如何从一个字符串中拉出短语

我如何为两者拉出"16"

Bar Foo Bar:Foo8:16 Foo Bar Bar foo barz

8:16 Foo Bar Bar foo barz

这是我尝试过的

String V,Line ="Bar Foo Bar: Foo8:16 Foo Bar Bar foo barz";
V = Line.substring(Line.indexOf("([0-9]+:[0-9]+)+")+1);
V = V.substring(V.indexOf(":")+1, V.indexOf(" "));
System.out.println(V);

这是我得到的错误

Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: -9
    at java.lang.String.substring(String.java:1955)  
    at Indexing.Index(Indexing.java:94)  
    at Indexing.main(Indexing.java:24)

我在http://regexr.com/上测试了正则表达式("([0-9] +:[0-9] +)+"),它正确地突出了"8:16"

您需要将捕获组放在第二个[0-9]+(或等效的\d+)上,并使用Matcher#find():

String value1 = "Bar Foo Bar: Foo8:16 Foo Bar Bar foo barz";
String pattern1 = "\\d+:(\\d+)"; // <= The first group is the \d+ in round brackets
Pattern ptrn = Pattern.compile(pattern1);
Matcher matcher = ptrn.matcher(value1);
if (matcher.find())
    System.out.println(matcher.group(1)); // <= Print the value captured by the first group
else
    System.out.println("false");

见演示