Java正则表达式如何检查该字符串不为空？

Question

There is regular expression for finding blank string and I want only negation. I also see this question but it does not work for java (see examples). Solution also not work for me (see 3-rd line in example).

For example

Pattern.compile("/^$|\\s+/").matcher(" ").matches() - false
Pattern.compile("/^$|\\s+/").matcher(" a").matches()- false
Pattern.compile("^(?=\\s*\\S).*$").matcher("\t\n a").matches() - false

return false in both cases.

PS If something is not clear ask me questions.

UPDATED

I want to use this regular expression in @Pattern annotation without creating custom annotation and programmatic validator for it. That's why I want a "plain" regexp solution without using find function.

Answer 1

It's not clear what you mean by negation.

If you mean "a string that contains at least one non-blank character," then you can use this:

Pattern.compile("\\S").matcher(str).find()

If it's really necessary to use matches , then you can do it with this.

Pattern.compile("\\A\\s*\\S.*\\Z").matcher(str).matches()

This just matches 0 or more spaces followed by a non-space followed by any characters at all up to the end of the string.

If you mean "a string that is all non-blank with at least one such character," then you can use this:

Pattern.compile("\\A\\S+\\Z").matcher(str).matches()

You need to study the Java regex syntax . In Java, regular expressions are compiled from strings, so there's no need for special delimiters like /.../ or %r{...} as you'll see in other languages.

Answer 2

How about this:

if(!string.trim().isEmpty()) {
    // do something
}

Answer 3

使用regex \\s ：空格字符： \\t\\n\\x0B\\f\\r 。

Pattern.compile("\\s")