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Java 尝试将新添加的文件(图像)的名称更改为与目录中现有文件(图像)相同的名称。不起作用_Java_File - Fatal编程技术网

Java 尝试将新添加的文件(图像)的名称更改为与目录中现有文件(图像)相同的名称。不起作用

Java 尝试将新添加的文件(图像)的名称更改为与目录中现有文件(图像)相同的名称。不起作用,java,file,Java,File,我创建了一个方法,它将为我提供一个保存在旧图像的DB中的1列中的路径,名为slika,我从中返回该字符串: String DBslika = null; try { Class.forName("com.mysql.jdbc.Driver"); con1 = java.sql.DriverManager.getConnection("jdbc:mysql://loca

我创建了一个方法,它将为我提供一个保存在旧图像的DB中的1列中的路径,名为
slika
,我从中返回该字符串:

        
       String DBslika = null;
        try {
            Class.forName("com.mysql.jdbc.Driver");  
            con1 = java.sql.DriverManager.getConnection("jdbc:mysql://localhost/studenti", "root", "");
            
            String b=String.valueOf(id);

            update = con1.prepareStatement("Select slika from studenti where ID="+b);
            ResultSet rs = get.executeQuery();
             
              while(rs.next())
             {
                     DBslika = rs.getString(1);
             }
              rs.close();
              
            
        } catch (ClassNotFoundException ex) {
            Logger.getLogger(StudentChange.class.getName()).log(Level.SEVERE, null, ex);
        } catch (SQLException ex) {
            Logger.getLogger(StudentChange.class.getName()).log(Level.SEVERE, null, ex);
        }
        
        return DBslika;
    }     
然后单击按钮选择并将新图像添加到原始图像辞职的目录中,我尝试添加新图像并将其名称更改为与旧图像名称相同,同时在过程中替换旧图像:

  private void DodajSlikuActionPerformed(java.awt.event.ActionEvent evt) {                                           
        
         //Old name of the existing path of the image
        
         String staraPutanja = ImePostojeceg();
         Path stariPut = Paths.get(staraPutanja);
         //Path fileStari = stariPut.getFileName();
         
         //New name of the path of new image
        
        String put = "C:"+File.separator+"Users"+File.separator+"boban_eyma1sy"+File.separator+"OneDrive"+File.separator+"Documents"+File.separator+"NetBeansProjects"+File.separator+"StudentInfo"+File.separator+"src"+File.separator+"Studenti"+File.separator+""; 
        JFileChooser chooser = new JFileChooser();
        chooser.showOpenDialog(null);
        File f = chooser.getSelectedFile(); 
        String filename = f.getAbsolutePath();
        skuplja= filename;
        Path path = Paths.get(skuplja); 
        Path pathEnd=null;
        try {
            
          pathEnd =   Files.move(path, path.resolveSibling(stariPut));
        } catch (IOException ex) {
            Logger.getLogger(StudentChange.class.getName()).log(Level.SEVERE, null, ex);
        }
        
        Path file = pathEnd.getFileName();
        String fromFile = skuplja;
        String toFile = put.concat(file.toString());
        
        Path source = Paths.get(fromFile);
        Path target = Paths.get(toFile);
        v= "src/Studenti/" + file.toString();
           
        try{
            Files.copy(source, target, StandardCopyOption.REPLACE_EXISTING);
        }catch(IOException e){
            e.printStackTrace();
        }
    }   
这将对旧图像进行处理,不会将其添加到新目录中,并且会以某种方式破坏它,并在它所在的桌面上以其名称给出数字82。它将有图像照片图标,并会要求寻找一个程序来打开它。我认为null正在做一些事情,但我看不到其他初始化它的方法。有什么建议吗