如何通过Java读取文件夹中的所有文件?这与哪个API无关。
当前回答
File directory = new File("/user/folder");
File[] myarray;
myarray=new File[10];
myarray=directory.listFiles();
for (int j = 0; j < myarray.length; j++)
{
File path=myarray[j];
FileReader fr = new FileReader(path);
BufferedReader br = new BufferedReader(fr);
String s = "";
while (br.ready()) {
s += br.readLine() + "\n";
}
}
其他回答
import java.io.File;
public class ReadFilesFromFolder {
public static File folder = new File("C:/Documents and Settings/My Documents/Downloads");
static String temp = "";
public static void main(String[] args) {
// TODO Auto-generated method stub
System.out.println("Reading files under the folder "+ folder.getAbsolutePath());
listFilesForFolder(folder);
}
public static void listFilesForFolder(final File folder) {
for (final File fileEntry : folder.listFiles()) {
if (fileEntry.isDirectory()) {
// System.out.println("Reading files under the folder "+folder.getAbsolutePath());
listFilesForFolder(fileEntry);
} else {
if (fileEntry.isFile()) {
temp = fileEntry.getName();
if ((temp.substring(temp.lastIndexOf('.') + 1, temp.length()).toLowerCase()).equals("txt"))
System.out.println("File= " + folder.getAbsolutePath()+ "\\" + fileEntry.getName());
}
}
}
}
}
/**
* Function to read all mp3 files from sdcard and store the details in an
* ArrayList
*/
public ArrayList<HashMap<String, String>> getPlayList()
{
ArrayList<HashMap<String, String>> songsList=new ArrayList<>();
File home = new File(MEDIA_PATH);
if (home.listFiles(new FileExtensionFilter()).length > 0) {
for (File file : home.listFiles(new FileExtensionFilter())) {
HashMap<String, String> song = new HashMap<String, String>();
song.put(
"songTitle",
file.getName().substring(0,
(file.getName().length() - 4)));
song.put("songPath", file.getPath());
// Adding each song to SongList
songsList.add(song);
}
}
// return songs list array
return songsList;
}
/**
* Class to filter files which have a .mp3 extension
* */
class FileExtensionFilter implements FilenameFilter
{
@Override
public boolean accept(File dir, String name) {
return (name.endsWith(".mp3") || name.endsWith(".MP3"));
}
}
你可以过滤任何文本文件或任何其他扩展..把它换成。mp3
虽然我同意Rich, Orian和其他人使用:
final File keysFileFolder = new File(<path>);
File[] fileslist = keysFileFolder.listFiles();
if(fileslist != null)
{
//Do your thing here...
}
出于某种原因,这里所有的例子都使用绝对路径(即从根目录开始,或者,对于windows..)
我想补充的是,也可以使用相对路径。 所以,如果你的pwd(当前目录/文件夹)是folder1,你想要解析folder1/子文件夹,你只需写(在上面的代码中代替):
final File keysFileFolder = new File("subfolder");
从Test文件夹中列出类路径内的文件
import java.io.File;
import java.io.IOException;
public class Hello {
public static void main(final String[] args) throws IOException {
System.out.println("List down all the files present on the server directory");
File file1 = new File("/prog/FileTest/src/Test");
File[] files = file1.listFiles();
if (null != files) {
for (int fileIntList = 0; fileIntList < files.length; fileIntList++) {
String ss = files[fileIntList].toString();
if (null != ss && ss.length() > 0) {
System.out.println("File: " + (fileIntList + 1) + " :" + ss.substring(ss.lastIndexOf("\\") + 1, ss.length()));
}
}
}
}
}
防止listFiles()函数的nullpointerexception,并递归地从子目录中获取所有文件。
public void listFilesForFolder(final File folder,List<File> fileList) {
File[] filesInFolder = folder.listFiles();
if (filesInFolder != null) {
for (final File fileEntry : filesInFolder) {
if (fileEntry.isDirectory()) {
listFilesForFolder(fileEntry,fileList);
} else {
fileList.add(fileEntry);
}
}
}
}
List<File> fileList = new List<File>();
final File folder = new File("/home/you/Desktop");
listFilesForFolder(folder);
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