RegEx a name with special characters in javascript - javascript

I'm relative new to RegEx and I've encountered a problem. I want to regex a name. I want it to be max 100 characters, contain at least 2 alphabetic characters and it will allow the character '-'.
I have no problem to only check for alphabetic characters or both alphabetic characters and hyphen but I dont't want a name that potantially can be '---------'.
My code without check for hyphens is
var nameRegExp = /^([a-z]){2,100}$/;
An explanation for the code is appreciated as well.
Thanks!

I guess
/^(?=.*[a-z].*[a-z])[a-z-]{1,100}$/
the lookahead part (^(?=.*[a-z].*[a-z])) checks if there are at least two letters. This pattern ("start of string, followed by...") is a common way to express additional conditions in regexes.
You can limit the number of - by adding a negative assertion, as #Mike pointed out:
/^(?=.*[a-z].*[a-z])(?!(?:.*-){11,})[a-z-]{1,100}$/ // max 10 dashes
however it might be easier to write an expression that would match "good" strings instead of trying to forbid "bad" ones. For example, this
/^[a-z]+(-[a-z]+)*$/
looks like a good approximation for a "name". It allows foo and foo-bar-baz, but not the stuff like ---- or foo----bar----.

To limit the number of - you could add a negative look-ahead, where the number 3 is one more than the maximum number you want to allow
/^(?!(?:[a-z]*-){3,})(?=-*[a-z]-*[a-z])[a-z-]{2,100}$/

Related

Javascript regex to make sure that string matches format x:y

I am trying to parse a string which has two numbers, both can be between 1 and 3 digits, and will have a colon in between. Here are some examples:
"1:1"
"1:12"
"12:1"
"123:12"
Also, the given string may also be invalid, and I need to detect if it is. My attempts so far to make sure the string is valid have looked like this: .match(/[1-9]\:[1-9]/);. But then I noticed that this wont work if a string such as this is inputted: "characters12:4characters". How would I go about validating the string to make sure it is in the format x:y?
Any help would be deeply appreciated.
Edit: numbers which contain 0 at the beginning is valid, but may not be given.
You may use
/^\d{1,3}:\d{1,3}$/
See the regex demo
Details
^ - start of a string
\d{1,3} - one, two or three digits (\d is a shorthand character class that matches any digit (it can also be written as a [0-9] character class) and {1,3} is a limited quantifier that matches1 to 3 consecutive occurrences of the quantified subpattern)
: - a colon
\d{1,3} - one, two or three digits
$ - end of the string.

Javascript regex: check if a string is alphanumeric AND it contains a letter (at least)

The string length can be 4 - 12 characters.
It may contain ONLY letters and numbers, but it has to contain at least 1 number.
And I need to solve this with a single regex pattern.
I tried something like:
/^(?=.*[a-z]*)(?=.*[0-9]+).{4,12}$/i
This won't accept less than 4 or more than 12 chars and it also checks if the string contains a number, but obviously it's not good because of the .* parts.
I wasn't able to figure out how to exclude all non-alphanumeric characters.
Any help would be appreciated!
Thanks in advance!
I think your pattern is close, but I would use this:
/^(?=.*[0-9])[a-z0-9]{4,12}$/i
The only lookahead you need is one which asserts that there is a single number. There is no requirement for there to be any letters, so don't bother adding an assertion for that. Then, match any alphanumeric character 4 to 12 times.
console.log(/^(?=.*[0-9])[a-z0-9]{4,12}$/i.test('abc'));
console.log(/^(?=.*[0-9])[a-z0-9]{4,12}$/i.test('123'));
console.log(/^(?=.*[0-9])[a-z0-9]{4,12}$/i.test('abcd'));
console.log(/^(?=.*[0-9])[a-z0-9]{4,12}$/i.test('Abc1'));

