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searching for a regex to convert
0.015.000 -> 15.000
0.150.000 -> 150.00
015.000 -> 15.00
You could try the below string.replace function. Use ^ to tell the regex engine to do the matching operation from the start. By putting 0 and . inside a character class would match either 0 or dot.
string.replace(/^[0.]+/, "")
Example:
> "0.015.000".replace(/^[0.]+/, "")
'15.000'
> "0.150.000".replace(/^[0.]+/, "")
'150.000'
> "015.000".replace(/^[0.]+/, "")
'15.000'
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i need a regex that fits these conditions
may contain letters and numbers. numbers are optional but must contain at least 1 letter
at least 2 characters
can contain ONLY the "-" character of special characters. this is optional
must begin with a letter
no whitespace
no turkish characters
How should i create a regex query according to these conditions?
I would create a function like so:
function isGood(string){
return /^[a-z][a-z0-9-]+$/i.test(string);
}
console.log(isGood('This should return false'))
console.log(isGood('#no'));
console.log(isGood('Nice'));
console.log(isGood('a'));
console.log(isGood('a!'));
console.log(isGood('great'));
console.log(isGood('This-should-also-pass-the-test'));
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I need to remove + or 00 from the beginning of a number in case they exist. So a number like +37253783478 would output 37253783478 and 0037253783478 would output 37253783478. What would the regex look like that matches this pattern?
EDIT: I've managed to remove the leading zeros using ^0+ but I can't figure out how to match both cases.
If I understand the requirement, the following will match both cases. Essentially, what you need to do is use the regex or operator |.
The following will remove all leading 0s
str.replace(/(^0+|^\+)/,'')
But if you just need to remove exactly two leading 0s, use this:
str.replace(/(^00|^\+)/,'')
And here it is in action on your examples:
let nums = ['+37253783478', '0037253783478', '0037253780478', '375378+0478'];
let replaced = nums.map(num => num.replace(/(^0+|^\+)/,''));
console.log(replaced);
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I have a string like this: *ngIf="thisThing > 0"
Using RegEx, I'd like to select/return ONLY the > or in some cases, <.
I have the following expression: (?<=")(?:.*?)(<|>)(?:.*?)(?=") but this still selects everything inside the quotes. I only want to match that >.
I've been testing it here: https://regex101.com/r/4KPfbT/1
I guess you're over-complexifying it. Why not this way:
\w*=".*([<>]).*"
It works, take a look:
console.log('*ngIf="thisThing > 0"'.match(/w*=".*([<>]).*"/)[1])
console.log('whatever="otherThing < 999"'.match(/w*=".*([<>]).*"/)[1])
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^[a-zA-Z]\w+#[a-zA-Z_]+?\.[a-zA-Z]{2,4}$
i had tried this regex for email but it allow following cases
123#mail.com
example.mail#mail.com
Here is the code which searches for the number at the start.
If number is matched it prints message in console.
let regex = /^[0-9]/;
let object = [{email:'123#mail.com'},{email:'example.mail#mail.com'}];
for(let i =0;i<object.length;i++){
if(object[i].email.match(regex)){
console.log('E-mail ',object[i].email,' is not valid.')
}
}
This is the used regex: ^[0-9]
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var input="gazanie, flower, blossom";
I need to replace input using JavaScript or jQuery.
Output should be like:
"#gazanie #flower #blossom"
"gazanie, flower, blossom".split(',').map((e)=>`#${e.trim()}`).join('');
THat's it :
split : to split string according to delimitor which is , & get array
map : to add prefix for each element which is #
trim : eliminate space if it is the 1st char or/and is the last char.
join : to return back from Array to String .
DEMO :
var input="gazanie, flower, blossom";
var output=input.split(',').map((e)=>`#${e.trim()}`).join(' ');
console.log(output);