I am trying to come up with a regular expression that will accept an input as long as it doesn't equal 0. To clarify, I mean only 0. An input that contains a 0 would still match, for instance 408.
This is what I have so far:
^[^0]
Use this:
^(?!0$).*
Essentially, it checks that the start of the string is not followed by a 0 and the end of the string. The only string that this is the case is "0", so all others will be matched.
However, if you're controlling the validation yourself, it would be easier to just forgo regex and check that the string is not equal to "0".
see this demo https://regex101.com/r/vS6vT3/1
/^[^0]{1}.*$/gm
or with Negative Lookahead
/^(?!0{1}$).*$/gm
Related
I need regex with following conditions
It should accept maximum of 5 digits then upto 3 decimal places
it can be negative
it can be zero
it can be only numbers (max. upto 5 digit place)
it can be null
I have tried following but its not, its not fulfilling all conditions
#"^([\-\+]?)\d{0,5}(.[0-9]{1,3})?)$"
E.g. maximum value can hold is from -99999.999 to 99999.999
Use this regex:
^[-+]?\d{0,5}(\.[0-9]{1,3})?$
I only made two changes here. First, you don't need to escape any characters inside a character class normally, except for opening and closing brackets, or possibly backslash itself. Hence, we can use [-+] to capture an initial plus or minus. Second, you need to escape the dot in your regex, to tell the engine that you want to match a literal dot.
However, I would probably phrase this regex as follows:
^[-+]?\d{1,5}(\.[0-9]{1,3})?$
This will match one to five digits, followed by an optional decimal point, followed by one to three digits.
Note that we want to capture things like:
0.123
But not
.123
i.e. we don't want to capture a leading decimal point should it not be prefixed by at least one number.
Demo here:
Regex101
I assume you're doing this in C# given the notation. Here's a little code you can use to test your expression, with two corrections:
You have to escape the dot, otherwise it means "any character". So, \. instead of .
There was an extraneous close parenthesis that prevented the expression from compiling
C#:
var expr = #"^([\-\+]?)\d{0,5}(\.[0-9]{1,3})?$";
var re = new Regex(expr);
string[] samples = {
"",
"0",
"1.1",
"1.12",
"1.123",
"12.3",
"12.34",
"12.345",
"123.4",
"12345.123",
".1",
".1234"
};
foreach(var s in samples) {
Console.WriteLine("Testing [{0}]: {1}", s, re.IsMatch(s) ? "PASS" : "FAIL");
}
Results:
Testing []: PASS
Testing [0]: PASS
Testing [1.1]: PASS
Testing [1.12]: PASS
Testing [1.123]: PASS
Testing [12.3]: PASS
Testing [12.34]: PASS
Testing [12.345]: PASS
Testing [123.4]: PASS
Testing [12345.123]: PASS
Testing [.1]: PASS
Testing [.1234]: FAIL
It should accept maximum of 5 digits
[0-9]{1,5}
then upto 3 decimal places
[0-9]{1,5}(\.[0-9]{1,3})?
it can be negative
[-]?[0-9]{1,5}(\.[0-9]{1,3})?
it can be zero
Already covered.
it can be only numbers (max. upto 5 digit place)
Already covered. 'Up to 5 digit place' contradicts your first rule, which allows 5.3.
it can be null
Not covered. I strongly suggest you remove this requirement. Even if you mean 'empty', as I sincerely hope you do, you should detect that case separately and beforehand, as you will certainly have to handle it differently.
Your regular expression contains ^ and $. I don't know why. There is nothing about start of line or end of line in the rules you specified. It also allows a leading +, which again isn't specified in your rules.
Consider a simple integer digit identifying expression like this:
[0-9]+ printf("Integer");
Now if i give 123 as an input it returns Integer, fair enough. Now if I give s123 as the input it prints out sInteger. The unmatched s is being printed by default ECHO that's cool with me. But why is Integer also printed. Shouldn't lex return just s? My input is considered as a whole string right? I mean s123 is considered as a 1 full input?. As soon as s is encountered which does not match [0-9]+ so it should just echo default unmatched value s123 but why sInteger?
