In detail, needs to match a string with the following pattern:
Always starts with plain English letters (i.e. A-Z, a-z)
The length of the above letter is between 1 to 3
Followed by any numbers of which the first digit is not 0 (i.e. AB01 is not valid but AB1 is)
Thanks.
How about:
[A-Za-z]{1,3}[1-9]\d*
which is a literal translation of your remarkably precise question.
Cameron is right, except you have to put in a starting limit. For future refernce, you should look into a good online regex tester. I like Derek Slager's.
^[A-Za-z]{1,3}[1-9]\d*
if you want to match the whole line, and not just the beginning, then
^[A-Za-z]{1,3}[1-9]\d*$
Related
I have a regex
/^([a-zA-Z0-9]+)$/
this just allows only alphanumerics but also if I insert only number(s) or only character(s) then also it accepts it. I want it to work like the field should accept only alphanumeric values but the value must contain at least both 1 character and 1 number.
Why not first apply the whole test, and then add individual tests for characters and numbers? Anyway, if you want to do it all in one regexp, use positive lookahead:
/^(?=.*[0-9])(?=.*[a-zA-Z])([a-zA-Z0-9]+)$/
This RE will do:
/^(?:[0-9]+[a-z]|[a-z]+[0-9])[a-z0-9]*$/i
Explanation of RE:
Match either of the following:
At least one number, then one letter or
At least one letter, then one number plus
Any remaining numbers and letters
(?:...) creates an unreferenced group
/i is the ignore-case flag, so that a-z == a-zA-Z.
I can see that other responders have given you a complete solution. Problem with regexes is that they can be difficult to maintain/understand.
An easier solution would be to retain your existing regex, then create two new regexes to test for your "at least one alphabetic" and "at least one numeric".
So, test for this :-
/^([a-zA-Z0-9]+)$/
Then this :-
/\d/
Then this :-
/[A-Z]/i
If your string passes all three regexes, you have the answer you need.
The accepted answers is not worked as it is not allow to enter special characters.
Its worked perfect for me.
^(?=.*[0-9])(?=.*[a-zA-Z])(?=\S+$).{6,20}$
one digit must
one character must (lower or upper)
every other things optional
Thank you.
While the accepted answer is correct, I find this regex a lot easier to read:
REGEX = "([A-Za-z]+[0-9]|[0-9]+[A-Za-z])[A-Za-z0-9]*"
This solution accepts at least 1 number and at least 1 character:
[^\w\d]*(([0-9]+.*[A-Za-z]+.*)|[A-Za-z]+.*([0-9]+.*))
And an idea with a negative check.
/^(?!\d*$|[a-z]*$)[a-z\d]+$/i
^(?! at start look ahead if string does not
\d*$ contain only digits | or
[a-z]*$ contain only letters
[a-z\d]+$ matches one or more letters or digits until $ end.
Have a look at this regex101 demo
(the i flag turns on caseless matching: a-z matches a-zA-Z)
Maybe a bit late, but this is my RE:
/^(\w*(\d+[a-zA-Z]|[a-zA-Z]+\d)\w*)+$/
Explanation:
\w* -> 0 or more alphanumeric digits, at the beginning
\d+[a-zA-Z]|[a-zA-Z]+\d -> a digit + a letter OR a letter + a digit
\w* -> 0 or more alphanumeric digits, again
I hope it was understandable
What about simply:
/[0-9][a-zA-Z]|[a-zA-Z][0-9]/
Worked like a charm for me...
Edit following comments:
Well, some shortsighting of my own late at night: apologies for the inconvenience...
The - incomplete - underlying idea was that only one "transition" from a digit to an alpha or from an alpha to a digit was needed somewhere to answer the question.
But next regex should do the job for a string only comprised of alphanumeric characters:
/^[0-9a-zA-Z]*([0-9][a-zA-Z]|[a-zA-Z][0-9])[0-9a-zA-Z]*$/
which in Javascript can be furthermore simplified as:
/^[0-9a-z]*([0-9][a-z]|[a-z][0-9])[0-9a-z]*$/i
In IMHO it's more straigthforward to read and understand than some other answers (no backtraking and the like).
Hope this helps.
If you need the digit to be at the end of any word, this worked for me:
/\b([a-zA-Z]+[0-9]+)\b/g
\b word boundary
[a-zA-Z] any letter
[0-9] any number
"+" unlimited search (show all results)
It's driving nuts.
The input strings are:
abc|qw|xzy mno
abc||xzy mno
abc|qw|xzy
abc|qw|
I need to extract the first word (if any) after the 2nd vertical bar, in all cases above xyz but in general words in multiple (natural) languages.
Also, all lines must be considered as a block so single line does not apply, iow, the EOL is the break to account for.
Thank you, guys.
You can use the following regexp with the RegexOptions.Multiline option.
