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Find 3 or more whitespaces with regex in C# [duplicate]

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Regex to validate string for having three non white-space characters
(2 answers)
Closed 3 years ago.
As said above, I want to find 3 or more whitespaces with regex in C#. Currently I tried:
\s{3,} and [ ]{3,} for Somestreet 155/ EG 47. Both didnt worked out. What did I do wrong?

This \s{3,} matches 3 or more whitespace in a row. You need for example this pattern \s.*\s.*\s to match a string with 3 whitespaces anywhere.
So this would match:
a b c d
a b c
a b
abc d e f
a
a b // ends in 1 space
// just 3 spaces
a // ends in 3 spaces

Linq is an alternative way to count spaces:
string source = "Somestreet 155/ EG 47";
bool result = source
.Where(c => c == ' ') // spaces only
.Skip(2) // skip 2 of them
.Any(); // do we have at least 1 more (i.e. 3d space?)
Edit: If you want not just spaces but whitespaces Where should be
...
.Where(c => char.IsWhiteSpace(c))
...

You could count the whitespace matches:
if (Regex.Matches(yourString, #"\s+").Count >= 3) {...}
The + makes sure that consecutive matches to \s only count once, so "Somestreet 155/ EG 47" has three matches but "Somestreet 155/ EG47" only has two.
If the string is long, then it could take more time than necessary to get all the matches then count them. An alternative is to get one match at a time and bail out early if the required number of matches has been met:
static bool MatchesAtLeast(string s, Regex re, int matchCount)
{
bool success = false;
int startPos = 0;
while (!success)
{
Match m = re.Match(s, startPos);
if (m.Success)
{
matchCount--;
success = (matchCount <= 0);
startPos = m.Index + m.Length;
if (startPos > s.Length - 2) { break; }
}
else { break; }
}
return success;
}
static void Main(string[] args)
{
Regex re = new Regex(#"\s+");
string s = "Somestreet 155/ EG\t47";
Console.WriteLine(MatchesAtLeast(s, re, 3)); // outputs True
Console.ReadLine();
}

Try ^\S*\s\S*\s\S*\s\S*$ instead.
\S matches non-whitespace characters, ^ matches beginnning of a string and $ matches end of a string.
Demo

Regex. Up to six digits after separator but not all zeros

I need input in this form:
first digit ist always 0
second is always . or ,
than up to six digits, but only digits not letters or other symbols
and not all zeros
private const string Pattern = #"^0(,|.)(?!0+$)+";
var regex = new Regex(Pattern, RegexOptions.IgnoreCase);
if (!regex.IsMatch(inputToCheck))
{...}
This works ok for all the conditions except one with digits only.
This input should be disabled too: "0,01a", "0.01a1", "0,01a0"
How can I extend my regex pattern to this condition?
Some examples of valid and invalid input.
Valid:
0,123456
0,01
0,010
0,2
invalid:
1,123456
2,123456
0,0
0,00
0,000
a
a,1
0,01a
0,01a1
0,01a0

I think you're on the right track. Here's my solution to this:
^0[,.](?!0+$)\d{1,6}$
This will make sure that the first digit is zero. It then checks that the next character is either a comma or a dot. Then, using a lookahead it ensures that the rest of the subject string is not entirely zeros. If this passes, it checks that the remaining characters are all digits.

You can use a regex like this:
^0[.,][1-9]{0,6}$
Of course this regex don't allow 0 after the , or .. If you want to allow 0 but restrict ending by 0 you can do:
^0[.,][0-9]{0,5}[1-9]$
And also you can shorten it a little to:
^0[.,]\d{0,5}[1-9]$

Try this expression
^^0(,|.)([0-9]{1,4}$)+

You shouldn't really be using regex to parse numbers, you can do it by just validating it as a number as so..
CultureInfo culture = new CultureInfo("de-DE");
string[] inputs = new string[]{"0,123456",
"0,01",
"1,123456",
"0,0"};
foreach(var input in inputs)
{
double val;
if(Double.TryParse(input, NumberStyles.Number, culture, out val)
&& Math.Round(val, 6) == val
&& val != 0.0
&& (int)val == 0)
Console.WriteLine("{0} is valid", input);
else
Console.WriteLine("{0} is invalid", input);
}
Output
0,123456 is valid
0,01 is valid
1,123456 is invalid
0,0 is invalid
IDEOne example

Regex Lookahead and lookbehind at most one digit

I'm looking for create RegEx pattern
8 characters [a-zA_Z]
must contains only one digit in any place of string
I created this pattern:
^(?=.*[0-9].*[0-9])[0-9a-zA-Z]{8}$
This pattern works fine but i want only one digit allowed. Example:
aaaaaaa6 match
aaa7aaaa match
aaa88aaa don't match
aaa884aa don't match
aaawwaaa don't match

You could instead use:
^(?=[0-9a-zA-Z]{8})[^\d]*\d[^\d]*$
The first part would assert that the match contains 8 alphabets or digits. Once this is ensured, the second part ensures that there is only one digit in the match.
EDIT: Explanation:
The anchors ^ and $ denote the start and end of string.
(?=[0-9a-zA-Z]{8}) asserts that the match contains 8 alphabets or digits.
[^\d]*\d[^\d]* would imply that there is only one digit character and remaining non-digit characters. Since we had already asserted that the input contains digits or alphabets, the non-digit characters here are alphabets.

