Checking if a string starts and ends with number characters using regex

问题

I'm trying

String string = "123456";    
if(string.startsWith("[0-9]") && string.endsWith("[0-9]")){
                        //code
                    }

And the if clause is never called.

回答1:

Don't use a regex:

Character.isDigit(string.charAt(0)) && 
                              Character.isDigit(string.charAt(string.length()-1))

(see Character.isDigit())

回答2:

The methods startsWith() and endsWith() in class String accept only String, not a regex.

回答3:

You can use:

String string = "123test123";
if(string.matches("\\d.*\\d"))
{
    // ...
}

回答4:

You can use the matches method on String thusly:

public static void main(String[] args) throws Exception {
    System.out.println("123456".matches("^\\d.*?\\d$"));
    System.out.println("123456A".matches("^\\d.*?\\d$"));
    System.out.println("A123456".matches("^\\d.*?\\d$"));
    System.out.println("A123456A".matches("^\\d.*?\\d$"));
}

Output:

true
false
false
false

回答5:

Please follow the code snippet.

String variableString = "012testString";
Character.isDigit(string.charAt(0)) && variableString.Any(c => char.IsUpper(c));

来源：https://stackoverflow.com/questions/18019584/checking-if-a-string-starts-and-ends-with-number-characters-using-regex