Question

我正在尝试确定网址是好还是坏。我无法通过HttpURLConnection超出主机为空的异常。

我的代码如下：

URL url;
UrlValidator validate = new UrlValidator();
if(validate.isValid(rs.getString("url"))){
    url = new URL(rs.getString("url"));
    System.out.println(url.getHost().length());
    if(url.getHost().length() == 0){//testing purposes
        System.out.println("BAD"); //testing purposes
        System.exit(0);//testing purposes
    }
    HttpURLConnection connection = (HttpURLConnection)  url.openConnection();
    if(connection != null){
        int rc = connection.getResponseCode();
        System.out.println(rc);
        /*
        connection.setRequestMethod("HEAD");
        connection.connect();
        String contentType = connection.getContentType();
        //if(contentType.contains("audio") || contentType.contains("mp3") || contentType.contains("mpeg")){
            System.out.println(contentType+" == "+rs.getString("url"));
        //}
    }

我继续得到这个例外

Exception in thread "main" java.lang.RuntimeException: java.lang.IllegalArgumentException: protocol = http host = null
at sun.net.www.protocol.http.HttpURLConnection.getInputStream(HttpURLConnection.java:1241)
at sun.net.www.protocol.http.HttpURLConnection.getHeaderField(HttpURLConnection.java:2696)
at java.net.HttpURLConnection.getResponseCode(HttpURLConnection.java:477)

我不介意例外，但我希望能够忽略它而且我不能。

这是我正在使用的导致错误的确切代码。

                    url = new URL("http://www.hulkshare.com/dl/epsbxff31peb/waka_flocka_flame_ft_drake-round_of_applause_%28remix%29.mp3");
                if(url.getHost() != null){
                    System.out.println("Good Url");
                    //System.exit(0);

                    HttpURLConnection connection = (HttpURLConnection)  url.openConnection();
                    if(connection != null){
                    int rc = connection.getResponseCode();
                    System.out.println(rc);

                    connection.setRequestMethod("HEAD");
                    connection.connect();
                    String contentType = connection.getContentType();
                        System.out.println(contentType);

                    }
                }

也许有人可以在他们的日食上测试它，看看他们是否可以复制错误，或者它只是我。

更新

这是我尝试在上述异常中返回false的方法。

    public static boolean isValidURL(){
    URL url;
    try {
        url = new URL("http://www.hulkshare.com/dl/epsbxff31peb/waka_flocka_flame_ft_drake-round_of_applause_%28remix%29.mp3");

        //System.out.println(url.getHost());
        if(url.getHost().length() == 0 || url.getHost() == null){
            System.out.println("BAD");
            //System.exit(0);
            return false;
        }else{
            HttpURLConnection connection = (HttpURLConnection)  url.openConnection();
            if(connection != null){
            int rc = connection.getResponseCode();
            System.out.println(rc);

            connection.setRequestMethod("HEAD");
            connection.connect();
            String contentType = connection.getContentType();
            //if(contentType.contains("audio") || contentType.contains("mp3") || contentType.contains("mpeg")){
                System.out.println(contentType);
            //}
            }
            return true;
        }
    } catch (IllegalArgumentException | IOException e) {
        //e.printStackTrace();
        return false;
    }
}

该方法继续抛出异常。我被困在这里，无法超越这个例外。

Answer 1

您的网址有问题。如果我使用http://www.example.com/而不是hulkshare网址运行您的代码，则会超过您坚持使用的那一行。（在您已经收到HTTP响应代码后，尝试将方法设置为HEAD时，最终会因为不同的原因而失败几行，而您无法做到这一点。）

如果我尝试在命令行上获取该文件，它也会失败：

$ wget "http://www.hulkshare.com/dl/epsbxff31peb/waka_flocka_flame_ft_drake-round_of_applause_%28remix%29.mp3"
--2015-01-21 11:05:11--  http://www.hulkshare.com/dl/epsbxff31peb/waka_flocka_flame_ft_drake-round_of_applause_%28remix%29.mp3
Resolving www.hulkshare.com... 109.201.151.6, 109.201.151.5, 109.201.151.3, ...
Connecting to www.hulkshare.com|109.201.151.6|:80... connected.
HTTP request sent, awaiting response... 302 Found
Location: //?force=1 [following]
http://?force=1: Invalid host name.

更新

我现在通过您的评论和问题编辑了解到，您真正想要做的就是捕获Exception并返回false。我怀疑你根本没有抓住RuntimeException被抛出的东西。只需在catch子句中添加更广泛的网络：

} catch (RuntimeException | IOException e) {
    return false;
}

无法通过host = null获取java异常

1 个答案:

更新