I want a Regex for my mongoose schema to test if a username contains only letters, numbers and underscore, dash or dot. What I got so far is
/[a-zA-Z0-9-_.]/
but somehow it lets pass everything.
Your regex is set to match a string if it contains ANY of the contained characters, but it doesn't make sure that the string is composed entirely of those characters.
For example, /[a-zA-Z0-9-_.]/.test("a&") returns true, because the string contains the letter a, regardless of the fact that it also includes &.
To make sure all characters are one of your desired characters, use a regex that matches the beginning of the string ^, then your desired characters followed by a quantifier + (a plus means one or more of the previous set, a * would mean zero or more), then end of string $. So:
const reg = /^[a-zA-Z0-9-_.]+$/
console.log(reg.test("")) // false
console.log(reg.test("I-am_valid.")) // true
console.log(reg.test("I-am_not&")) // false
Try like this with start(^) and end($),
^[a-zA-Z0-9-_.]+$
See demo : https://regex101.com/r/6v0nNT/3
/^([a-zA-Z0-9]|[-_\.])*$/
This regex should work.
^ matches at the beginning of the string. $ matches at the end of the string. This means it checks for the entire string.
The * allows it to match any number of characters or sequences of characters. This is required to match the entire password.
Now the parentheses are required for this as there is a | (or) used here. The first stretch was something you already included, and it is for capital/lowercase letters, and numbers. The second area of brackets are used for the other characters. The . must be escaped with a backslash, as it is a reserved character in regex, used for denoting that something can be any character.
Related
I'm developing a pattern that validates string if it does not contain more then two matches of #. here is code:
^[^\!|\#|\$|\%|\^|\&|\*|\+][\w]*([\w ]+\#[\w ]*){0,2}$
[^!|\#|\$|\%|\^|\&|*|+]
this is group of not acceptable symbols.
additionally, the pattern should validate string in case if it contains other symbols( - _ , . / ). each symbol should have it's own counter and should not match in any position more than two times.
for example if i have s string like this:
Mcloud dr. #33/#45, some text, some text
it should be valid. but in this case should not:
Mcloud dr. ###33/$#45, ####, ----
What would you suggest ?
Given that you want to match alphanumerics characters and some special symbols ()-_,./ You have to mention them in a character class like this.
Regex: ^(?!.*([(),.#/-])\1)([\w (),.#/-]+)$
Explanation:
(?!.*([(),.#/-])\1) asserts that there shouldn't be more than one character mentioned in character class. This asserts from beginning of string to end.
([\w (),.#/-]+) matches the rest of the string for allowed characters from beginning to end.
Regex101 Demo
I want to check by regex if:
String contains number
String does not contain special characters (!<>?=+#{}_$%)
Now it looks like:
^[^!<>?=+#{}_$%]+$
How should I edit this regex to check if there is number anywhere in the string (it must contain it)?
you can add [0-9]+ or \d+ into your regex, like this:
^[^!<>?=+#{}_$%]*[0-9]+[^!<>?=+#{}_$%]*$
or
^[^!<>?=+#{}_$%]*\d+[^!<>?=+#{}_$%]*$
different between [0-9] and \d see here
Just look ahead for the digit:
var re = /^(?=.*\d)[^!<>?=+#{}_$%]+$/;
console.log(re.test('bob'));
console.log(re.test('bob1'));
console.log(re.test('bob#'))
The (?=.*\d) part is the lookahead for a single digit somewhere in the input.
You only needed to add the number check, is that right? You can do it like so:
/^(?=.*\d)[^!<>?=+#{}_$%]+$/
We do a lookahead (like peeking at the following characters without moving where we are in the string) to check to see if there is at least one number anywhere in the string. Then we do our normal check to see if none of the characters are those symbols, moving through the string as we go.
Just as a note: If you want to match newlines (a.k.a. line breaks), then you can change the dot . into [\W\w]. This matches any character whatsoever. You can do this in a number of ways, but they're all pretty much as clunky as each other, so it's up to you.
Good Day,
I'm creating a javascript function to exclude certain codes from the database.
For example, the following are valid codes:
FF99
1G499
RNDD
In other words, I want the codes to consist of only alphanumerics. I opened up the console in Chrome and tried:
> re = new RegExp('\w+')
> re.test('whatever')
true
> re.test('what???')
true
> var test = new RegExp('^\w+')
> test.test('what')
true
> test.test('what999')
true
> test.test('what???')
true
So I know that \w can either be a 0-9, a-z, A-Z. I don't know why the regex is passing if I enter '?' when they shouldn't.
