Question

I’m testing if a string is valid/invalid by checking if certain characters exists. Strings that contain ; # / % = | + \ ” < > are deemed invalid. I’ve the following implementation in Java currently, but I’d prefer a more elegant regex solution.

public boolean isStringValid(String name) {
    if (   name.contains(";")
        || name.contains("#")
        || name.contains("/")
        || name.contains("%")
        || name.contains("=")
        || name.contains("|")
        || name.contains("+")
        || name.contains("\\")
        || name.contains("\"")
        || name.contains("<")
        || name.contains(">")) {
        return false;
    }
    else {
        return true;
    }
}

What I’ve done is changed it to the following,

public boolean isNameValid(String name) {
    return !Pattern.matches(".*(;|#|/|%|=|\\||\\+|\\\\|\"|<|>)+.*", name);
}

but I can’t seem to get the regex string right. The original regex string before adding in all the Java escape characters is as follows,

.*(;|#|/|%|=|\||\+|\\|"|<|>)+.*

Using character classes like [A-z] doesn’t seem to be an option because a name like “d@vik” is supposed to be considered valid in my case.

Answer 1

You are almost right. Just use find() method instead of matches(). And compile pattern only once. This is the most expensive operation. And you can simplify you pattern using [ ]: in this case you do not have to write | between subpatterns you are looking for:

private static Pattern validator = Pattern.compile("[;%#=\\+]"); // etc: write all characters you need.

Now re-write you isNameValid() as following:

public boolean isNameValid(String name) {
    return !validator.find();
}

BTW pay attention on backslash. If you want your pattern to include backslash it should be written 4 times: twice for regex escaping and twice for java escaping.