java.io.IOException: 无效的 Http 响应
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java.io.IOException: Invalid Http response
提问by AlexP
Now before you say there are questions like this I'd like to point out I've looked over most of them without any luck. Also I'm a first timer here so be gentle.
现在，在你说有这样的问题之前，我想指出我已经查看了其中的大部分，但没有任何运气。另外，我是第一次来这里，所以要温柔。
I have this annoyance right now in my current program:
我现在在我当前的程序中有这个烦恼：
Basically this part of my program uses a search engine to find torrent files.
基本上我程序的这部分使用搜索引擎来查找 torrent 文件。
public static ArrayList<String> search(String args) throws IOException {
args = args.replace(":", "");
ArrayList<String> list = new ArrayList<String>();
URL url = new URL("http://pirateproxy.net/search/" + args + "/");
URLConnection con = url.openConnection();
BufferedReader br = new BufferedReader(new InputStreamReader(con.getInputStream())); <---- THIS
}
public static void main(String[] args) {
try {
search("The Hobbit: The Desolation of Smaug");
} catch (IOException e) {
e.printStackTrace();
}
}
THE ERROR:
错误：
java.io.IOException: Invalid Http response
at sun.net.www.protocol.http.HttpURLConnection.getInputStream(Unknown Source)
at service.ServiceDownloader.search(ServiceDownloader.java:20)
at service.ServiceDownloader.main(ServiceDownloader.java:45)
Now the fun part is that it ONLY goes wrong for this movie ("The Hobbit: The Desolation of Smaug"), every other movie works perfectly. I don't understand this. Please do help. (Also I have removed every unnecessary code from the search method)
现在有趣的是，它只在这部电影中出错（“霍比特人：史矛革的荒凉”），其他每一部电影都完美无缺。我不明白这个。请帮忙。（我还从搜索方法中删除了所有不必要的代码）
If I did not put enough information here please ask me for more.
如果我没有在这里提供足够的信息，请向我索取更多信息。
采纳答案by Abimaran Kugathasan
You should URL encode the String The Hobbit: The Desolation of Smaug, since you have special character there. Ex : space.
您应该对 String 进行 URL 编码The Hobbit: The Desolation of Smaug，因为那里有特殊字符。例如：空间。
回答by John Stoner
I suspect it tripped on the colon (:) not the space. Are there other titles with a colon?
我怀疑它在冒号 (:) 而不是空格上绊倒了。还有其他带冒号的标题吗？
回答by svarog
Instead of concatenating strings and needlessly creating interim strings and failing because the url is not encoded, you can use the built in UriBuilderto generate a valid URL
您可以使用内置的UriBuilder来生成有效的 URL，而不是连接字符串和不必要地创建临时字符串并因为 url 未编码而失败
URL path = UriBuilder.fromPath("http://pirateproxy.net")
.path("search")
.path("some movie ")
.build()
.toURL();
// http://pirateproxy.net/search/some%20movie%20
