I am trying to build an Ajax RegularExpressionValidator to look for strings in a TextBox that are missing a dash (-) surrounded by characters.
User is suppose to enter values like so
AA-BBBBB
1-223344
Basically, any number of alpha-numeric chars surrounding a dash (-). Could contain multiple dashes, but I only care that it has at least one dash.
Thanks for your time.
Be kind with your laughter as I don't know what I am doing with RegEx, but it would appear to be the best way to tackle this.
Based on what I read, i tried this
^[\S-\S]+$
and even gave this a shot
^[*-*]+$
Try these:
^\w+-\w+$ // Alphanumeric and underscore
^[A-Za-z\d]+-[A-Za-z\d]+$ // Only alphanumeric
^[A-Za-z\d]+-[A-Za-z\d]+$ // Only alphanumeric upper case
// Only alpha OR only numeric (upper case) (Exclusive Or)
^([A-Z]+-[A-Z]+)|(\d+-\d+)$
See test in Rubular.
Something like this
# ^[^\W_]+-+[^\W_]+$
^ # Beginning of string
[^\W_]+ # 1 or more alphanum
-+ # 1 or more dash '-'
[^\W_]+ # 1 or more alphanum
$ # End of string
Related
This may be a very simple question but I have not much experience with regex expressions. This page is a good source of regex expressions but could not figure out how to include them into my following code:
data %>% filter(grepl("^A01H1", icl))
Question
I would like to extract the values in one column of my data frame starting with this A01H1 up to 2 more digits, for example A01H100, A01H140, A01H110. I could not find a solution despite my few attempts:
Attempts
I looked at this question from which I used ^A01H1[0-9].{2} to select up tot two more digits.
I tried with adding any character ^A01H1[0-9][0-9][x-y] to stop after two digits.
Any help would be much appreciated :)
You can use "^A01H1\\d{1,2}$".
The first part ("^A01H1"), you figured out yourself, so what are we doing in the second part ("\\d{1,2}$")?
\d includes all digits and is equivalent to [0-9], since we are working in R you need to escape \ and thus we use \\d
{1,2} indicates we want to have 1 or 2 matches of \\d
$ specifies the end of the string, so nothing should come afterwards and this prevents to match more than 2 digits
It looks as if you want to match a part of a string that starts with A01H1, then contains 1 or 2 digits and then is not followed with any digit.
You may use
^A01H1\d{1,2}(?!\d)
See the regex demo. If there can be no text after two digits at all, replace (?!\d) with $.
Details
^ - start of strinmg
A01H1 - literal string
\d{1,2} - one to two digits
(?!\d) - no digit allowed immediately to the right
$ - end of string
In R, you could use it like
grepl("^A01H1\\d{1,2}(?!\\d)", icl, perl=TRUE)
Or, with the string end anchor,
grepl("^A01H1\\d{1,2}$", icl)
Note the perl=TRUE is only necessary when using PCRE specific syntax like (?!\d), a negative lookahead.
I try to find a regex that matches the string only if the string does not end with at least three '0' or more. Intuitively, I tried:
.*[^0]{3,}$
But this does not match when there one or two zeroes at the end of the string.
If you have to do it without lookbehind assertions (i. e. in JavaScript):
^(?:.{0,2}|.*(?!000).{3})$
Otherwise, use hsz's answer.
Explanation:
^ # Start of string
(?: # Either match...
.{0,2} # a string of up to two characters
| # or
.* # any string
(?!000) # (unless followed by three zeroes)
.{3} # followed by three characters
) # End of alternation
$ # End of string
You can try using a negative look-behind, i.e.:
(?<!000)$
Tests:
Test Target String Matches
1 654153640 Yes
2 5646549800 Yes
3 848461158000 No
4 84681840000 No
5 35450008748 Yes
Please keep in mind that negative look-behinds aren't supported in every language, however.
What wrong with the no-look-behind, more general-purpose ^(.(?!.*0{3,}$))*$?
The general pattern is ^(.(?!.* + not-ending-with-pattern + $))*$. You don't have to reverse engineer the state machine like Tim's answer does; you just insert the pattern you don't want to match at the end.
This is one of those things that RegExes aren't that great at, because the string isn't very regular (whatever that means). The only way I could come up with was to give it every possibility.
.*[^0]..$|.*.[^0].$|.*..[^0]$
which simplifies to
.*([^0]|[^0].|[^0]..)$
That's fine if you only want strings not ending in three 0s, but strings not ending in ten 0s would be long. But thankfully, this string is a bit more regular than some of these sorts of combinations, and you can simplify it further.
