使用JAVA中的查询参数从链接下载图像

在java代码的帮助下，我正在从link下载图像

public static BufferedImage ImageDownloader(String urlString){
    BufferedImage image = null;
    try {
        URL url = new URL(urlString.replace(" ","%20"));
        URLConnection connection = url.openConnection();
        connection.setRequestProperty("User-Agent", "Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.11 (KHTML, like Gecko) Chrome/23.0.1271.95 Safari/537.11");
        connection.connect();
        InputStream inputStream = connection.getInputStream();
        image = ImageIO.read(inputStream);
    } catch (IOException e) {
        e.printStackTrace();
    }
    return image;
}

上面的代码可以完美地下载图像，但它不能用这样的链接下载图像

我知道我可以删除该查询参数并更新url，但还有比这更好的解决方案吗？

使用

transferFrom（）

---

只是不要设置用户代理：

public static BufferedImage ImageDownloader(String urlString){
    BufferedImage image = null;
    try {
        String cleanUrl = urlString.replace(" ","%20");
        URL url = new URL(cleanUrl);
        URLConnection connection = url.openConnection();
        connection.connect();
        InputStream inputStream = connection.getInputStream();
        image = ImageIO.read(inputStream);
    } catch (IOException e) {
        e.printStackTrace();
    }
    return image;
}

或者：

public static BufferedImage ImageDownloader(String urlString){
    BufferedImage image = null;
    try {
        String cleanUrl = urlString.replace(" ","%20");
        URL url = new URL(cleanUrl);
        image = ImageIO.read(url.openStream());
    } catch (IOException e) {
        e.printStackTrace();
    }
    return image;
}

或者：

public static BufferedImage ImageDownloader(String urlString){
    BufferedImage image = null;
    try {
        URL url = new URL(urlString.replace(" ","%20"));
        image = ImageIO.read(url);
    } catch (IOException e) {
        e.printStackTrace();
    }
    return image;
}

public static BufferedImage ImageDownloader(String urlString){
    BufferedImage image = null;
    try {
        URL url = new URL(urlString.replace(" ","%20"));
        image = ImageIO.read(url);
    } catch (IOException e) {
        e.printStackTrace();
    }
    return image;
}