111111111 - Invalid
A121278237 - Invalid
7777777777 - In开发者_开发知识库valid
121263263 - Valid
111111112 - Valid
^([0-9])(?!\1+$)[0-9]+$
should work. It needs a string of at least two digits to match successfully.
Explanation:
Match a digit and capture it into backreference #1:
([0-9])
Assert that it's impossible to match a string of any length (>1) of the same digit that was just matched, followed by the end of the string:
(?!\1+$)
Then match any string of digits until the end of the string:
[0-9]+$
EDIT: Of course, in Java you need to escape the backslash inside a string ("\\"
).
- take a [0-9] regex and throw away strings that not only contain digits.
- take the first character, and use it as a regex [C]+ to see if the string contains any other digits.
Building on Tim's answer, you eliminate the requirement of "at least two digits" by adding an or clause.
^([0-9])(?!\1+$)[0-9]+$|^[0-9]$
For example:
String regex = "^([0-9])(?!\\1+$)[0-9]+$|^[0-9]$";
boolean a = "12".matches(regex);
System.out.println("a: " + a);
boolean b = "11".matches(regex);
System.out.println("b: " + b);
boolean c = "1".matches(regex);
System.out.println("c: " + c);
returns
a: true
b: false
c: true
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