Java:将文件名拆分为基本文件名和扩展名

问题描述：

有没有比之类的更好的方法来获取文件基名和扩展名

Is there a better way to get file basename and extension than something like

File f = ...
String name = f.getName();
int dot = name.lastIndexOf('.');
String base = (dot == -1) ? name : name.substring(0, dot);
String extension = (dot == -1) ? "" : name.substring(dot+1);

答

我知道其他人提到了 String.split，但这里有一个变体，它只产生两个标记(基础和扩展):

I know others have mentioned String.split, but here is a variant that only yields two tokens (the base and the extension):

String[] tokens = fileName.split("\.(?=[^\.]+$)");

例如:

"test.cool.awesome.txt".split("\.(?=[^\.]+$)");

产量:

["test.cool.awesome", "txt"]

正则表达式告诉 Java 在任何句点之后进行拆分，然后是任意数量的非句点，然后是输入的结尾.只有一个句点符合这个定义(即最后句点).

The regular expression tells Java to split on any period that is followed by any number of non-periods, followed by the end of input. There is only one period that matches this definition (namely, the last period).

~~技术上~~ 从正则上来说，这种技术被称为零宽度正向预测.

~~Technically~~ Regexically speaking, this technique is called zero-width positive lookahead.

顺便说一句，如果您想拆分路径并获取完整的文件名，包括但不限于点扩展名，请使用带正斜杠的路径，

BTW, if you want to split a path and get the full filename including but not limited to the dot extension, using a path with forward slashes,

    String[] tokens = dir.split(".+?/(?=[^/]+$)");

例如:

    String dir = "/foo/bar/bam/boozled"; 
    String[] tokens = dir.split(".+?/(?=[^/]+$)");
    // [ "/foo/bar/bam/" "boozled" ]

Java:将文件名拆分为基本文件名和扩展名

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