This question already has an answer here:
javascript regexp replace not working, but string replace works
(1 answer)
Closed 1 year ago.
Hello team I am new to JS so I am trying to use RegEx with replacing to take input from the user and replace it if it doesn't match the RegEx I have to be able to put 7 digits or 6 digits followed with one letter currently I am doing this
someID.replace('^(([0-9]{1,7})|([0-9]{1,6}[a-zA-Z]{1}))$')
I am not able to replace the current string with the RegEx expression if I enter
12345678900 it remain the same in that situation I need to be 1234567 after the replace or if I have 12345678asd to be 123456a. How can I achieve that by only replace function and a RegEx expresion
You need to use a different regex and a dirrent replace function.
You will also need to get rid of $ if you want to be able to successfully match the string, without worrying about how it ends.
const sampleIDs = [
"123456789000",
"123456abc",
];
sampleIDs.forEach(id => {
const clean = id.match(/^\d{6}[\d\D]/);
console.log(clean[0]);
});
Related
This question already has answers here:
Regular expression to stop at first match
(9 answers)
Closed 7 months ago.
I am trying to perform this transformation to a string (using javascript):
Input:
[hello]{world}and[good]{night}
Output:
<span class="top">hello<span class="bottom">world</span></span>and<span class="top">good<span class="bottom">night</span></span>
To do that I am using the following regex:
text.replace(/\[(.*)\]\{(.*)\}/gim, "<span class='top'>$1<span class='bottom'>$2</span></span>")
It works correctly when only setting one occurrence of the pattern in a string [hello]{world}
But if I add a string with more than one, the regex matches the first [] and the last {} instead, and prints this:
<span class='top'>hello]{world}and[good<span class='bottom'>night</span></span>
How can I tell regex to match the first pattern and the second pattern instead of matching it as one bigger pattern?
Note that between the [] and {} I expect there to be no text. So [hello]world and good{night} should not be matched.
You need to put ? after .* to make the quantifier lazy, instead of greedy.
const text = '[hello]{world}and[good]{night}'
const result = text.replace(/\[(.*?)\]\{(.*?)\}/gim, "<span class='top'>$1<span class='bottom'>$2</span></span>")
console.log(result)
This question already has answers here:
Getting content between curly braces in JavaScript with regex
(5 answers)
Closed 2 years ago.
I need to match a specific regex syntax and split them so that we can match them to an equivalent value from a dictionary.
Input:
{Expr "string"}
{Expr "string"}{Expr}
Current code:
value.match(/\{.*\}$/g)
Desired Output:
[{Expr "string"}]
[{Expr "string"},{Expr}]
Use a non-greedy quantifier .*?. And don't use $, because that forces it to match all the way to the end of the string.
value = '{Expr "string"}{Expr}'
console.log(value.match(/\{.*?\}/g));
One option, assuming your version of JavaScript support it, would be to split the input on the following regex pattern:
(?<=\})(?=\{)
This says to split at each }{ junction between two terms.
var input = "{Expr \"string\"}{Expr}";
var parts = input.split(/(?<=\})(?=\{)/);
console.log(parts);
This question already has answers here:
How to detect exact length in regex
(8 answers)
Closed 4 years ago.
So I am trying to use a regular expression to check against strings but it doesn't seem to be working properly.
Basically I want it to match a alpha-numeric string that is exactly 3 characters long. The expression I am using below does not seem to be working for this:
const msg = message.content;
const regex = /[A-Za-z0-9]{3}/g;
if (msg.match(regex)) {
// Do something
}
Am I doing something wrong? Any help would be appreciated. Thanks in advance.
You need to add ^ and $ for the start-of-string anchor and end-of-string anchor, respectively - otherwise, for example, for #123, the 123 will match, and it will pass the regex. You also might consider using the i flag rather than repeat A-Za-z, and you can use \d instead of 0-9.
It looks like you just want to check whether the string passes the regex's test or not, in which case .test (evaluates to a boolean) might be a bit more appropriate than .match. Also, either way, there's no need for the global flag if you're just checking whether a string passes a regex:
const regex = /^[a-z\d]{3}$/i;
if (regex.test(msg)) {
// do something
}
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, '');
This question already has answers here:
Is there a RegExp.escape function in JavaScript?
(18 answers)
Closed 4 years ago.
I'm trying to create a dynamic regex to select URL's based on a segment or the whole URL.
For example, I need to get var.match(/http:\/\/www.something.com\/something/)
The text inside the match() needs to be converted so that special characters have \ in front of them such for example "\/". I was not able to find a function that converts the URL to do this? Is there one?
If not, what characters require a \ in front?
I use this to escape a string when generating a dynamic regex:
var specials = /[*.+?|^$()\[\]{}\\]/g;
var url_re = RegExp(url.replace(specials, "\\$&"));
( ) [ ] ? * ^ $ \ . + | and in your case, / since you're using that as the delimiter in the match.
Further info mostly to pre-empt comments and downvotes: I don't really know where - come from as a special character. It's only special when inside character class brackets [ and ] which you're already escaping. If you want to include characters that are sometimes special (which the OP doesn't) that would include look-ahead/behind characters as well, which include =, < and >.