Mule 读取超时异常

Question

我有一个 esb，我从中进行 web 服务调用，大多数情况下都可以正常工作，但有时我会遇到以下异常

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异常之后，有时http出站调用仍然成功，有时没有成功

为什么这不一致？

mule http 连接器上的某些配置是否有可能帮助这种异常情况保持一致？

我要问的是...如何在引发读取超时异常后停止处理 http 出站请求？

流程如下所示

<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

您可以在不同的地方配置，从而增加 HTTP 传输的超时：

• Response time out on the endpoints,
• Connection and socket timeouts on the connector。

但这只是进一步推动问题：增加超时可能会暂时解决您的问题，但您仍然面临失败的风险。

为了正确处理它，我认为您应该严格检查每个 HTTP 出站端点之后的响应状态代码，如果状态代码不是您所期望的，则可能使用 filter 来中断流程。

此外，在服务器接收到 HTTP 请求之后、响应返回到 Mule 之前，您很可能会遇到响应超时。在这种情况下，就 Mule 而言，调用失败，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（或者它认为失败）。

Answer 2

检查httpconnection中的服务器SO_TIMEOUT，设置为0

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