If you want a non regex solution, I wrote this for a small project :
public static bool ContainsOneDigit(string s)
{
if (String.IsNullOrWhiteSpace(s) || s.Length != 8)
return false;
int nb = 0;
foreach (char c in s)
{
if (!Char.IsLetterOrDigit(c))
return false;
if (c >= '0' && c <= '9') // just thought, I could use Char.IsDigit() here ...
nb++;
}
return nb == 1;
}

Regex for alphanumeric, at least 1 number and special chars

I am trying to find a regex which will give me the following validation:
string should contain at least 1 digit and at least 1 special character. Does allow alphanumeric.
I tried the following but this fails:
#"^[a-zA-Z0-9##$%&*+\-_(),+':;?.,!\[\]\s\\/]+$]"
I tried "password1$" but that failed
I also tried "Password1!" but that also failed.
ideas?
UPDATE
Need the solution to work with C# - currently the suggestions posted as of Oct 22 2013 do not appear to work.

Try this:
Regex rxPassword = new Regex( #"
^ # start-of-line, followed by
[a-zA-Z0-9!##]+ # a sequence of one or more characters drawn from the set consisting of ASCII letters, digits or the punctuation characters ! # and #
(<=[0-9]) # at least one of which is a decimal digit
(<=[!##]) # at least one of which is one of the special characters
(<=[a-zA-Z]) # at least one of which is an upper- or lower-case letter
$ # followed by end-of-line
" , RegexOptions.IgnorePatternWhitespace ) ;
The construct (<=regular-expression) is a zero-width positive look-behind assertion.

Sometimes it's a lot simpler to do things one step at a time. The static constructor builds the escaped character class characters from a simple list of allowed special characters. The built-in Regex.Escape method doesn't work here.
public static class PasswordValidator {
private const string ALLOWED_SPECIAL_CHARS = #"##$%&*+_()':;?.,![]\-";
private static string ESCAPED_SPECIAL_CHARS;
static PasswordValidator() {
var escapedChars = new List<char>();
foreach (char c in ALLOWED_SPECIAL_CHARS) {
if (c == '[' || c == ']' || c == '\\' || c == '-')
escapedChars.AddRange(new[] { '\\', c });
else
escapedChars.Add(c);
}
ESCAPED_SPECIAL_CHARS = new string(escapedChars.ToArray());
}
public static bool IsValidPassword(string input) {
// Length requirement?
if (input.Length < 8) return false;
// First just check for a digit
if (!Regex.IsMatch(input, #"\d")) return false;
// Then check for special character
if (!Regex.IsMatch(input, "[" + ESCAPED_SPECIAL_CHARS + "]")) return false;
// Require a letter?
if (!Regex.IsMatch(input, "[a-zA-Z]")) return false;
// DON'T allow anything else:
if (Regex.IsMatch(input, #"[^a-zA-Z\d" + ESCAPED_SPECIAL_CHARS + "]")) return false;
return true;
}
}

This may be work, there are two possible, the digit before special char or the digit after the special char. You should use DOTALL(the dot point all char)
^((.*?[0-9].*?[##$%&*+\-_(),+':;?.,!\[\]\s\\/].*)|(.*?[##$%&*+\-_(),+':;?.,!\[\]\s\\/].*?[0-9].*))$

This worked for me:
#"(?=^[!##$%\^&*()_-+=[{]};:<>|./?a-zA-Z\d]{8,}$)(?=([!##$%\^&*()_-+=[{]};:<>|./?a-zA-Z\d]\W+){1,})(?=[^0-9][0-9])[!##$%\^&*()_-+=[{]};:<>|./?a-zA-Z\d]*$"
alphanumeric, at least 1 numeric, and special character with a min length of 8

This should do the work
(?:(?=.*[0-9]+)(?=.*[a-zA-Z]+)(?=.*[##$%&*+\-_(),+':;?.,!\[\]\s\\/]+))+
Tested with javascript, not sure about c#, may need some little adjust.
What it does is use anticipated positive lookahead to find the required elements of the password.
EDIT
Regular expression is designed to test if there are matches. Since all the patterns are lookahead, no real characters get captured and matches are empty, but if the expression "match", then the password is valid.
But, since the question is C# (sorry, i don't know c#, just improvising and adapting samples)
string input = "password1!";
string pattern = #"^(?:(?=.*[0-9]+)(?=.*[a-zA-Z]+)(?=.*[##$%&*+\-_(),+':;?.,!\[\]\s\\/]+))+.*$";
Regex rgx = new Regex(pattern, RegexOptions.None);
MatchCollection matches = rgx.Matches(input);
if (matches.Count > 0) {
Console.WriteLine("{0} ({1} matches):", input, matches.Count);
foreach (Match match in matches)
Console.WriteLine(" " + match.Value);
}
Adding start of line, and a .*$ to the end, the expression will match if the password is valid. And the match value will be the password. (i guess)