.*[^0].{0,2}$
I need to use an asp:RegularExpressionValidator control in my asp.net site and I want to restrict the characters to digits and semi-colon only.
So it needs to be able to take any amount of digits and ;
Valid values:
123
123;456
123;456;789;....
What is the regex for that?
Try this simple regex: [0-9;]+ or [\d;]+
\d match a digit [0-9]
; the literal character ;
+ means that it match between one character and unlimited times, as many times as possible,
If you want to guarantee that at least numbers are present in your expression you could do this too:
#npinti has a valid point better will be: ^[\d;]+$
where ^ indicates the begging of your expression and $ the end of it.
Online Demo
Updated::
Password strength:
Contain characters from three of the following four categories:
English uppercase characters (A through Z)
English lowercase characters (a through z)
Base 10 digits (0 through 9)
Non-alphabetic characters (for example, !, $, #, %
IS it possible to compare two fields value(entered) with regex...if yes then please add onr another condition to above list.
compare password with username entered they must be different
EDIT: This answer was written before the question was edited. It originally included the requirement to not include the user's account name, and be at least 8 characters long.
Given that you need to use the user's account name as part of it anyway, is there any reason you particularly want to do this as a regular expression? You may want to use regular expressions to express the patterns for the four categories (although there are other ways of doing it too) but I would write the rules out separately. For example:
// Categories is a list of regexes in this case. You could easily change
// it to anything else.
int categories = Categories.Count(regex => regex.IsMatch(password));
bool valid = password.IndexOf(name, StringComparison.OrdinalIgnoreCase) == -1
&& password.Length >= 8
&& categories >= 3;
If you need to do it in one expression it should be something like this:
^(?:(?=.*[a-z])(?=.*[A-Z])(?=.*[0-9])|(?=.*[a-z])(?=.*[A-Z])(?=.*[!%,.;:])|(?=.*[a-z])(?=.*[0-9])(?=.*[!%,.;:])|(?=.*[A-Z])(?=.*[0-9])(?=.*[!%,.;:])).{8,}$
See it here on Regexr
Positive lookaheads (the (?=.*[a-z])) are used to check if the string contains the character group you want.
The problem here is, you want 3 out of 4, that means you have to make an alternation with all the allowed combinations.
The last part .{8,} is then matching the string and checking for at least 8 characters.
^ and $ are anchors, that anchor the pattern to the start and the end of the string.
[!%,.;:] is a character class, here you can add all the characters you want to include. Maybe its simpler to use a Unicode script like \p{P} for all punctuation characters. For more details see here on regular-expresssions.info
Update
compare password with username entered they must be different
normally you should be able to build up your regular expression using string concatenation. I have no idea how it is in your case where you put the regex ...
Something like this (pseudo)
String Username = "FooBar";
regex = "^(?:(?=.*[a-z])(?=.*[A-Z])(?=.*[0-9])|(?=.*[a-z])(?=.*[A-Z])(?=.*[!%,.;:])|(?=.*[a-z])(?=.*[0-9])(?=.*[!%,.;:])|(?=.*[A-Z])(?=.*[0-9])(?=.*[!%,.;:]))(?i)(?!.*" + Username + ").+$";
I used here also an inline modifier (?i) to match it case independent. The (?!.* is the start of negative lookahead, meaning the string should not contain ...
Hi I'm trying to right a regular expression that will take a string and ensure it starts with an 'R' and is followed by 4 numeric digits then anything
eg. RXXXX.................
Can anybody help me with this? This is for ASP.NET
You want it to be at the beginning of the line, not anywhere. Also, for efficiency, you dont want the .+ or .* at the end because that will match unnecessary characters. So the following regex is what you really want:
^R\d{4}
This should do it...
^R\d{4}.*$
\d{4} matches 4 digits
.* is simply a way to match any character 0 or more times
the beginning ^ and end $ anchors ensure that nothing precedes or follows
As Vincent suggested, for your specific task it could even be simplified to this...
^R\d{4}
Because as you stated, it doesn't really matter what follows.
/^R\d{4}.*/ and set the case insensitive option unless you only want capital R's
^R\d{4}.*
The caret ^ matches the position before the first character in the string.
\d matches any numeric character (it's the same as [0-9])
{4} indicates that there must be exactly 4 numbers, and
.* matches 0 or more other characters
To use:
string input = "R0012 etc..";
Match match = Regex.Match(input, #"^R\d{4}.*", RexOptions.IgnoreCase);
if (match.Success)
{
// Success!
}
Note the use of RexOptions.IgnoreCase to ignore the case of the letter R (so it'll match strings which start with r. Leave this out if you don't want to undertake a case insensitive match.