(?<=^(?:[^|]*\|){2})\w+
(?<= begins a positive lookbehind, so this matches a word that must be preceded by the beginning of the line followed by two pipe-delimited sequences.
I would like to find the match for:
\024jack3hall2\c$
\024jack3hall02\c$
\024jack3hall12\c$
but not for:
\024jack3hall023\c$
difference is the number of digits in the end part. I would like to have only 1 or 2, not 3.
my try:
\\\\024[a-zA-Z0-9]+[0-9]{1,2}\\[a-zA-Z]{1}\$(?!.)
I tried only on http://regexr.com/ but will implement in C#.
Is it possible to edit my try or I have to write several separate checks?
Why is
{1,2}
not working? \024jack3hall12343\c$ is also matching,
From the examples you have shown, something as simple as:
[^\d](\d{1,2})\\
Should work. It will match 1 or 2 digits followed by a \ so long as it isn't proceeded by another digit.
The matched digits are in a capture group if you need them (or you can just remove the brackets if you don't need that).
As for your original effort, right here:
\\\\024[a-zA-Z0-9]+[0-9]{1,2}
You are matching 1 or more from the range a-z, A-Z or 0-9. So that will match your extra digits if they come at the end of that pattern.
Answer:
\\\\024[a-zA-Z0-9]+[^\d](\d{1,2})\\[a-zA-Z]{1}\$(?!.)
I believe you were not escaping backslash properly.
Here is the correct regex:
\\024[a-zA-Z0-9]+[0-9]{1,2}\\[a-zA-Z]{1}\$(?!.)
Can someone please validate this for me (newbie of regex match cons).
Rather than asking the question, I am writing this:
Regex rgx = new Regex (#"^{3}[a-zA-Z0-9](\d{5})|{3}[a-zA-Z0-9](\d{9})$"
Can someone telll me if it's OK...
The accounts I am trying to match are either of:
1. BAA89345 (8 chars)
2. 12345678 (8 chars)
3. 123456789112 (12 chars)
Thanks in advance.
You can use a Regex tester. Plenty of free ones online. My Regex Tester is my current favorite.
Is the value with 3 characters then followed by digits always starting with three... can it start with less than or more than three. What are these mins and max chars prior to the digits if they can be.
You need to place your quantifiers after the characters they are supposed to quantify. Also, character classes need to be wrapped in square brackets. This should work:
#"^(?:[a-zA-Z0-9]{3}|\d{3}\d{4})\d{5}$"
There are several good, automated regex testers out there. You may want to check out regexpal.
Although that may be a perfectly valid match, I would suggest rewriting it as:
^([a-zA-Z]{3}\d{5}|\d{8}|\d{12})$
which requires the string to match one of:
[a-zA-Z]{3}\d{5} three alpha and five numbers
\d{8} 8 digits or
\d{12} twelve digits.
Makes it easier to read, too...
I'm not 100% on your objective, but there are a few problems I can see right off the bat.
When you list the acceptable characters to match, like with a-zA-Z0-9, you need to put it inside brackets, like [a-zA-Z0-9] Using a ^ at the beginning will negate the contained characters, e.g. `[^a-zA-Z0-9]
Word characters can be matched like \w, which is equivalent to [a-zA-Z0-9_].
Quantifiers need to appear at the end of the match expression. So, instead of {3}[a-zA-Z0-9], you would need to write [a-zA-Z0-9]{3} (assuming you want to match three instances of a character that matches [a-zA-Z0-9]
I am new to regex (15 minutes of experience) so I can't figure this one out. I just want something that will match an alphanumeric string with no spaces in it. For example:
"ThisIsMyName" should match, but
"This Is My Name" should not match.
^[a-zA-Z0-9]+$ will match any letters and any numbers with no spaces (or any punctuation) in the string. It will also require at least one alphanumeric character. This uses a character class for the matching. Breakdown:
^ #Match the beginning of the string
[ #Start of a character class
a-z #The range of lowercase letters
A-Z #The range of uppercase letters
0-9 #The digits 0-9
] #End of the character class
+ #Repeat the previous one or more times
$ #End of string
Further, if you want to "capture" the match so that it can be referenced later, you can surround the regex in parens (a capture group), like so:
^([a-zA-Z0-9]+)$
Even further: since you tagged this with C#, MSDN has a little howto for using regular expressions in .NET. It can be found here. You can also note the fact that if you run the regex with the RegexOptions.IgnoreCase flag then you can simplify it to:
^([a-z0-9])+$
this will match any sequence of non-space characters:
\S+
Take a look at this link for a good basic Regex information source: http://regexlib.com/CheatSheet.aspx
They also have a handy testing tool that I use quite a bit: http://regexlib.com/RETester.aspx
That said, #eldarerathis' or #Nicolas Bottarini's answers should work for you.
I have just written a blog entry about regex, maybe it's something you may find useful:)
http://blogs.appframe.com/erikv/2010-09-23-Regular-Expression
Try using this regex to see if it works: (\w+)