I haven't used regular expressions at all, so I'm having difficulty troubleshooting. I want the regex to match only when the contained string is all numbers; but with the two examples below it is matching a string that contains all numbers plus an equals sign like "1234=4321". I'm sure there's a way to change this behavior, but as I said, I've never really done much with regular expressions.
string compare = "1234=4321";
Regex regex = new Regex(@"[\d]");
if (regex.IsMatch(compare))
{
//true
}
regex = new Regex("[0-9]");
if (regex.IsMatch(compare))
{
//true
}
In case it matters, I'm using C# and .NET2.0.
Use the beginning and end anchors.
Regex regex = new Regex(@"^\d$");
Use "^\d+$"
if you need to match more than one digit.
Note that "\d"
will match [0-9]
and other digit characters like the Eastern Arabic numerals ٠١٢٣٤٥٦٧٨٩
. Use "^[0-9]+$"
to restrict matches to just the Arabic numerals 0 - 9.
If you need to include any numeric representations other than just digits (like decimal values for starters), then see @tchrist's comprehensive guide to parsing numbers with regular expressions.
Your regex will match anything that contains a number, you want to use anchors to match the whole string and then match one or more numbers:
regex = new Regex("^[0-9]+$");
The ^
will anchor the beginning of the string, the $
will anchor the end of the string, and the +
will match one or more of what precedes it (a number in this case).
If you need to tolerate decimal point and thousand marker
var regex = new Regex(@"^-?[0-9][0-9,\.]+$");
You will need a "-", if the number can go negative.
-
) and period (.
) to occur more than once.
-
and .
optional via ?
. -?\d+(?:\.\d+)?
would match integers or decimals. (The ?:
in the parens just makes the parens a non-capturing group and used to only group for clarity.)
^-?[0-9][0-9,\.]*$
to avoid failing for the basic case. + is replaced with *
This works with integers and decimal numbers. It doesn't match if the number has the coma thousand separator ,
"^-?\\d*(\\.\\d+)?$"
some strings that matches with this:
894
923.21
76876876
.32
-894
-923.21
-76876876
-.32
some strings that doesn't:
hello
9bye
hello9bye
888,323
5,434.3
-8,336.09
87078.
It is matching because it is finding "a match" not a match of the full string. You can fix this by changing your regexp to specifically look for the beginning and end of the string.
^\d+$
Perhaps my method will help you.
public static bool IsNumber(string s)
{
return s.All(char.IsDigit);
}
Char.IsDigit
returns true
for any character that is a member of the UnicodeCategory.DecimalDigitNumber category. This may not be what the OP wants. Also see Why Char.IsDigit returns true for chars which can't be parsed to int?.
If you need to check if all the digits are number (0-9) or not,
^[0-9]+$
Matches
1425
0142
0
1
And does not match
154a25
1234=3254
Sorry for ugly formatting. For any number of digits:
[0-9]*
For one or more digit:
[0-9]+
^\d+$, which is "start of string", "1 or more digits", "end of string" in English.
Here is my working one:
^(-?[1-9]+\\d*([.]\\d+)?)$|^(-?0[.]\\d*[1-9]+)$|^0$
And some tests
Positive tests:
string []goodNumbers={"3","-3","0","0.0","1.0","0.1","0.0001","-555","94549870965"};
Negative tests:
string []badNums={"a",""," ","-","001","-00.2","000.5",".3","3."," -1","--1","-.1","-0"};
Checked not only for C#, but also with Java, Javascript and PHP
Use beginning and end anchors.
Regex regex = new Regex(@"^\d$");
Use "^\d+$"
if you need to match more than one digit.
While non of the above solutions was fitting my purpose, this worked for me.
var pattern = @"^(-?[1-9]+\d*([.]\d+)?)$|^(-?0[.]\d*[1-9]+)$|^0$|^0.0$";
return Regex.Match(value, pattern, RegexOptions.IgnoreCase).Success;
Example of valid values:
"3",
"-3",
"0",
"0.0",
"1.0",
"0.7",
"690.7",
"0.0001",
"-555",
"945465464654"
Example of not valid values:
"a",
"",
" ",
".",
"-",
"001",
"00.2",
"000.5",
".3",
"3.",
" -1",
"--1",
"-.1",
"-0",
"00099",
"099"
Another way: If you like to match international numbers such as Persian or Arabic, so you can use following expression:
Regex = new Regex(@"^[\p{N}]+$");
To match literal period character use:
Regex = new Regex(@"^[\p{N}\.]+$");
Regex for integer and floating point numbers:
^[+-]?\d*\.\d+$|^[+-]?\d+(\.\d*)?$
A number can start with a period (without leading digits(s)), and a number can end with a period (without trailing digits(s)). Above regex will recognize both as correct numbers.
A . (period) itself without any digits is not a correct number. That's why we need two regex parts there (separated with a "|").
Hope this helps.
I think that this one is the simplest one and it accepts European and USA way of writing numbers e.g. USA 10,555.12 European 10.555,12 Also this one does not allow several commas or dots one after each other e.g. 10..22 or 10,.22 In addition to this numbers like .55 or ,55 would pass. This may be handy.
^([,|.]?[0-9])+$
10'555.12
(and our "Bundeskanzlei" 10 555,12
).
console.log(/^(0|[1-9][0-9]*)$/.test(3000)) // true
If you want to extract only numbers from a string the pattern "\d+" should help.
To check string is uint, ulong or contains only digits one .(dot) and digits Sample inputs
Regex rx = new Regex(@"^([1-9]\d*(\.)\d*|0?(\.)\d*[1-9]\d*|[1-9]\d*)$");
string text = "12.0";
var result = rx.IsMatch(text);
Console.WriteLine(result);
Samples
123 => True
123.1 => True
0.123 => True
.123 => True
0.2 => True
3452.434.43=> False
2342f43.34 => False
svasad.324 => False
3215.afa => False
The following regex accepts only numbers (also floating point) in both English and Arabic (Persian) languages (just like Windows calculator):
^((([0\u0660\u06F0]|([1-9\u0661-\u0669\u06F1-\u06F9][0\u0660\u06F0]*?)+)(\.)[0-9\u0660-\u0669\u06F0-\u06F9]+)|(([0\u0660\u06F0]?|([1-9\u0661-\u0669\u06F1-\u06F9][0\u0660\u06F0]*?)+))|\b)$
The above regex accepts the following patterns:
11
1.2
0.3
۱۲
۱.۳
۰.۲
۲.۷
The above regex doesn't accept the following patterns:
3.
.3
0..3
.۱۲
Regex regex = new Regex ("^[0-9]{1,4}=[0-9]{1,4]$")
Success story sharing
"^\d{6}$"
. Some other countries have more complicated rules for zip codes, so regex solutions for multiple countries can get pretty complicated too. Check out some of the examples at stackoverflow.com/q/578406/1288[0-9]
instead of\d
. It makes RegEx easier to read.