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.
Related
Goal: to make a generalized Regular Expression for a fixed length string inside a larger string. This string has a specified padding character, followed by an integer counter that increments. Ideally, there would be some way to say, "I want this group to be of length 10 and contain only one type of character followed by a different character."
I am trying to match this within variable data (could be numbers could be letters could be symbols):
The padding-characters + numbers add up to a specified length, here would be 5.
These are the allowed padding + number combinations.
$$$$1
$$$12
$$123
$1234
Here is an example:
<variable-data> <padding-characters> <numbers> <variable-data>
............... .................... ddddddddd ...............
(where periods are any characters and 'd' is any digit)
Example Data:
ABC $$$$ 1 $!#
Example Regex:
ABC\$*\d+\$!#
Match:
ABC$$$$1$!#
ABC$$$12$!#
ABC$$123$!#
ABC$1234$!#
ABC12345$!#
No Match:
ABC$$123456789$!#
ABC1$2$34$!#
Regex101
What I've Tried:
ABC(?=.{5})\$*\d+\$!#
This does not work because it still matches into the next digits because of \d+. Another thing I tried was
ABC(?=[\$\d]{5}[^\$\d])(\$*\d+)\$!#
Which aims to stop looking after it encounters a non-digit or non $, but that's not helpful since the next part of the string COULD start with a $ or a digit.
The easiest Regex to solve this:
(\$\$\$\$\d|\$\$\$\d\d|\$\$\d\d\d|\$\d\d\d\d|\d\d\d\d\d)
But I am trying to make this more generalized, and there can be a variable amount of padding E.G.
$$$$$$$$$1
$$$$$$$$12
...
You could look ahead to check that you don't have an inverted sequence of padding character and digit within the scope of the next 5 characters, and then require and capture 5 characters that are only digits and padding characters:
ABC(?!.{0,3}\d\$)([\$\d]{5})\$!#
If you need at least one digit, then:
ABC(?!.{0,3}\d\$)([\$\d]{4}\d)\$!#
ABC(?=.{5}\$!#)\$*\d+\$!#
This is very similar to your first attempt, but with the slight difference that the lookahead also contains the terminating string. This gives it something to anchor to, to make sure the regex doesn't match anything more.
I am trying to understand regex, for digits of length 10 I can simply do
/^[0-9]{10}$/
for hyphen only I can do
/^[-]$/
combining the two using group expression will result in
/^([0-9]{10})|([-])$/
This expression does not work as intended, it somehow will match part of the string instead of not match at all if the string is invalid.
How do I make the regex expression that accepts only "-" or 10 digits?
It would have worked fine to combine your two regexps exactly as you had them. In other words, just use the alternation/pipe operator to combine
/^[0-9]{10}$/
and
/^[-]$/
as is, directly into
/^[0-9]{10}$|^[-]$/
↑↑↑↑↑↑↑↑↑↑↑ ↑↑↑↑↑ YOUR ORIGINAL REGEXPS, COMBINED AS IS WITH |
This can be represented as
and that would have worked fine. As others have pointed out, you don't need to specify the hyphen in a character class, so
/^[0-9]{10}$|^-$/
↑ SIMPLIFY [-] TO JUST -
Now, we notice that each of the two alternatives has a ^ at the beginning and a $ at the end. That is a bit duplicative, and it also makes it little harder to see immediately that the regexp is always matching things from beginning to end. Therefore, we can rewrite this, as explained in other answers, by taking the ^ and $ out of both sub-regexps, and combine their contents using the grouping operator ():
/^([0-9]{10}|-)$/
↑↑↑↑↑↑↑↑↑↑↑↑↑ GROUP REGEXP CONTENTS WITH PARENS, WITH ANCHORS OUTSIDE
The corresponding visualization is
That would also work fine, but you could use \d instead of [0-9], so the final, simplest version is:
/^(\d{10}|-)$/
↑↑ USE \d FOR DIGITS
and this visualizes as
If for some reason you don't want to "capture" the group, use (?:, as in
/^(?:\d{10}|-)$/
↑↑ DON'T CAPTURE THE GROUP
and the visualization now shows that group is not captured:
By the way, in your original attempt to combine the two regexps, I noticed that you parenthesized them as in
/^([0-9]{10})|([-])$/
↑↑↑↑↑↑↑↑↑↑↑ ↑↑↑↑↑ YOU PARENTHESIZED THE SUB-REGEXPS
But actually this is not necessary, because the pipe (alternation, of "or") operator has low precedence already (actually it has the lowest precedence of any regexp operator); "low precedence" means it will apply only after things on both side are already processed, so what you wrote here is identical to
/^[0-9]{10}|[-]$/
which, however, still won't work for the reasons mentioned in other answers, as is clear from its visualization:
How do I make the regex expression that accepts only "-" or 10 digits?
