我很难过......假设我有这个目录树:
{someRoot}/
{someRoot}/bar/
{someRoot}/bar/file1.txt
{someRoot}/foo/
{someRoot}/foo/baz/
{someRoot}/foo/baz/file3.txt
{someRoot}/foo/abracadabra.txt
{someRoot}/foo/file2.txt
{someRoot}/aardvark.txt
{someRoot}/food.txt
{someRoot}/zebra.txt
你会注意到订购。将此 order1 称为。在每个阶段,目录首先出现在文件之前。 (注意: bar/file1.txt
在foo
之前,所以在全球范围内,目录并非都在所有文件之前。)
如果我枚举这个目录树,然后递归枚举子目录,我将得到以下List<File>
,订购 order2 。
{someRoot}/
{someRoot}/aardvark.txt
{someRoot}/bar/
{someRoot}/foo/
{someRoot}/food.txt
{someRoot}/zebra.txt
{someRoot}/bar/file1.txt
{someRoot}/foo/abracadabra.txt
{someRoot}/foo/baz/
{someRoot}/foo/file2.txt
{someRoot}/foo/baz/file3.txt
如果我创建直截了当的Comparator<File>
:
Comparator<File> fc = new Comparator<File>(){
@Override public int compare(File o1, File o2) {
return o1.compareTo(o2);
}
};
我排序,我从词典排序中得到这个排序( order3 ):
{someRoot}
{someRoot}/aardvark.txt
{someRoot}/bar
{someRoot}/bar/file1.txt
{someRoot}/foo
{someRoot}/food.txt
{someRoot}/foo/abracadabra.txt
{someRoot}/foo/baz
{someRoot}/foo/baz/file3.txt
{someRoot}/foo/file2.txt
{someRoot}/zebra.txt
但我不想要这个排序(有问题:请注意food.txt
介于目录foo
及其子项目之间),我想要 order1 。我怎样才能写一个比较器来帮助我呢?
答案 0 :(得分:4)
这适用于我的测试。
new Comparator<File>() {
@Override
public int compare(File first, File second) {
if (first.isDirectory() && second.isDirectory())
return first.compareTo(second);
if (first.isDirectory())
return this.compareToFile(first, second);
if (second.isDirectory())
return -(this.compareToFile(second, first));
return this.compareFiles(first, second);
}
private int compareFiles(File first, File second) {
File firstParentFile = first.getParentFile();
File secondParentFile = second.getParentFile();
if (isSubDir(firstParentFile, secondParentFile))
return -1;
if (isSubDir(secondParentFile, firstParentFile))
return 1;
return first.compareTo(second);
}
private int compareToFile(File directory, File file) {
File fileParent = file.getParentFile();
if (directory.equals(fileParent))
return -1;
if (isSubDir(directory, fileParent))
return -1;
return directory.compareTo(file);
}
private boolean isSubDir(File directory, File subDir) {
for (File parentDir = directory.getParentFile(); parentDir != null; parentDir = parentDir.getParentFile()) {
if (subDir.equals(parentDir)) {
return true;
}
}
return false;
}
答案 1 :(得分:2)
这个递归可以根据需要获取已排序的文件树:
public static void main(String[] args) {
List<File> files = getFileTree(new File("."));
for (File f : files)
System.out.println(f);
}
private static List<File> getFileTree(File file) {
List<File> files = new LinkedList<File>();
files.add(file);
if (file.isDirectory()) {
File[] current = file.listFiles();
Arrays.sort(current, new Comparator<File>() {
@Override
public int compare(File o1, File o2) {
if (o1.isDirectory())
return o2.isDirectory() ? o1.compareTo(o2) : -1;
else if (o2.isDirectory())
return 1;
return o1.compareTo(o2);
}
});
for (File f : current)
files.addAll(getFileTree(f));
}
return files;
}
答案 2 :(得分:0)
修改强> 您必须比较第一级的基本目录和第二级的文件名
import java.io.File;
import java.io.IOException;
import java.util.Comparator;
