I try to create a regex for a String which is NotBlank and cannot contain "<". My question is what Im doing wrong thank you.

"(\\A(?!\\s*\\Z))|([^<] )"

Edit Maybe this way how to combine this regex

^[^<] $

with this regex

\\A(?!\\s*\\Z). 

CodePudding user response：

With regex, you can use

\A(?!\s \z)[^<] \z
(?U)\A(?!\s \z)[^<] \z

The (?U) is only necessary when you expect any Unicode chars in the input.

In Java, when used with matches, the anchors on both ends are implicit:

text.matches("(?U)(?!\\s \\z)[^<] ")

The regex in matches is executed once and requires the full string match. Here, it matches

• \A - (implicit in matches) - start of string
• (?U) - Pattern.UNICODE_CHARACTER_CLASS option enabled so that \s could match any Unicode whitespaces
• (?!\\s \\z) - until the very end of string, there should be no one or more whitespaces
• [^<] - one or more chars other than <
• \z - (implicit in matches) - end of string.

See the Java test:

String texts[]  = {"Abc <<", "     ", "", "abc 123"};
Pattern p = Pattern.compile("(?U)(?!\\s \\z)[^<] ");
for(String text : texts)
{
    Matcher m = p.matcher(text);
    System.out.println("'"   text   "' => "   m.matches());
}

Output:

'Abc <<' => false
'     ' => false
'' => false
'abc 123' => true

See an online regex test (modified to fit the single multiline string demo environment so as not to cross over line boundaries.)

CodePudding user response：

You can try to use this regex:

[^<\s] 

Any char that is not "<", for 1 or more times.

Here is the example to test it: https://regex101.com/r/9ptt15/2

However, you can try to solve it without a regular expression:

boolean isValid = s != null && !s.isEmpty() && s.indexOf(" ") == -1 && s.indexOf("<") == -1;