如何根據最後修改

Question

按asc順序對文件進行排序我正在從目錄中獲取所有文件，然後根據需要從中選擇文件並將它們存儲在數組中，現在我想對最後修改的最後一個文件進行排序。下面是我使用

public static int GetFilesCount(File folderPath,int count,String type,Context context) 
{ 
    BackupCount=count; 
    BackupFolderPath=folderPath; 
    Backuptype=type; 
    con=context; 
    DatabaseHandler objhandler; 
    Cursor  cursor=null; 
    int total = 0; 
    String ext=""; 

    // Check files count set by user 

    File[] fList = folderPath.listFiles(); 
    ArrayList<String> myfiles = new ArrayList<String>(); 

    for (File file : fList){ 
     if (file.isFile()){ 
      try { 
       String FileName=file.getName(); 
       ext=GetFileExtension(FileName); 
       if(ext.equals("db")) 
       { 
        objhandler=new DatabaseHandler(context, folderPath+File.separator+FileName, null); 
        database= objhandler.openDataBase(); 
        String selectQuery = "SELECT * FROM "+ type + " LIMIT 1"; 
        cursor = database.rawQuery(selectQuery, null); 
        Integer ColCount=cursor.getColumnCount(); 
        if(cursor.getCount()>0) 
        { 
         if(Backuptype.equals("SMS")) 
         { 
          if(ColCount.equals(9)) 
          { 
           myfiles.add(FileName); 

           total++; 
          } 
         } 
         else if(Backuptype.equals("CallLogs")) 
         { 
          if(ColCount.equals(6)) 
          { 
           myfiles.add(FileName); 
           total++; 
          } 
         } 
         else if(Backuptype.equals("Contacts")) 
         { 
          if(ColCount.equals(9)) 
          { 
           myfiles.add(FileName); 
           total++; 
          } 
         } 
        }  
        if(total>count) 
        { 
         // String[] listFiles=new String[myfiles.size()]; 
         // listFiles = myfiles.toArray(listFiles); 
         // File[] f = null; 
         for(int i=0;i<=myfiles.size();i++) 
         { 
          // f[i]=new File(folderPath+File.separator+myfiles.get(i)); 
          System.out.println("Total SMS Files: "+myfiles.size()); 
          System.out.println("file in folder: "+myfiles.get(i).toString()); 
         } 



         /*Arrays.sort(f, new Comparator<File>(){ 
          public int compare(File f1, File f2) 
          { 
          return Long.valueOf(f1.lastModified()).compareTo(f2.lastModified()); 
          } });*/ 

         System.out.println("file in folder: "+myfiles.size()); 
         // Deletefile(folderPath+File.separator+myfiles.get(0)); 
        } 
       } 


      }catch (Exception e) { 
       e.printStackTrace(); 
       // TODO: handle exception 
      }finally{ 
       cursor.close(); 
       database.close(); 
      } 
     } 
    } 
    return 1; 
} 

Answer 1

嘗試這部分代碼合併到你的代碼的代碼：

final List<File> files = new ArrayList<File>(); 
    Collections.sort(files, new Comparator<File>() 
    { 

     @Override 
     public int compare(final File o1, final File o2) 
     { 
     return o1.lastModified() >= o2.lastModified() ? 1 : -1; 
     } 

    }); 

Answer 2

從您的評論：

我很困惑如何把選定的文件文件[]

因爲你有：

// File[] f = null; 

我覺得這是你缺少

File[] f = new File[myfiles.size()]; // init the array which should hold the files 
for(int i = 0; i < myfiles.size(); i++) { 
    files[i] = new File(folderPath+File.separator+myfiles.get(i)); 
} 

那麼CA使用Arrays.sort方法

Arrays.sort(files, new Comparator<File>() { 
    @Override 
    public int compare(File o1, File o2) { 
     return Long.valueOf(o1.lastModified()).compareTo(o2.lastModified()); 
    } 
}); 

的部分更重要的是，你應該List工作，並使用Collections.sort

List<File> f = new ArrayList<>(); 
for(int i = 0; i < myfiles.size(); i++) { 
    f.add(new File(folderPath+File.separator+myfiles.get(i))); 
} 

Collections.sort(f, new Comparator<File>() { 
    @Override 
    public int compare(File o1, File o2) { 
     return Long.valueOf(o1.lastModified()).compareTo(o2.lastModified()); 
    } 
});