[英]How to pull phrase out of a string
我如何為兩者拉出“16”
這是我嘗試過的
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);
這是我得到的錯誤
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/上測試了正則表達式(“([0-9] +:[0-9] +)+”),它正確地突出了“8:16”
您需要將捕獲組放在第二個[0-9]+
(或等效的\\d+
)上並使用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");
見演示
String.indexOf(String str)不接受正則表達式。 它需要一個字符串。
你可以這樣做:
String V, Line = "Bar Foo Bar: Foo8:16 Foo Bar Bar foo barz";
V = Line.substring(Line.indexOf("16"), Line.indexOf("16") + 2);
System.out.println(V);
或者為了使它看起來更整潔,你可以替換這一行:
V = Line.substring(Line.indexOf("16"), Line.indexOf("16") + 2);
有:
int index = Line.indexOf("16");
V = Line.substring(index, index + 2);
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