java中一个用例的正则表达式，从DANNO开始，以integer结束

我试图创建一个正则表达式来从字符串中获取一个数字。我使用并使用了以下正则表达式

（DANNO）[^0-9]*[0-9]*

，这在他们的控制台上运行得很好，但当我尝试在java代码中使用时，它只是在某些情况下为我提供了DANNO

以下是一些示例字符串以及预期结果：

N. DANNO: 1234567890 => DANNO: 1234567890
DANNO N° 1234567890 => DANNO N° 1234567890
ORDINE N  1234567890 N  DANNO 1234567890 N  CLIENTE 123456 => DANNO 1234567890
N°DANNO1234567890 => DANNO1234567890
DANNON°1234567890 => DANNON°1234567890

---

使用：

\bDANNO\D*（\D+）

作为正则表达式并抓取捕获的组#1@anubhava令人惊讶的是，这起作用了。非常感谢你。

public static void main(String[] args) throws ParseException {
    final String regex = "(DANNO)[^0-9]*[0-9]*";
    final String string = "N. DANNO: 1234567890\n"
         + "DANNO N° 1234567890\n"
         + "ORDINE N  1234567890 N  DANNO 1234567890 N  CLIENTE 123456\n"
         + "N°DANNO1234567890\n"
         + "DANNON°1234567890";

    final Pattern pattern = Pattern.compile(regex, Pattern.MULTILINE);
    final Matcher matcher = pattern.matcher(string);

    while (matcher.find()) {
        System.out.println("Full match: " + matcher.group(0)); //Prints full serach result
        for (int i = 1; i <= matcher.groupCount(); i++) {
            System.out.println("Group " + i + ": " + matcher.group(i)); //Prints DANO
        }
    }
}

Full match: DANNO: 1234567890
Group 1: DANNO
Full match: DANNO N° 1234567890
Group 1: DANNO
Full match: DANNO 1234567890
Group 1: DANNO
Full match: DANNO1234567890
Group 1: DANNO
Full match: DANNON°1234567890
Group 1: DANNO