regex for specific phone number arrangement

i'm currently using this regex
"/^([0-9\(\)\/\+ \-]*)$/",
which is fine,but the problem is i am also using a masking script,
which produces this line automatically,
(___) ___-____
and it messes up my validation, what regex code can allow me to verify only this type of input from the use
(999) 999-9999
and also not accept a "blank" input field from user when entered. any length is fine, as long as it only accepts this inputs that i mentioned above.
This should work:
^\(\d{3}\)\s{0,1}\d{3}-\d{3}$
Breaking this regexp:
\(\d{3}\) matches only three numbers between brackets.
\s{0,1} matches only 0 or 1 space.
\d{3}-\d{3} matches only three numbers followed by '-' and then three other numbers.
First, when asking about regular expressions, you should always say which language or tool you are using because that affects what features are available and which characters need to be quoted with backslash. I'll assume you're asking about JavaScript based on your question's tags.
You say any length is fine. I shall take that to mean that each sequence of consecutive digits can contain any number of digits from one to infinity. I shall assume there's exactly one space and exactly one dash. On that basis, your RE is:
/^\(\d+\) \d+-\d+$/
If, as is more likely, you want to limit the lengths of the digit sequences, you would say something like:
/^\(\d{3,4}\) \d{3}-\d{1,10}$/
(three or four digits, exactly three digits, one to ten digits).
I have omitted any capturing parentheses (...) , which are a bit redundant if you're capturing the whole string ^(....)$ .
Here's a concise summary of JavaScript regex syntax:
http://www.regextester.com/jssyntax.html
Formatting and validation are two very different things. If you try to mix them, you will fail.
That being said, before performing validation you should strip all formatting characters from your string, then validate the content.
// remove everything that isn't a digit
var strippedNumber = value.replace(/\D/g, '');
if (strippedNumber.length === 10) {
// valid phone number
}

Regular Expression with optional elements in input string in javascript

Can anyone give me the regular expression for currency which have the following formats :
1000 - valid
1,000 - valid
1,000.00 or 1000.00 - valid.
This means, the number May or May Not contain a comma(,) separator every 3 digits.
The number May Or May Not contain a dot (.), and if it carries a dot(.) it should show atleast 1 number after the decimal place.
And lastly it should be numerical characters only. If I need to make my question clear kindly suggest.
/^\d{1,3}(?:(?:,\d{3})*|(?:\d{3})*)(?:\.\d{1,2})?$/
"Between one and three digits, then any number of groups of three digits prefixed by a comma or any number of groups of three digits not prefixed by said comma (disallowing a mix of the two kinds of groups), then an optional group of one or two digits prefixed by a dot."
Note: This regex assumes that you want to validate an entire string against the criteria outlined in your question. If you want to use it to find such numbers in a longer string, you will need to remove the ^ and $ from the beginning and end of the expression.
Something like so should work: (,?\d{3})+(\.\d{2})?. The regex will attempt to match a sequence of 3 digits precedeed by an optional comma, which is then, finally followed by an optional decimal point and 2 digits.
Please refer to this tutorial for more information.
EDIT: As per the comment below, the above regex can fail. I would recommend first using this regular expression: ^[\d][\d.,]+$ to make sure that you only have digits, thousand and decimal seperators. This regular expression will also make sure that the number starts with a digit, not with anything else. You could most likely have one regular expression which does everything, but it will most likely be quite complex.

Regex to validate brazilian money using Javascript

I need to validate some form fileds that contain brazilian money (its name is "Real") using Javascript. It has the following format:
0,01
0,12
1,23
12,34
123,45
1.234,56
12.235,67
123.456,78
1.234.567,89
12.345.678,90
123.456.789,01
1.234.567.890,12
My regex knowledge is weak, can somebody help me please?
Does this do what you want?
^\d{1,3}(?:\.\d{3})*,\d{2}$
That says "1 to 3 digits, optionally followed by any number of groups of three digits preceded by a period, followed by a comma and two more digits." If you want to allow the leading whitespace present in your example, add \s* to the front:
^\s*\d{1,3}(?:\.\d{3})*,\d{2}$
EDIT: As #ElRonnoco pointed out, the above regular expression accepts leading zeroes (e.g. 010.100,00). To disallow those, you may use this longer version:
^\s*(?:[1-9]\d{0,2}(?:\.\d{3})*|0),\d{2}$
EDIT 2 The above regular expressions all match a string containing a single monetary amount and nothing else. It's not clear from the question if that's the intent.
EDIT 3 To allow numbers that have no decimal part, or only one decimal digit, change it like this:
^\s*(?:[1-9]\d{0,2}(?:\.\d{3})*|0)(?:,\d{1,2})?$
I would give this regex a try:
\d+(?:\.\d{3})*?,\d{2}
What it says is:
- match digits until
a. a dot followed by 3 digits is found (and this step can be repeated several times)
b. or a comma followed by 2 digits is found
EDIT:
- thanks for the comments, I forgot about the constraint for the first value
updated regex
\d{1,3}(?:\.\d{3})*?,\d{2}
Complementing Mark's reply:
Who needs "." in the string and not "," to count cents. And need find the values in middle a text :
(?:[1-9]\d{0,2}(?:\,\d{3})*|0)(?:.\d{1,2})?
https://regexr.com/61166

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