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Reference - What does this regex mean?
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I have a regular expression in a code(written by someone else), I am trying to understand what that expression means.
var decimal = /^\d[0,1]+(\.\d[1,4])?$/;
Can anyone explain to me what it does...
In order:
^ - Match the beginning of the input
\d - A digit (0-9)
[0,1]+ - One or more occurrences of the characters 0, ,, or 1 —but see note below, this is probably not what the author meant to do
( - The beginning of a capture group
\. - A literal . (without the backslash, it would mean something special)
\d - A digit
[1,4] - Exactly one of the characters 1, ,, or 4 —but see note below, this is probably not what the author meant to do
) - The end of the capture group
? - Inidicates that the entire capture gruop is optional (zero or once)
$ - Match the end of the input
Re the [0,1]+ and [1,4], the expression was probably supposed to have {0,1} and {1,4} instead, which mean:
{0,1} - match what came before either zero times or once (note that you have to remove the + that was after the [0,1])
{1,4} - match what came before 1, 2, 3, or 4 times
Here's an explanation on regex101.com
If we speculate that they probably meant this
/^\d{0,1}(\.\d{1,4})?$/
...then in prose it means: Match any number that may or may not have one leading digit, and then may or may not have a decimal point followed by one to four digits. But it's still got issues, not least that the string "" matches it, and (depending on what you're doing) you probably want to support values equal to or greater than 2, which that expression doesn't.
Basically: If it's meant to validate a decimal, throw it away, and search for something that does a better job, such as this if you really want at most four digits of precision and you want to capture the fractional portion (as your original does):
/^(?:0|[1-9]\d*)(\.\d{1,4})?$/
If you want to allow any level of precision:
/^(?:0|[1-9]\d*)(\.\d+)?$/
If you don't need the capture group:
/^(?:0|[1-9]\d*)(?:\.\d{1,4})?$/ // Only allow 1-4 digits of precision
/^(?:0|[1-9]\d*)(?:\.\d+)?$/ // Allow any number of digits of precision
That last is probably what I'd go with. Note that it doesn't allow leading zeros you wouldn't normally write (e.g., it disallows 02.345). If you want to allow them, then just /^\d*(?:\.\d+)?$/.
The crucial parts:
^: Beginning of input
\d: A digit
[0,1]+: One or more occurences of 0 or 1 or ,
(\.\d[1,4])?: An optional capture group, containing: a . literal, a digit, and a 1 or 4 or ,
$: End of input
The full story can be found here.
So some allowed input is:
80.94
41111111.44
4,,,1.44
30
Related
I am trying to parse a string which has two numbers, both can be between 1 and 3 digits, and will have a colon in between. Here are some examples:
"1:1"
"1:12"
"12:1"
"123:12"
Also, the given string may also be invalid, and I need to detect if it is. My attempts so far to make sure the string is valid have looked like this: .match(/[1-9]\:[1-9]/);. But then I noticed that this wont work if a string such as this is inputted: "characters12:4characters". How would I go about validating the string to make sure it is in the format x:y?
Any help would be deeply appreciated.
Edit: numbers which contain 0 at the beginning is valid, but may not be given.
You may use
/^\d{1,3}:\d{1,3}$/
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.
I am trying to write a regular expression for a bit of javascript code that takes a user's input of a mobile number and in one regular expression, performs the following checks:
Starts with 07
Contains only numbers, whitespace or dashes
Contains exactly 11 numbers
Is this possible to do in just one regular expression and if so, how please?
I don't think it is possible with one regex, but it is possible by testing for two conditions:
if(/^07[\d\- ]+$/.test(str) && str.replace(/[^\d]/g, "").length === 11) {
//string matches conditions
}
Explanation of the regex:
^: Anchor that means "match start of string".
07: Match the string 07. Together with the above, it means that the string must start with 07.
[: Beginning of a character class i.e., a set of characters that we want to allow
\d: Match a digit (equivalent to 0-9).
\-:
" ": Match whitespace (markdown doesn't let me show a single space as code)
]: End of character class.
+: One or more of the previous.
$: Anchor that means "match end of string". Together with the ^, this basically means that this regex must apply to the entire string.
So here we check to see that the string matches the general format (starts with 07 and contains only digits, dashes or spaces) and we also make sure that we have 11 numbers in total inside the string. We do this by getting of anything that is not a digit and then checking to see that the length of the string is equal to 11.
Since #Vivin throws out the challenge :
/^07([-\s]*\d){9}[-\s]*$/
^07 : begin with digits 07
( : start group
[-\s]* : any number of - or whitespace
\d : exactly one digit
){9} : exactly 9 copies of this group (11 digits including 07)
[-\s]* : optional trailing spaces or -
$ : end of string
Of course a more useful way might be as follows
if ((telNo = telNo.replace (/[-\s]+/g, '')).match (/^07\d{9}$/)) {
....
}
which has the advantage (?) of leaving just the digits in telNo
Thank you all for trying, but after a good while trying different ideas, I finally found a working "single" regular expression:
07((?:\s|-)*\d(?:\s|-)*){9}
This make sure that it starts with 07, only contains digits, whitespace or dashes, and only 11 of them (9 plus the first 2) are numbers.
Sorry to have wasted your time.
Explanation:
() - include in capture
(?:) - do not include in capture
\s - whitespace
| - or
- - dash
* - zero or more
\d - digits only
{9} - exactly nine of what is captured
I'm trying to match a string using regex (of which I am new to) but I can't get it to match.