You can use:
/^([0-9]{10}|-)$/
RegEx Demo
Your regex is just asserting presence of hyphen in the end due to misplacements of parentheses.
Here is the effective breakdown of OP's regex:
^([0-9]{10}) # matches 10 digits at start
| # OR
([-])$ # matches hyphen at end
which will cause OP's regex to match any input starting with 10 digits or ending with hyphen making these invalid inputs also a valid match:
1234567890111
1234----
------------------
1234567890--------
To get the regex expression that accepts only "-" or 10 digits - change your regexp as shown below:
^(\d{10}|-)$
DEMO link
The problem with your regex is it's looking for strings either
starting with 10 digits i.e. ^([0-9]{10}) or
ends with "-" - i.e. ([-])$
You needs an addtional wrapping ^( .. )$ to get this work. i.e.
/^(([0-9]{10})|([-]))$/
Better yet /^([0-9]{10}|-)$/ since [-] and - are both the same.
I am buried in a RegExp hell and can't find way out, please help me.
I need RegExp that matches only numbers (at least 1 number) and one of this characters: <, >, = (exactly one of them one time).
My reg. expression looks like this:
^[0-9]+$|^[=<>]{1}$
And I thought it should match when my string containts one or more digits and exactly 1 special character defined by me. But it doesn't act correctly. I think there might be problem with my start/end of string definition but Im not sure about that.
Examples that should pass include:
<1
=2
22>
>1
=00123456789
Examples that should not pass this reg. exp.:
<<2
==222
<>=2
I thought it should match when my string containts one or more digits and exactly 1 special character
No, the original pattern matches a string contains one or more digits or exactly 1 special character. For example it will match 123 and = but not 123=.
Try this pattern:
^\d+[=<>]$
This will match that consists of one or more digits, followed by exactly one special character. For example, this will match 123= but not 123 or =.
If you want your special character to appear before the number, use a pattern like this instead:
^[=<>]\d+$
This will match =123 but not 123 or =.
Update
Given the examples you provided, it looks like you want to match any string which contains one or more digits and exactly one special character either at the beginning or the end. In that case use this pattern:
^([=<>]\d+|\d+[=<>])$
This will match <1, =2, 22>, and >1, but not 123 or =.
Just use [0-9]+[=<>]
Here are visualizers of your regexp and this one:
http://www.regexper.com/#%5E%5B0-9%5D%2B%24%7C%5E%5B%3D%3C%3E%5D%7B1%7D%24
http://www.regexper.com/#%5B0-9%5D%2B%5B%3D%3C%3E%5D
Your regex says:
1 or more numbers OR 1 symbol
Also, the ^ and $ means the whole string, not contains. if you want a contains, drop them. I don't know if you have a space between the number and symbol, so put in a conditional space:
[0-9]+\s?[=<>]{1}
This should work.
^[0-9]+[=<>]$
1 or more digits followed by "=<>".
Try this regex:
^\d+[=<>]$
Description
This one:
/^\d+[<>=]$|^[<>=]\d+$/
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.
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.