I need to create a RegEx to verify the string input to a text box is exactly 2 digits long and is within the range 01-25 or 99.
I am new to creating my own RegEx and came up with:
[01-25,99]{2}
This verifies the number is 2 digits long, but it finds many matches outside of the indicated range.
Character classes can contain character ranges, not value ranges.
Try this alternation:
0[1-9]|1\d|2[0-5]|99
If you want to limit the entire input to be such, wrap in ^ and $:
^0[1-9]|1\d|2[0-5]|99$
This should work:
# ^(?:0[1-9]|1\d|2[0-5]|99)$
^
(?:
0 [1-9]
| 1 \d
| 2 [0-5]
| 99
)
$
Maybe something as dumb as (0[1-9]|1[0-9]|2[0-5]|99)
the way that regex works is as individual numbers, not as a range like you would assume. This means that your pattern :
[01-25,99]
would actually match anything that has a 0,1-2,5, a comma, or a 9 (saying it twice doesn't make a difference), and putting `{2} means that you are matching any string that has both of these in a row.
since it was answered while I was typing this, look at the pattern sln posted for one that should work.
Also like this
[0][1-9]|[1][0-9]|[2][0-5]|[99]
Related
sorry for the novice question by I've looked and can't seem to find a question that addressed this.
I want a regex that describes all strings over L={0,1} ending with an even number of 0s.
Examples: 00, 0000, 100, 0100, 001100... basically anything starting with 0 or 1 and ending with an even number of 0s
This is what I've got so far: ((0|1)*1)00+ but this doesn't allow me to get 00 since that must be a 1 always. I can't find a way to put as many 0s as I want at the beginning without having to put that 1.
Thanks a lot.
You could write the pattern as:
^([01]*1)?(00)+$
^ Start of string
( Capture group
[01]*1 Match zero or more repetitions of either 0 or 1 followed by matching 1
)? Close the group and make it optional using ?
(00)+ Match one or more repetitions of 00
$ End of string
See a Regex demo.
If supported, you can also use non capture groups (?:
An even number of 0s is (00)*. It needs to be at the end, so that part of the regex will be (00)*$.
What precedes that even number of 0s? Either nothing or an arbitrary sequence of 0s and 1s ending with a 1. So that's (|[01]*1).
Putting it together, we have:
^(|[01]*1)(00)*$
(I'm assuming extended regex syntax, where (, ), and | don't have to be escaped. Adjust the syntax as needed.)
I have not tested this.
For the example i have these four ip address:
10.100.0.11; wrong
10.100.1.12; good
10.100.11.4; good
10.100.44.1; wrong
The task has simple rules. In the 3rd place cant be 0, and the 4rd place cant be a solo 1.
I need to select they from an ip table in different routers and i know only this rules.
My solution:
^(10.100.[1-9]{1,3}.[023456789]{1,3})$
but in this case every number with 1 like 10, 100 etc is missing, so in this way this solution is wrong.
^(10.100.[1-9]{1,3}.[1-9]{2,3})$
This solve the problem of the single 1, but make another one.
From the rules you have given, this regex should work:
10\.100\.([123456789]\d*|\d{2,})\.([^1]$|\d{2,})
it also matches 3rd position number containing a 0 but not in the first place.
so 10.100.10.4 will match as well as 10.100.02.4
I don't know if it's the intended behavior since I'm not familiar with ip adress.
The last part \.([^1]$|\d{2,}) reads like this:
"after the 3rd dot is either
a character which is not 1 followed by the end of the line
or two or more digits"
If you want to avoid malformed string containing non-digit character like 10.100.12.a to be match you should replace [^1] by [023456789] or lazier (and therefore better ;) by [02-9]
I use https://regex101.com to debug regex. It's just awesome.
Here is your regex if you want to play with it
You might use
^10\.100\.[1-9]{1,3}\.(?:[02-9]|\d{2,3})$
The pattern matches
^ Start of string
10\.100\. Match 10.100. (note to escape the dot to match it literally)
[1-9]{1,3} Match 3 times a digit 1-9
\. Match a dot
(?: Non capture group
[02-9] Match a digit 0 or 2-9
| Or
\d{2,3} Match 2 or 3 digits 0-9
) Close the group
$ End of string
Regex demo
I'm looking to build a regex that matches the following group of numbers:
10xxxxxxx
1116xxxxx
143xxxxxx
146xxxxxx
149xxxxxx
159xxxxxx
16xxxxxxx
(note the length is always 9)
where x is any digit. My best attempt yielded this:
/^1[01456][1369]*[6]*[0-9]$/
However, I can't get the length of the string to always be 9. Any ideas?
Edit: Maybe I wasn't clear enough, it needs to match those 7 cases, and ONLY those, inclusively and exclusively.