These should be accepted:
GT-00-TRE
KK-10-HUH
JU-05-OPR
These should not:
HTH-00-AS
HM-99-ASD
NM-05-AK
So the pattern goes 2 letters, hyphen, 2 digits (between 00 and 11 inclusive), hyphen, 3 letters.
So far the best I can come up with is:
var thePattern = /^[a-z]{2}[-][00-11][-][a-z]{3}$/gi;
I can't help but feel that I'm pretty close.
Can anyone give me any pointers?
Thanks.
This should be what you need:
var thePattern = /^[a-z]{2}[-](0\d|1[0-1])[-][a-z]{3}$/gi;
In order to do a range 00-11, you have to say "(0 followed by 0-9) or (1 followed by 0 or 1)". This is because specifying a range within [] only works for single digits. Luckily your case is pretty simple, otherwise it could get quite complex to work around that.
Your regex is OK, but for one thing: the digits matching is a bit more complex
(0\d|10|11)
you want to match a zero followed by a digit (\d) OR (|) a ten OR a eleven.
Something in square brackets represents just a single character in a range. [0-5] means any single digit between 0 and 5, [a-q] means any lowercase letter from a to q. There's no such thing as [00-11] because it would require to work on more than one character at a time.
I need a regular expression that validate for a number with length 4, 5, 6
I used ^[0-9]{4} to validate for a number of 4, but I do not know how to include validation for 5 and 6.
Try this:
^[0-9]{4,6}$
{4,6} = between 4 and 6 characters, inclusive.
[0-9]{4,6} can be shortened to \d{4,6}
Be aware that, as written, Peter's solution will "accept" 0000. If you want to validate numbers between 1000 and 999999, then that is another problem :-)
^[1-9][0-9]{3,5}$
for example will block inserting 0 at the beginning of the string.
If you want to accept 0 padding, but only up to a lengh of 6, so that 001000 is valid, then it becomes more complex. If we use look-ahead then we can write something like
^(?=[0-9]{4,6}$)0*[1-9][0-9]{3,}$
This first checks if the string is long 4-6 (?=[0-9]{4,6}$), then skips the 0s 0*and search for a non-zero [1-9] followed by at least 3 digits [0-9]{3,}.
If the language you use accepts {}, you can use [0-9]{4,6}.
If not, you'll have to use [0-9][0-9][0-9][0-9][0-9]?[0-9]?.
To match standalone 4-6-digit numbers, you may use
^\d{4,6}$ // If full string match is expected
\b\d{4,6}\b // If no letters/digits/underscores are expected on both ends
(?<!\d)\d{4,6}(?!\d) // If no digits are expected on both ends, but letters/_ are allowed
(^|\D)(\d{4,6})(?!\d) // Same as above, in case lookbehinds are not supported (get Group 2 value)
See Regex #1 - Regex #2 - Regex #3 and Regex #4 demos.
Details:
^ - start of string
\b - a word boundary
(?<!\d) - a negative lookbehind that fails the match if there is a digit immediately to the left of the current location
(^|\D) - a capturing group matching either start of string or a non-digit char
\d{4,6} - four, five or six digits
(?!\d) - a negative lookahead that fails the match if there is a digit immediately to the right of the current location
$ - end of string
I want to allow people to enter times into a textbox in various formats. One of the formats would be either:
2h for 2 hours, or
2.5h for 2 and a half hours
I want to use a regex to recognise the pattern but it's not picking it up for some reason:
I have:
var hourRegex = /^\d{1,2}[\.\d+]?[h|H]$/;
which works for 2h, but not for 2.5h.
I thought that this regex would mean - Start at the beginning of the string, have one or two digits, then have none or one decimal points which if present must be followed by one or more digits then have a h or a H and then it must be the end of the string.
I have tried the regex tool here but no luck.
/^\d{1,2}(?:\.\d+)?h$/i; Use parentheses instead of square braces.
Start at the beginning
One or two digits
Optional: a dot followed by at least one digit
End with a h
Case insensitive
RegExp tuturial
[...] - square braces mean: anything which is within the provided range.
[^...] means: Match a character which is not within the provided range
(...) - parentheses mean: Group me. Optionally, the first characters of a group can start with:
?: - Don't reference me (me, I = group)
?= - Don't include me in the match, though I have to be here
?! - I may not show up at this point
{a,b}, {a,} means: At least a, maximum b characters. Omitting b = Infinity
+ means: at least one time, match as much as possible equivalen to {1,}
* means: match as much as possible equivalent to {0,}
+? and *? have the same effect as previously described, with one difference: Match as less as possible
Examples
[a-z] One character, any character between a, b, c, ..., z
(a-z) Match "a-z", and group it
[^0-9] Match any non-number character
See also
MDN: Regular Expressions - A more detailed guide
The trouble is here :
[\.\d+]
you can not use character classes inside brackets.
Use this instead:
(\.[0-9]+)?
You've confused your square brackets with your parenthesis. Square brackets look for a single match of any contained character, whereas parenthesis look for a match of the entire enclosed pattern.
Your issue lies in [\.\d+]? It's looking for . or 0-9 or +.
Instead you should try:
/^\d{1,2}(\.\d+)?(h|H)$/
Although that will still allow users to enter invalid numbers, such as 99.3 which is probably not the expected behavior.