Question

我有一个esb，我从中进行了一次web服务调用，大部分时间都可以正常工作，但有时我得到以下异常

java.net.SocketTimeoutException: Read timed out
    at java.net.SocketInputStream.socketRead0(Native Method)
    at java.net.SocketInputStream.read(SocketInputStream.java:129)
    at java.io.BufferedInputStream.fill(BufferedInputStream.java:218)
    + 3 more (set debug level logging or '-Dmule.verbose.exceptions=true' for everything)

奇怪的是，在我得到这个例外之后，有时候http出站呼叫仍然成功，有时候没有成功

为什么这不一致？

是否有可能mule http连接器上的某些配置可以帮助此异常情况一致地运行？

我要问的是......如何在抛出读取超时异常后停止处理http出站请求？

流程如下图所示

<queued-asynchronous-processing-strategy name="allow2Threads" maxThreads="2"/>

<flow name="TestFlow" processingStrategy="allow2Threads">
     <vm:inbound-endpoint path="performWebserviceLogic" exchange-pattern="one-way" />

    ....  some transformation logic
    ....
    <http:outbound-endpoint address="http://localhost:8080/firstwebservicecall" responseTimeout="65000" exchange-pattern="request-response"/>
    ....
    ....  some transformation logic on response...
    <http:outbound-endpoint address="http://localhost:8080/secondWeberviceCall" responseTimeout="20000" exchange-pattern="request-response"/>
    ......some transformation logic on response...

    <catch-exception-strategy>
        <choice>
            <when expression="#[groovy:message.getExceptionPayload().getRootException.getMessage().equals('Read timed out') and message.getSessionProperty('typeOfCall').equals('firstWeberviceCall')]">
                    .... unreliable ...result... as firstWeberviceCall may succeed even after the control comes here
                    and if we process http://localhost:8080/firstwebservicecall .. the transaction takes place twice... as already it succeeded above even after an exception is thrown
            </when>
            <when expression="#[groovy:message.getExceptionPayload().getRootException.getMessage().equals('Read timed out') and message.getSessionProperty('typeOfCall').equals('secondWeberviceCall')]">
                    ..... reliable ... if control comes here and if we process http://localhost:8080/secondWeberviceCall .. the transaction takes place only once
            </when>
            <when expression="#[groovy:message.getExceptionPayload().getRootException.getMessage().equals('Connect timed out') and message.getSessionProperty('typeOfCall').equals('firstWeberviceCall')]">
                ....reliable
            </when>
            <when expression="#[groovy:message.getExceptionPayload().getRootException.getMessage().equals('Connect timed out') and message.getSessionProperty('typeOfCall').equals('secondWeberviceCall')]">
                ....reliable
            </when>
        </choice>
    </catch-exception-strategy>   
</flow>

Answer 1

您可以配置，从而增加HTTP传输在不同位置的超时：

这只会进一步解决问题：增加超时可能会暂时解决您的问题，但您仍然会遇到失败。

为了正确处理它，我认为你应该在每个HTTP出站端点之后严格检查响应状态代码，如果状态代码不符合预期，可以使用filter来打破流程。

此外，在服务器收到HTTP请求之后以及响应返回Mule之前，您可以获得响应超时。在这种情况下，就Mule而言，呼叫失败并且必须重试。这意味着远程服务必须是幂等的，即客户端应该能够安全地重试任何失败的操作（或者它认为已经失败）。

Answer 2

检查httpconnection中的服务器SO_TIMEOUT，将其设置为0

检查 - https://www.mulesoft.org/jira/browse/MULE-6331

Mule Read超时异常

2 个答案: