Java Regex：检查句子是否仅包含字母和数字

Question

My following program prints weird results that I don't understand, I guess it is due to my lack of great understanding on Java Regex. So I wish to split the testStr by period first, then check if each sentence contains alphabet or numbers. But surprisingly, I got the following output, which is opposite to my wish:

blah blah1 is not a character!
 blah blah2 is not a character!
 blah blah3 is not a character!
 ??** is not a character!     // only this output is expected

my code below:

String testStr = "blah blah1. blah blah2. blah blah3. ??**...";
String[] myStrArray = testStr.split("[.]");

System.out.println("length of myStrArray is: " + myStrArray.length);

for (String str : myStrArray) {
    if (!Pattern.matches("\\w+", str)) {
        System.out.println(str + " is not a character!");
        continue;
    }

    System.out.println("got a meaningful sentence " + str.trim());

}

Answer 1

Your program splits string using dot as a separator, so you get:

blah blah1 
blah blah2 
blah blah3 
??**...

Then you try to match each line using regex \\w+ . Please note that call of match() is equivalent to call of find() but with regex that includes ^ and $ , ie think that your regex is ^\\w+$ .

I think that now it is obvious that any one of your strings does not match this pattern because 3 first strings contain space and the last does not contain neither alphabet characters nor digits.

Answer 2

Change your regex to: ^[a-zA-Z0-9\\s]+$ it'll allow only characters, numbers and spaces as required. pay attention that part of the "magic" is the use of ^ and $ which force a full match (from beginning to end).

Further, the reason I've used a-zA-Z0-9 instead of \\w is that \\w includes _ which doesn't fit the requirements.

Answer 3

You can use a character set. Change the regex you are using ( "\\\\w+" ) to this:

"[\\s&&[^\\W_]]"

This will allow alphanumerals ( [^\\W_] => a-zA-Z0-9 ) and whitespaces ( \\s ) to be matched, instead of only word characters.