将文件名添加到java中的数组列表
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Adding file names to an array list in java
提问by Chris G
I am currently needing to load the contents of a folders filenames to an arraylist I have but I am unsure as how to do this.
我目前需要将文件夹文件名的内容加载到我拥有的数组列表中，但我不确定如何执行此操作。
To put it into perspective I have a folder with One.txt, Two.txt, Three.txt etc. I want to be able to load this list into an arraylist so that if I was to check the arraylist its contents would be :
为了正确看待它，我有一个包含 One.txt、Two.txt、Three.txt 等的文件夹。我希望能够将这个列表加载到一个数组列表中，这样如果我要检查数组列表，它的内容将是：
arraylist[0] = One
数组列表[0] = 一
arraylist[1] = Two
数组列表[1] = 2
arraylist[3] = Three
数组列表[3] = 三
If anyone could give me any insight into this it would be much appreciated.
如果有人能让我对此有任何见解，将不胜感激。
回答by Hyman
You can try with
你可以试试
File folder = new File("myfolder");
if (folder.isDirectory())
{
// you can get all the names
String[] fileNames = folder.list();
// you can get directly files objects
File[] files = folder.listFiles();
}
回答by Ziv
See the Jave File API, particularly the list()function.
请参阅 Java 文件 API，尤其是list()函数。
File my_dir = new File("DirectoryName");
assert(my_dir.exists()); // the directory exists
assert(my_dir.isDirectory()); // and is actually a directory
String[] filenames_in_dir = my_dir.list();
Now filenames_in_diris an array of all the filenames, which is almost precisely what you want.
现在filenames_in_dir是所有文件名的数组，这几乎正是您想要的。
If you want to strip the extensions off the individual filenames, that's basic string handling - iterate over the filenames and take care of each one.
如果您想从单个文件名中去除扩展名，那就是基本的字符串处理 - 遍历文件名并处理每个文件名。
回答by cletus
File dir = new File("/some/path/name");
List<String> list = new ArrayList<String>();
if (dir.isDirectory()) {
String[] files = dir.list();
Pattern p = Pattern.compile("^(.*?)\.txt$");
for (String file : files) {
Matcher m = p.matcher(file);
if (m.matches()) {
list.add(m.group(1));
}
}
}
回答by kingpin
Have a look at java.io.File it gives you all the control you may need. Here is a URL for it http://java.sun.com/j2se/1.5.0/docs/api/java/io/File.html
看看 java.io.File 它为您提供了您可能需要的所有控制。这是它的 URL http://java.sun.com/j2se/1.5.0/docs/api/java/io/File.html
回答by polygenelubricants
Here's a solution that uses java.io.File.list(FilenameFilter). It keeps the .txtsuffix; you can strip these easily if you really need to.
这是一个使用java.io.File.list(FilenameFilter). 它保留.txt后缀；如果你真的需要，你可以很容易地剥离这些。
File dir = new File(".");
List<String> list = Arrays.asList(dir.list(
new FilenameFilter() {
@Override public boolean accept(File dir, String name) {
return name.endsWith(".txt");
}
}
));
System.out.println(list);
回答by Rose Blax
Here's My answer, I've used this before personally to get all the filenames,
to be used in a loadsave function in one of my games.
这是我的答案，我之前亲自使用过它来获取所有文件名，
以便在我的一款游戏的加载保存功能中使用。
public void getFiles(String path){
//Put filenames in arraylist<string>
File dir = new File(path);
ArrayList<String> filenames = new ArrayList<String>();
for(File file : dir.listFiles()){
savefiles.add(file.getName());
}
//Check if the files are in the arraylist
for (int i = 0; i < savefiles.size(); i++){
String s = savefiles.get(i);
System.out.println("File "+i+" : "+s);
}
System.out.println("\n");
}
I hope that may have been of use to you C: - Hugs rose
我希望这可能对你有用 C: - 拥抱玫瑰