.Net Removing all the first 0 of a string

I got the following :
01.05.03
I need to convert that to 1.5.3
The problem is I cannot only trim the 0 because if I got :
01.05.10
I need to convert that to 1.5.10
So, what's the better way to solve that problem ? Regex ? If so, any regex example doing that ?

Expanding on the answer of #FrustratedWithFormsDesigner:
string Strip0s(string s)
{
return string.Join<int>(".", from x in s.Split('.') select int.Parse(x));
}

Regex-replace
(?<=^|\.)0+
with the empty string. The regex is:
(?<= # begin positive look-behind (i.e. "a position preceded by")
^|\. # the start of the string or a literal dot †
) # end positive look-behind
0+ # one or more "0" characters
† note that not all regex flavors support variable-length look-behind, but .NET does.
If you expect this kind of input: "00.03.03" and want to to keep the leading zero in this case (like "0.3.3"), use this expression instead:
(?<=^|\.)0+(?=\d)
and again replace with the empty string.
From the comments (thanks Kobi): There is a more concise expression that does not require look-behind and is equivalent to my second suggestion:
\b0+(?=\d)
which is
\b # a word boundary (a position between a word char and a non-word char)
0+ # one or more "0" characters
(?=\d) # positive look-ahead: a position that's followed by a digit
This works because the 0 happens to be a word character, so word boundaries can be used to find the first 0 in a row. It is a more compatible expression, because many regex flavors do not support variable-length look-behind, and some (like JavaScript) no look-behind at all.

You could split the string on ., then trim the leading 0s on the results of the split, then merge them back together.
I don't know of a way to do this in a single operation, but you could write a function that hides this and makes it look like a single operation. ;)
UPDATE:
I didn't even think of the other guy's regex. Yeah, that will probably do it in a single operation.

Here's another way you could do what FrustratedWithFormsDesigner suggests:
string s = "01.05.10";
string s2 = string.Join(
".",
s.Split('.')
.Select(str => str.TrimStart('0'))
.ToArray()
);
This is almost the same as dtb's answer, but doesn't require that the substrings be valid integers (it would also work with, e.g., "000A.007.0HHIMARK").
UPDATE: If you'd want any strings consisting of all 0s in the input string to be output as a single 0, you could use this:
string s2 = string.Join(
".",
s.Split('.')
.Select(str => TrimLeadingZeros(str))
.ToArray()
);
public static string TrimLeadingZeros(string text) {
int number;
if (int.TryParse(text, out number))
return number.ToString();
else
return text.TrimStart('0');
}
Example input/output:
00.00.000A.007.0HHIMARK // input
0.0.A.7.HHIMARK // output

There's also the old-school way which probably has better performance characteristics than most other solutions mentioned. Something like:
static public string NormalizeVersionString(string versionString)
{
if(versionString == null)
throw new NullArgumentException("versionString");
bool insideNumber = false;
StringBuilder sb = new StringBuilder(versionString.Length);
foreach(char c in versionString)
{
if(c == '.')
{
sb.Append('.');
insideNumber = false;
}
else if(c >= '1' && c <= '9')
{
sb.Append(c);
insideNumber = true;
}
else if(c == '0')
{
if(insideNumber)
sb.Append('0');
}
}
return sb.ToString();
}

string s = "01.05.10";
string newS = s.Replace(".0", ".");
newS = newS.StartsWith("0") ? newS.Substring(1, newS.Length - 1) : newS;
Console.WriteLine(newS);
NOTE: You will have to thoroughly check for possible input combination.

This looks like it is a date format, if so I would use Date processing code
DateTime time = DateTime.Parse("01.02.03");
String newFormat = time.ToString("d.M.yy");
or even better
String newFormat = time.ToShortDateString();
which will respect you and your clients culture setting.
If this data is not a date then don't use this :)

I had a similar requirement to parse a string with street adresses, where some of the house numbers had leading zeroes and I needed to remove them while keeping the rest of the text intact, so I slightly edited the accepted answer to meet my requirements, maybe someone finds it useful. Basically doing the same as accepted answer, with the difference that I am checking if the string part can be parsed as an integer, and defaulting to the string value when false;
string Strip0s(string s)
{
int outputValue;
return
string.Join(" ",
from x in s.Split(new[] { ' ' })
select int.TryParse(x, out outputValue) ? outputValue.ToString() : x);
}
Input: "Islands Brygge 34 B 07 TV"
Output: "Islands Brygge 34 B 7 TV"