public class FileComparator implements Comparator<File> {
@Override
public int compare(File o1, File o2) {
int pathCompare = this.getPath(o1).compareTo(this.getPath(o2));
if (pathCompare != 0) {
return subPathCompare;
} else {
if (o1.isDirectory() && !o2.isDirectory()) {
return -1;
} else if (!o1.isDirectory() && o2.isDirectory()) {
return 1;
} else {
return o1.compareTo(o2);
}
}
}
//maybe there is a better way to speparete path and file name, but it works
private String getPath(File file) {
if (file.isFile()) {
return file.getParent().toLowerCase();
} else {
return file.getPath().toLowerCase();
}
}
}
如果路径和文件的分离是必要的,因为如果只做一个:
if (o1.isDirectory() && !o2.isDirectory()) {
return -1;
} else if (!o1.isDirectory() && o2.isDirectory()) {
return 1;
} else {
return o1.compareTo(o2);
}
结果不是案例1,而是另一个案例,其中所有directorys都是第一个项目,文件是最后一个:
\dir1\
\dir2\
\dir3\
\dir4\
\dir1\file1\
\dir1\file2\
\dir2\file1\
\dir2\file2\
所以我的解决方案是首先检查两个比较文件是否属于同一目录。
答案 3 :(得分:0)
Comparator<File> fc = new Comparator<File>(){
@Override public int compare(File o1, File o2) {
boolean isFirstDirectory = o1.isDirectory();
boolean isSecondDirectory = o2.isDirectory();
if(!isFirstDirectory && !isSecondDirectory) {
return o1.compareTo(o2);
} else if (isFirstDirectory && !isSecondDirectory){
return (int) 1;
} else if(isSecondDirectory && !isFirstDirectory) {
return -1;
} else {
return o1.compareTo(o2);
}
}
};
答案 4 :(得分:0)
我确实找到了一种方法来实现我想要的伪相反(目录在文件之后,没有检查文件和目录名在根级别的边缘情况)。
......不,将它转换成我想要的东西并不容易,因为我正在使用目录的词典排序,它将不同级别的目录排序和同一级别的目录排序结合起来。
static class FileDirectoryPriorityComparator implements Comparator<File>
{
@Override public int compare(File f1, File f2) {
if (f1.isDirectory())
{
if (f2.isDirectory())
{
return f1.compareTo(f2);
}
else
{
return compareDirFile(f1, f2);
}
}
else
{
if (f2.isDirectory())
{
return -compareDirFile(f2, f1);
}
// f1 and f2 are both files
else
{
return compareFiles(f1, f2);
}
}
}
private int compareDirFile(File dir, File file) {
/*
* If dir compares differently to file's parent, use that ordering.
* Otherwise, dir comes before file.
*/
File fparent = file.getParentFile();
if (fparent == null)
return -1;
int i = dir.compareTo(fparent);
if (i != 0)
return i;
return -1;
}
private int compareFiles(File f1, File f2) {
/*
* If f1's parent compares differently to f2's parent, use that ordering.
* Otherwise use default ordering.
*/
File fp1 = f1.getParentFile();
File fp2 = f2.getParentFile();
if (fp1 == null)
{
if (fp2 != null)
{
return 1;
}
}
else
{
if (fp2 == null)
{
return -1;
}
else
{
int i = fp1.compareTo(fp2);
if (i != 0)
return i;
}
}
return f1.compareTo(f2);
}
}
答案 5 :(得分:0)
当然为时已晚。在任何情况下,这里都有一个Comparator
实现,按照SO中的要求进行排序。
public class DirectoryBeforeFileComparator implements Comparator<File> {
@Override
public int compare(File o1, File o2) {
if (o1.isDirectory() && !o2.isDirectory()) {
// directory before non-directory.
return -1;
}
if (!o1.isDirectory() && o2.isDirectory()) {
// non-directory after directory
return 1;
}
// compare two pathnames lexicographically
return o1.compareTo(o2);
}
}