0
我有很多url文件需要通過文件名進行過濾。我有一個應用程序需要在這些文件名稱中查找的單詞列表。我曾嘗試使用contains()函數,但它一直要求CharSequence。我曾嘗試將數組列表轉換爲CharSequence列表,但仍然無效。JAVA過濾目錄中的文件
下面是代碼:
import java.io.File;
import java.util.ArrayList;
import java.util.Arrays;
public class Filter {
public static File folder = new File("C:/Users/blah/blah);
static String temp = "";
public static void main(String[] args) {
// TODO code application logic here
System.out.println("Reading files under the folder "+ folder.getAbsolutePath());
listFilesForFolder(folder);
}
public static void listFilesForFolder(final File folder){
ArrayList<String> aa = new ArrayList<String>
(Arrays.asList("one","two","three","four","five","six","seven"));
CharSequence[] cs = aa.toArray(new CharSequence[aa.size()]);
for(final File fileEntry : folder.listFiles()){
if(fileEntry.isDirectory()){
listFilesForFolder(fileEntry);
} else {
if (fileEntry.isFile()){
temp = fileEntry.getName();
if((temp.substring(temp.lastIndexOf('.') + 1,
temp.length()).toLowerCase()).equals("url"))
System.out.println("File = " + folder.getAbsolutePath()+ "\\" + fileEntry.getName());
}
}
}
}
}
將'folder.listFiles()'存儲在一個變量中並迭代它,否則會爲每個循環創建一個新的文件列表,這會減慢您的應用程序。 – BackSlash
你的數組怎麼可以charsequance?你寧願不將數組內部的內容轉換爲charsequence? – Zavior