Am I missing something?
You're misinterpreting your results. The regexp \w+ means "one or more word characters". It doesn't specify where in the string these characters can be found. In all of your tests, the provided string contains at least one word character, so they all pass.
What you're meaning to do is ensure the string contains only alphanumerics. Try the following regex:
^\w+$
Broken down, this means:
^ = match start of string
\w = letters or digits
+ = one or more of the previous element (in this case, the set) (this is greedy)
$ = match the end of the string
In English, this means, "between the start and end of the string, only alphanumeric characters will match. Match as many as possible, or none at all"
Documentation on ^ and $
Note: if you write your regex as a string, you need to escape the \ like so:
new RegExp("^\\w+$")
Otherwise, JavaScript will interpret \w as an escape sequence, which it is not. You can also use the syntax
new RegExp(/^\w+$/)
in which case you don't need to escape the \, since it isn't a string.
Test is returning true because the pattern \w+ is matching the alphanumeric part in your test strings.
re.test('what???') for example will return true because it matches what.
If you only want to match strings consisting of only alphanumeric characters, you should use something like ^\w+$ .
I want to extract a string from another using JavaScript / RegExp.
Here is what I got:
var string = "wp-button wp-image-45 wp-label";
string.match(/(?:(?:.*)?\s+)?(wp-image-([0-9]+))(:?\s(?:.*)?)?/);
// returnes: ["wp-button ", "wp-image-45", "45", undefined]
I just want to have "wp-image-45", so:
(Optional) any character
(Optional) followed by whitespace
(Required) followed by "wp-image-"
(Required) followed by any number
(Optional) followed by whitespacy
(Optional) followed by any character
What is missing here? Is it just some kind of bracketing or more?
I also tried
string.match(/(?:(?:.*)?\s+)?(?=(wp-image-([0-9]+)))(?=(:?\s(?:.*)?)?)/)
Edit: In the end I just want to have the number. But I'd also make this step in between.
Regexps are not required to start matching at the beginning of the string, so your attempts to match whitespace and any character aren't necessary. Also, "any character" includes whitespace (except newlines in certain modes).
This should be all you need:
string.match(/\bwp-image-(\d+)\b/)
This will capture, for example, "wp-image-123" into matching group 0, and "123" into matching group 1.
\b means "word boundary", which ensures that you won't match "abcwp-image-123def". A word boundary is defined as any place where a non-word character is followed by a word character, or vice versa. A word character is consists of a letter, a number or an underscore.
Also, I used \d instead of [0-9] simply out of convenience. They have slightly different meaning (\d also matches characters considered numbers in other languages), but that won't make a difference in your case.
If all of that surrounding stuff is optional and all you want is the number then there's no point to matching for any of that stuff except for that "wp-image-" prefix, just do:
var string = "wp-button wp-image-45 wp-label";
string.match(/wp-image-([0-9]+)/);
This code is used to get rid of mime type from rawdata.but I can not understand how it works
content.replace(/^[^,]*,/ , '')
it seems quite different from java.... any help will be appreciated.
Your mime-type probably is seperated by a comma , and at the beginning of your raw data.
This regex says take everything from the beginning (^) that is NOT a comma ([^,]*) (the star makes it as many characters until there is a comma) and take the comma itself (,). Then replace it by nothing ('').
This one only gets the first appearence because it is marked by the beginning ^ that it must be at the beginning of the string.
The first thing you need to know is that there are regex literals in JavaScript, constructed by pairs of slashes. So like "..." is a string, /.../ is a regex. That's actually the only difference your code shows as compared to a Java regex.
Then, [abc] within a regex is called a character class, meaning "one character out of a, b or c". Conversely, [^abc] is a negated character class, meaning "one character except a, b or c".
So your sample means:
/ # Start of regex literal
^ # Start the match at the start of the string
[^,]* # Match any number of characters except commas
, # Match a comma
/ # End of regex literal
The regular expression is the text between the two forward slashes, the first carat (^) means at the begining of the string, the brackets mean a character class, the carat inside the brackets means any character except a comma, then asterisk after the closing bracket means match zero or more of the character defined by the character class (which again is any character except the comma), and then finally the last comma means match the comma after all this. Then its used in a replace function so the matching result will be replaced with the second parameter, in your case: an empty string.
Basically it matches the first characters up to and including the first comma in the 'content' variable and then replaces it with an empty string.