How about:
^1(?:[06]\d{2}|116|4[369]\d|59\d)\d{5}$
use this pattern
^1[01456](16|3\d|6\d|9\d|\d\d)\d{5}$
Is this what you want?
^(?=[0-9]{9}$)(?:10|1116|143|146|149|159|16)
Demo
This starts by looking at the beginning of the string for exactly 9 digits using a positive lookahead anchored to the end of the string. Then we look for any of your 7 specific groups of numbers that the string can start with.
You can use this regex:
/^1[01456][1369][0-9]{6}$/
Since 3 digits are already matched by first 3 patterns 1, [01456] and [1369] so last one must match exact 6 characters to enforce it a 9 digit input.
I am trying to create a regex to have a string only contain 0-9 as the characters and it must be at least 1 char in length and no more than 45. so example would be 00303039 would be a match, and 039330a29 would not.
So far this is what I have but I am not sure that it is correct
[0-9]{1,45}
I have also tried
^[0-9]{45}*$
but that does not seem to work either. I am not very familiar with regex so any help would be great. Thanks!
You are almost there, all you need is start anchor (^) and end anchor ($):
^[0-9]{1,45}$
\d is short for the character class [0-9]. You can use that as:
^\d{1,45}$
The anchors force the pattern to match entire input, not just a part of it.
Your regex [0-9]{1,45} looks for 1 to 45 digits, so string like foo1 also get matched as it contains 1.
^[0-9]{1,45} looks for 1 to 45 digits but these digits must be at the beginning of the input. It matches 123 but also 123foo
[0-9]{1,45}$ looks for 1 to 45 digits but these digits must be at the end of the input. It matches 123 but also foo123
^[0-9]{1,45}$ looks for 1 to 45 digits but these digits must be both at the start and at the end of the input, effectively it should be entire input.
The first matches any number of digits within your string (allows other characters too, i.e.: "039330a29"). The second allows only 45 digits (and not less). So just take the better from both:
^\d{1,45}$
where \d is the same like [0-9].
Use this regular expression if you don't want to start with zero:
^[1-9]([0-9]{1,45}$)
If you don't mind starting with zero, use:
^[0-9]{1,45}$
codaddict has provided the right answer. As for what you've tried, I'll explain why they don't make the cut:
[0-9]{1,45} is almost there, however it matches a 1-to-45-digit string even if it occurs within another longer string containing other characters. Hence you need ^ and $ to restrict it to an exact match.
^[0-9]{45}*$ matches an exactly-45-digit string, repeated 0 or any number of times (*). That means the length of the string can only be 0 or a multiple of 45 (90, 135, 180...).
A combination of both attempts is probably what you need:
^[0-9]{1,45}$
^[0-9]{1,45}$ is correct.
Rails doesnt like the using of ^ and $ for some security reasons , probably its better to use \A and \z to set the beginning and the end of the string
For this case word boundary (\b) can also be used instead of start anchor (^) and end anchor ($):
\b\d{1,45}\b
\b is a position between \w and \W (non-word char), or at the beginning or end of a string.
I have two regular expressions that validate the values entered.
One that allows any length of Alpha-Numeric value:
#"^\s*(?<ALPHA>[A-Z0-9]+)\s*"
And the other only allows numerical values:
#"^\s*(?<NUM>[0-9]{10})"
How can I get a numerical string of the length of 11 not to be catched by the NUM regex.
I think what you're trying to say is that you don't want to allow any more than 10 digits. So, just add a $ at the end to specify the end of the regex.
Example: #"^\s*(?[0-9]{10})$"
Here's my original answer, but I think I read you too exact.
string myRegexString = `#"(?!(^\d{11}$)` ... your regex here ... )";
That reads "while ahead is not, start, 11 digits, end"
If it's single line, you could specify that your match must happen at the end of the line, like this in .net ...
^\s*([0-9]{10})\z
That will accept 1234567890 but reject 12345678901.
Do you mean you want to match up to 10 digits? Try this:
#"^\s*[0-9]{1,10}\s*$"
If you are trying to match only numbers that are 10 digits long, just add a trailing anchor using $, like this:
^\s*(?:[0-9]{10})\s*$
That will match any number that is exactly 10 digits long (with optional space on either side).
var pattern =/\b[0-9]{10}$\b/;
// the b modifier is used for boundary and $ is used for exact length
Match something non-numeric after the length 10 string. My regex-foo isn't that good, but I think you've got it setup there to catch a numeric string of exactly length 10, but since you don't match anything after that, a length 11 string would also match. Try matching beyond the end of the number and you'll be good.
This should match only 10 digits and allow arbitrary numbers of whitespaces before and after the digits.
Non-capturing version: (only matches, the matched digits are not stored)
^\s*(?:\d{10})\s*$
Capturing version: (the matched digits are available in subgroup 1, as $1 or \1)
^\s*(\d{10})\s*$
You could try alternation?
^\s*(?\d{1,10}|\d{12,})