像树命令一样打印目录 [英] print directory like tree command
本文介绍了像树命令一样打印目录的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我必须打印目录树.我有以下代码:
I have to print a directory tree. I have the following code:
public static void main(String[] args) throws Exception
{
File root = new File("C:\Users\User\Desktop\example");
if (!root.isDirectory())
{
throw new IllegalArgumentException("Invalid directory: " + root.getName());
}
int level = 0;
System.out.println(renderFolder(root, level, new StringBuilder(), false));
}
private static StringBuilder renderFolder(File folder, int level, StringBuilder sb, boolean isLast)
{
indent(sb, level, isLast).append(folder.getName()).append("
");
File[] objects = folder.listFiles();
for (int i = 0; i < objects.length; i++)
{
boolean last = ((i + 1) == objects.length);
if (objects[i].isDirectory())
{
renderFolder(objects[i], level + 1, sb, last);
}
}
return sb;
}
private static StringBuilder indent(StringBuilder sb, int level, boolean isLast)
{
for (int i = 1; i < level; i++)
{
sb.append("u2502 ");
}
if (level > 0)
{
sb.append(isLast
? "u2514u2500u2500"
: "u251cu2500u2500");
}
return sb;
}
使用此输出:
example
├──a
└──b
│ └──b1
│ │ ├──b11
│ │ │ └──b111
│ │ └──b12
│ │ │ ├──b121
│ │ │ └──b122
│ │ │ │ └──b1221
我没能去除无用的深纹.如何做到这一点?
另外我的回答是:如何保持线路连接(与 cmd 的树命令一样多)?"
I didn't manage to remove the useless deep lines. How to do this?
Also my answer is: "How can I keep lines connected(as much as the cmd's tree command)?"
推荐答案
试试这个,我在你的方法中添加了另一个参数,称为hierarchyTree 来指示是否在某个级别,我们应该打印 |或空白区域.
Try this one, I added another parameter into your methods called hierarchyTree to indicate if at a certain level, we should print | or empty space.
用作堆栈,可以根据递归级别轻松维护文件夹级别.
It is used as a stack to easily maintain the folder levels based on the recursion level.
private static StringBuilder renderFolder(File folder, int level, StringBuilder sb, boolean isLast, List<Boolean> hierarchyTree) {
indent(sb, level, isLast, hierarchyTree).append(folder.getName()).append("
");
File[] objects = folder.listFiles(new FilenameFilter() {
@Override
public boolean accept(File current, String name) {
return new File(current, name).isDirectory();
}
});
for (int i = 0; i < objects.length; i++) {
boolean last = ((i + 1) == objects.length);
// this means if the current folder will still need to print subfolders at this level, if yes, then we need to continue print |
hierarchyTree.add(i != objects.length - 1);
renderFolder(objects[i], level + 1, sb, last, hierarchyTree);
// pop the last value as we return from a lower level to a higher level
hierarchyTree.remove(hierarchyTree.size() - 1);
}
return sb;
}
private static StringBuilder indent(StringBuilder sb, int level, boolean isLast, List<Boolean> hierarchyTree) {
String indentContent = "u2502 ";
for (int i = 0; i < hierarchyTree.size() - 1; ++i) {
// determines if we need to print | at this level to show the tree structure
// i.e. if this folder has a sibling foler that is going to be printed later
if (hierarchyTree.get(i)) {
sb.append(indentContent);
} else {
sb.append(" "); // otherwise print empty space
}
}
if (level > 0) {
sb.append(isLast
? "u2514u2500u2500"
: "u251cu2500u2500");
}
return sb;
}
示例结果:
example
├──a
├──b
│ └──b1
│ ├──b11
│ │ └──b111
│ └──b12
│ ├──b121
│ └──b122
│ └──b1221
└──c
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