我无法找到适当的答案,以确定以下错误的含义:
java.net.SocketException: Software caused connection abort: recv failed
笔记:
相关代码:
BufferedReader reader;
try {
URL url = new URL(URI);
reader = new BufferedReader(new InputStreamReader(url.openStream())));
} catch( MalformedURLException e ) {
throw new IOException("Expecting a well-formed URL: " + e);
}//end try: Have a stream
String buffer;
StringBuilder result = new StringBuilder();
while( null != (buffer = reader.readLine()) ) {
result.append(buffer);
}//end while: Got the contents.
reader.close();
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Ada*_*amC 25
这通常意味着存在网络错误,例如TCP超时.我首先在连接上放置一个嗅探器(wireshark),看看你是否能看到任何问题.如果存在TCP错误,您应该能够看到它.此外,如果适用,您可以检查路由器日志.如果无处不在任何地方,那么这就是这类错误的另一个来源.
rus*_*tyx 10
当连接突然关闭时(当发送缓冲区中仍有数据时重置TCP连接时)会发生此错误.这种情况非常类似于更常见的"由同行重置连接".它可能在通过Internet连接时偶尔发生,但如果时间正确则会系统地发生(例如,在localhost上保持活动连接).
HTTP客户端应该只是重新打开连接并重试请求.重要的是要理解当连接处于这种状态时,除了关闭它之外没有其他方法可以解决.任何发送或接收的尝试都会产生相同的错误.
不要使用URL.open(),使用Apache-Commons HttpClient,它具有重试机制,连接池,保持活动和许多其他功能.
样品用法:
HttpClient httpClient = HttpClients.custom()
.setConnectionTimeToLive(20, TimeUnit.SECONDS)
.setMaxConnTotal(400).setMaxConnPerRoute(400)
.setDefaultRequestConfig(RequestConfig.custom()
.setSocketTimeout(30000).setConnectTimeout(5000).build())
.setRetryHandler(new DefaultHttpRequestRetryHandler(5, true))
.build();
// the httpClient should be re-used because it is pooled and thread-safe.
HttpGet request = new HttpGet(uri);
HttpResponse response = httpClient.execute(request);
reader = new BufferedReader(new InputStreamReader(response.getEntity().getContent()));
// handle response ...
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