matches in one go.
If you want a real string-based replace — for example because the match-string is dynamic and might contain characters that have a special meaning in regexen — the JavaScript idiom for that is:
var id= 'c_'+date.split('/').join('');
You can use:
String.prototype.replaceAll = function(search, replace) {
if (replace === undefined) {
return this.toString();
}
return this.split(search).join(replace);
}
This question already has answers here:
Parse query string in JavaScript [duplicate]
(11 answers)
Closed 8 years ago.
So let's say I have this HTML link.
<a id="avId" href="http://www.whatever.com/user=74853380">Link</a>
And I have this JavaScript
av = document.getElementById('avId').getAttribute('href')
Which returns:
"http://www.whatever.com/user=74853380"
How do I extract 74853380 specifically from the resulting string?
There are a couple ways you could do this.
1.) Using substr and indexOf to extract it
var str = "www.something.com/user=123123123";
str.substr(str.indexOf('=') + 1, str.length);
2.) Using regex
var str = var str = "www.something.com/user=123123123";
// You can make this more specific for your query string, hence the '=' and group
str.match(/=(\d+)/)[1];
You could also split on the = character and take the second value in the resulting array. Your best bet is probably regex since it is much more robust. Splitting on a character or using substr and indexOf is likely to fail if your query string becomes more complex. Regex can also capture multiple groups if you need it to.
You can use regular expression:
var exp = /\d+/;
var str = "http://www.whatever.com/user=74853380";
console.log(str.match(exp));
Explanation:
/\d+/ - means "one or more digits"
Another case when you need find more than one number
"http://www.whatever.com/user=74853380/question/123123123"
You can use g flag.
var exp = /\d+/g;
var str = "http://www.whatever.com/user=74853380/question/123123123";
console.log(str.match(exp));
You can play with regular expressions
Well, you could split() it for a one liner answer.
var x = parseInt(av.split("=")[1],10); //convert to int if needed
This question already has answers here:
Replace method doesn't work
(4 answers)
Closed 4 years ago.
I'm trying to further my understanding of regular expressions in JavaScript.
So I have a form that allows a user to provide any string of characters. I'd like to take that string and remove any character that isn't a number, parenthesis, +, -, *, /, or ^. I'm trying to write a negating regex to grab anything that isn't valid and remove it. So far the code concerning this issue looks like this:
var pattern = /[^-\+\(\)\*\/\^0-9]*/g;
function validate (form) {
var string = form.input.value;
string.replace(pattern, '');
alert(string);
};
This regex works as intended on http://www.infobyip.com/regularexpressioncalculator.php regex tester, but always alerts with the exact string I supply without making any changes in the calculator. Any advice or pointers would be greatly appreciated.
The replace method doesn't modify the string. It creates a new string with the result of the replacement and returns it. You need to assign the result of the replacement back to the variable:
string = string.replace(pattern, '');