Java 如何遍历文件夹中的所有文件（如果文件名未知）？
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How to loop through all the files in a folder (if the names of the files are unknown)?
提问by Buras
There is a folder: C:\\Users\..myfolder
有一个文件夹： C:\\Users\..myfolder
It contains .pdffiles (or any other, say .csv). I cannot change the names of those files, and I do not know the number of those files. I need to loop all of the files one by one.
How can I do this?
它包含.pdf文件（或任何其他文件，例如 .csv）。我无法更改这些文件的名称，而且我不知道这些文件的数量。我需要一个一个地循环所有文件。我怎样才能做到这一点？
(I know how to do this if I knew the names)
（如果我知道名字，我就知道该怎么做）
采纳答案by Boris the Spider
Just use File.listFiles
只需使用 File.listFiles
final File file = new File("whatever");
for(final File child : file.listFiles()) {
//do stuff
}
You can use the FileNameExtensionFilterto filter your files too
您也可以使用FileNameExtensionFilter来过滤文件
final FileNameExtensionFilter extensionFilter = new FileNameExtensionFilter("N/A", "pdf", "csv"//, whatever other extensions you want);
final File file = new File("whatever");
for (final File child : file.listFiles()) {
if(extensionFilter.accept(child)) {
//do stuff
}
}
Annoyingly FileNameExtensionFiltercomes from the javax.swingpackage so cannot be used directly in the listFiles()api, it is still more convenient than implementing a file extension filter yourself.
烦人的FileNameExtensionFilter是来自javax.swing包所以不能直接在listFiles()api中使用，还是比自己实现文件扩展名过滤器方便多了。
回答by user2827100
File.listFiles()gives you an array of files in a folder. You can then split the filenames to get the extension and check if it is .pdf.
File.listFiles()为您提供文件夹中的文件数组。然后，您可以拆分文件名以获取扩展名并检查它是否为 .pdf.
File[] files = new File("C:\Users\..myfolder").listFiles();
for (File file : files) {
if (!file.isFile()) continue;
String[] bits = file.getName().split(".");
if (bits.length > 0 && bits[bits.length - 1].equalsIgnoreCase("pdf")) {
// Do stuff with the file
}
}
回答by Ali Hashemi
You can use Java.io.File.listFiles()method to get a list of all files and folders inside a folder.
您可以使用Java.io.File.listFiles()方法来获取文件夹内所有文件和文件夹的列表。
回答by Sotirios Delimanolis
So you can have more options, try the Java 7 NIOway of doing this
所以你可以有更多的选择，试试Java 7 NIO 的方式
public static void main(String[] args) throws Exception {
try (DirectoryStream<Path> files = Files.newDirectoryStream(Paths.get("/"))) {
for (Path path : files) {
System.out.println(path.toString());
}
}
}
You can also provide a filter for the paths in the form of a DirectoryStream.Filterimplementation
您还可以以DirectoryStream.Filter实现的形式为路径提供过滤器
public static void main(String[] args) throws Exception {
try (DirectoryStream<Path> files = Files.newDirectoryStream(Paths.get("/"),
new DirectoryStream.Filter<Path>() {
@Override
public boolean accept(Path entry) throws IOException {
return true; // or whatever you want
}
})
) {
for (Path path : files) {
System.out.println(path.toString());
}
}
}
Obviously you can extract the anonymous class to an actual class declaration.
显然，您可以将匿名类提取到实际的类声明中。
Note that this solution cannot return nulllike the listFiles()solution.
请注意，此解决方案不能null像listFiles()解决方案一样返回。
For a recursive solution, check out the FileVisitorinterface. For path matching, use the PathMatcherinterface along with FileSystemsand FileSystem. There are examples floating around Stackoverflow.
对于递归解决方案，请查看FileVisitor接口。对于路径匹配，将PathMatcher接口与FileSystems和一起使用FileSystem。Stackoverflow 中有很多例子。
