閱碼場第1期:性能工程
2023-08-26 21:33 作者:bili_57619593086 | 我要投稿
列出文件和目錄
File dir = new File("directoryName");
String[] children = dir.list();
if (children == null) {
? // Either dir does not exist or is not a directory
} else {
? for (int i=0; i < children.length; i++) {
? ? // Get filename of file or directory
? ? String filename = children[i];
? }
}
// It is also possible to filter the list of returned files.
// This example does not return any files that start with `.'.
FilenameFilter filter = new FilenameFilter() {
? public boolean accept(File dir, String name) {
? ? return !name.startsWith(".");
? }
};
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