即使值为空，如何使程序继续运行

Question

我在 java 和 MySQL 中有一个数据库程序。但是问题是：当SQL中IMAGE的数据为空时，程序停止加载图片。它显示 "java.lang.IllegalArgumentException: input == null!"。我的问题是如何将图标图片设置为空并继续。

  try{
        PreparedStatement ps = conn.prepareStatement(President);
        ResultSet rs = ps.executeQuery();
        if(rs.next()){
        rs.absolute(1);
        FirstName1.setText(rs.getString("cFirstname"));
        LastName1.setText(rs.getString("CLastname"));
        MiddleName1.setText(rs.getString("cmiddlename"));
        BufferedImage im1 = ImageIO.read(rs.getBinaryStream("cimage"));
        ImageIcon image1 = new ImageIcon(im1);
        Image img1 = image1.getImage();
        Image newImage = img1.getScaledInstance(144, 144,     Image.SCALE_SMOOTH);
        ImageIcon after = new ImageIcon(newImage);
        btnPic1.setIcon(after);
        btnPic1.setText("");
        FirstName1.setVisible(true);
        LastName1.setVisible(true);
        btnPic1.setVisible(true);
        MiddleName1.setVisible(true);
        }

在我的数据中，此处存储的图像为空。

if(rs.next()){
        rs.absolute(2);
        FirstName2.setText(rs.getString("cFirstname"));
        LastName2.setText(rs.getString("CLastname"));
        MiddleName2.setText(rs.getString("cmiddlename"));
        BufferedImage im2 = ImageIO.read(rs.getBinaryStream("cimage"));
        ImageIcon image2 = new ImageIcon(im2);
        Image img2 = image2.getImage();
        Image newImage2 = img2.getScaledInstance(144, 144, Image.SCALE_SMOOTH);
        ImageIcon after2 = new ImageIcon(newImage2);
        btnPic2.setIcon(after2);
        btnPic2.setText("");
        FirstName2.setVisible(true);
        LastName2.setVisible(true);
        btnPic2.setVisible(true);
        MiddleName2.setVisible(true);
        }

那么我的程序就不会这样做了。

if(rs.next()){
        rs.absolute(3);
        FirstName3.setText(rs.getString("cFirstname"));
        LastName3.setText(rs.getString("CLastname"));
        MiddleName3.setText(rs.getString("cmiddlename"));
        BufferedImage im3 = ImageIO.read(rs.getBinaryStream("cimage"));
        ImageIcon image3 = new ImageIcon(im3);
        Image img3 = image3.getImage();
        Image newImage3 = img3.getScaledInstance(144, 144, Image.SCALE_SMOOTH);
        ImageIcon after3 = new ImageIcon(newImage3);
        btnPic3.setIcon(after3);
        btnPic3.setText("");
        FirstName3.setVisible(true);
        LastName3.setVisible(true);
        btnPic3.setVisible(true);
        MiddleName3.setVisible(true);
        }

Answer 1

哪一行抛出 IllegalArgumentException？也许我过于简单化了，但一个简单的解决方案可能是您在该行之前立即检查空图像变量，如果该变量为空，则不要使用它，并跳过该行。更好的解决方案是找出图像为空的原因并首先防止这种情况发生。

Answer 2

结果代码应该是这样的

if(rs.next()){
        rs.absolute(2);
        FirstName2.setText(rs.getString("cFirstname"));
        LastName2.setText(rs.getString("CLastname"));
        MiddleName2.setText(rs.getString("cmiddlename"));
            if(rs.getBinaryStream("cimage")!=null){
            BufferedImage im2 = ImageIO.read(rs.getBinaryStream("cimage"));
            ImageIcon image2 = new ImageIcon(im2);
            Image img2 = image2.getImage();
            Image newImage2 = img2.getScaledInstance(144, 144, Image.SCALE_SMOOTH);
            ImageIcon after2 = new ImageIcon(newImage2);
        btnPic2.setIcon(after2);
        btnPic2.setText("");}
            else{
            btnPic2.setText("No Picture");     
            }
        FirstName2.setVisible(true);
        LastName2.setVisible(true);
        btnPic2.setVisible(true);
        MiddleName2.setVisible(true);
        }

添加 "if null" 语句。虽然在这个使用。我反转它。感谢您的评论和回答。我已经能够想出这个解决方案。

即使值为空，如何使程序继续运行

How to make the program continues even if the value is null

java

mysql

null

image

imageicon