The string s123 is being matched by the regex [0-9]+. If you want to match strings which consist of only integers, you should try ^[0-9]+$.
I am trying to validate a comma separated list for numbers 1-8.
i.e. 2,4,6,8,1 is valid input.
I tried [0-8,]* but it seems to accept 1234 as valid. It is not requiring a comma and it is letting me type in a number larger than 8. I am not sure why.
[0-8,]* will match zero or more consecutive instances of 0 through 8 or ,, anywhere in your string. You want something more like this:
^[1-8](,[1-8])*$
^ matches the start of the string, and $ matches the end, ensuring that you're examining the entire string. It will match a single digit, plus zero or more instances of a comma followed by a digit after it.
/^\d+(,\d+)*$/
for at least one digit, otherwise you will accept 1,,,,,4
[0-9]+(,[0-9]+)+
This works better for me for comma separated numbers in general, like: 1,234,933
You can try with this Regex:
^[1-8](,[1-8])+$
If you are using python and looking to find out all possible matching strings like
XX,XX,XXX or X,XX,XXX
or 12,000, 1,20,000 using regex
string = "I spent 1,20,000 on new project "
re.findall(r'(\b[1-8]*(,[0-9]*[0-9])+\b)', string, re.IGNORECASE)
Result will be ---> [('1,20,000', ',000')]
You need a number + comma combination that can repeat:
^[1-8](,[1-8])*$
If you don't want remembering parentheses add ?: to the parens, like so:
^[1-8](?:,[1-8])*$
I'm trying to make a Regular Expression that captures the following:
- XX or XX:XX, up to 6 repetitions (XX:XX:XX:XX:XX:XX), where X is a hexadecimal number.
In other words, I'm trying to capture MAC addresses than can range from 1 to 6 bytes.
regex = re.compile("^([0-9a-fA-F]{2})(?:(?:\:([0-9a-fA-F]{2})){0,5})$")
The problem is that if I enter for example "11:22:33", it only captures the first match and the last, which results in ["11", "22"].
The question: is there any method that {0,5} character will let me catch all repetitions, and not the last one?
Thanks!
Not in Python, no. But you can first check the correct format with your regex, and then simply split the string at ::
result = s.split(':')
Also note that you should always write regular expressions as raw strings (otherwise you get problems with escaping). And your outer non-capturing group does nothing.
Technically there is a way to do it with regex only, but the regex is quite horrible:
r"^([0-9a-fA-F]{2})(?:([0-9a-fA-F]{2}))?(?:([0-9a-fA-F]{2}))?(?:([0-9a-fA-F]{2}))?(?:([0-9a-fA-F]{2}))?(?:([0-9a-fA-F]{2}))?$"
But here you would always get six captures, just that some might be empty.
I need to validate an input on a form. I'm expecting the input to be a number between 1 to 19 digits. The input can also start with zeros. However, I want to validate that they are not all zeros. I've got a regex that will ensure that the input is numeric and between 1 and 19 numbers.
^\d[1,19]$
But I can't figure out how to include a check that the entire string is not all zeros. I tried this
^(![0]{1,19})(\d[1,19])$
but it fails on 0000000000000000001 because it's allowing a variable number of zeros.
How do I check that the entire string is NOT zeros?
Thanks.
I'm trying to do this in a ASP.NET RegularExpressionValidator so I was hoping for a single expression. I have other options, so I'm not out of luck if this can't be done.
^(?!0+$)\d{1,19}$
Just do a negative lookahead:
(?!^0+$)(^\d{1,19})
This works fine in Perl.
(?!0+$) is a lookahead directive. The ?! is the negative lookahead command to search for 1 or more 0's to the end of the string. If that matches, then the characters are consumed, leaving the regular digit search of \d{1,19}.
Boost Perl Regexp has a good discussion of perl regexp as recognized by Boost.
you don't need RegEx for that
ulong test;
string number = "1234567890123456789";
if (ulong.TryParse(number, out test) && (test < 9999999999999999999ul) && (test > 0))
Console.WriteLine(test);