Regex: how to know that string contains at least 2 upper case letters?

Question

How to know that string contains at least 2 upper case letters? For example these are valid strings "Lazy Cat", "NOt very lazy cat". Working with Java 1.7.

Answer 1

这个正则表达式有效。

string.matches(".*[A-Z].*[A-Z].*")

Answer 2

Try with following regex:

"^(.*?[A-Z]){2,}.*$"

or

"^(.*?[A-Z]){2,}"

Answer 3

I'll now show you a full solution, I'll guide you.

If you don't want to use regex, you can simply loop on the String, chat by char and check whether it's an upper case:

for (int i=0;i<myStr.length();i++)
{
     //as @sanbhat suggested, use Character#isUpperCase on each character..
}

Answer 4

You have somany regex answers right now,

Go for it if you don't want to use **regex** ,

String someString = "abcDS";
int upperCount = 0;
for (char c : someString.toCharArray()) {
    if (Character.isUpperCase(c)) {
        upperFound++;
    }
}
// upperFound  here weather >2 or not

Answer 5

试试这个：

string.matches("[A-Z]+.*[A-Z]+");