如果内部包含数字，Java如何排序字符串？

Question

如何对Apache源档案名称（字符串）进行排序。 我尝试使用Jsoup下面的代码，但不返回给定的预期结果。 如何解决这个问题？

public static void getApacheArchives() throws IOException{
    String url = "https://archive.apache.org/dist/httpd/"; // or whatever goes here
    Document document = Jsoup.connect(url).followRedirects(false).timeout(60000/*wait up to 60 sec for response*/).get();

    Elements anchors = document.body().getAllElements().select("a");

    Collections.sort(anchors, new Comparator<Element>() {
        @Override
        public int compare(Element e1, Element e2) {
            return e1.text().compareTo(e2.text());
        }
    });

    for (int i = 0; i < anchors.size(); i++) {
        Element a = anchors.get(i);
        if ( 
                ( a.text().matches( "(apache_)[1].[0-9].[0-9]{1,2}.(tar.gz)" ) )
                ||
                ( a.text().matches( "(httpd-)[0-9]{1,2}.[0-9]{1,2}.[0-9]{1,2}.(tar.gz)") )
            ){
            System.out.println(a.text());
        }
    }
}

此代码返回以下结果：
... ... httpd-2.3.6.tar.gz
httpd-2.3.8.tar.gz
httpd-2.4.1.tar.gz
httpd-2.4.10.tar.gz
httpd-2.4.12.tar.gz
httpd-2.4.16.tar.gz
httpd-2.4.17.tar.gz
httpd-2.4.18.tar.gz
httpd-2.4.2.tar.gz
httpd-2.4.20.tar.gz
httpd-2.4.3.tar.gz
httpd-2.4.4.tar.gz
httpd-2.4.6.tar.gz
httpd-2.4.7.ta​​r.gz
httpd-2.4.9.tar.gz
...

但预期结果如下：
... ... httpd-2.3.6.tar.gz
httpd-2.3.8.tar.gz
httpd-2.4.1.tar.gz
httpd-2.4.2.tar.gz
httpd-2.4.3.tar.gz
httpd-2.4.4.tar.gz
httpd-2.4.6.tar.gz
httpd-2.4.7.ta​​r.gz
httpd-2.4.9.tar.gz
httpd-2.4.10.tar.gz
httpd-2.4.12.tar.gz
httpd-2.4.16.tar.gz
httpd-2.4.17.tar.gz
httpd-2.4.18.tar.gz
httpd-2.4.20.tar.gz
...

Answer 1

谢谢Tom，我找到了解决问题的方法！ Sorting Strings that contains number in Java

一些修改：

Collections.sort(anchors, new Comparator<Element>() {
        public int compare(Element o1, Element o2) {
            return extractInt(o1.text()) - extractInt(o2.text());
        }

        int extractInt(String s) {
            String num = s.replaceAll("\\D", "");
            // return 0 if no digits found
            return num.isEmpty() ? 0 : Integer.parseInt(num);
        }
    });