验证文本的正则表达式，其中包含某些特殊字符

Question

I am trying to create a regular expression which will validate a text. Conditions are as follows:

1. Must contain at least one alphanumeric character [A-Za-z0-9]

2. Can contain allowed special character like [@#]. This is optional.

3. No other characters are allowed other than the above mentioned chars[includes @ and #].
4. Text length should be less than 8.

valid text: A, A@, @A, A@a@, @@@a etc

invalid text : @, @#, a:**, A@%, AAAAAAAAA(9 characters) etc

I have tried the below regex but it is partly working:

(?=. [\\w])(?=. [@#])?.*{0,8}

Answer 1

^(?=.*[a-zA-Z0-9])[A-Za-z0-9@#]{0,8}$

Try this.See demo.

The lookahead will make sure there is atleast one character.

Answer 2

Use the below regex in matches method.

"(?=.*?[A-Za-z0-9])[A-Za-z0-9@#]{1,8}"

(?=.*?[A-Za-z0-9]) asserts that the string must contain atleast one letter or digit , so there must be a single character present in the match. That's why i defined [A-Za-z0-9@#]{1,8} instead of [A-Za-z0-9@#]{0,8}

String[] check = {"A", "A@", "@A", "A@a@", "@@@a", "@", "@#", "a:**", "A@%", "AAAAAAAAA"};
for(String i: check)
{

    System.out.println(i.matches("(?=.*?[A-Za-z0-9])[A-Za-z0-9@#]{1,8}"));
}

Output:

true
true
true
true
true
false
false
false
false
false

Answer 3

您可以使用：

^(?=.*?[A-Za-z0-9])[@#A-Za-z0-9]{1,8}$

Answer 4

You can try regex like this :

public static void main(String[] args) {
    String s = "aaaaaaa";
    System.out.println(s.matches("(?=.*[A-Za-z0-9])[A-Za-z0-9@#]{1,7}"));
}

O/P : true