0

点击收听查看我的流程图片

我的测试骡子应用程序。我使用 curl json 将数据发布到 http-end-point 并将 json 转换为对象并在我的数据库中插入名称（#[message.payload.name]）

我使用 catch-exception-strategy 保留错误日志（案例唯一名称）

我想用字段 #[message.payload.name] 中的条件更新我的表 - 意思是 Update mytable set Firstname = #[message.payload.name]

但在 catch-exception-strategy 中无法访问 #[message.payload.name]

我的配置

  <?xml version="1.0" encoding="UTF-8"?>

<mule xmlns:jdbc-ee="http://www.mulesoft.org/schema/mule/ee/jdbc" xmlns:data-mapper="http://www.mulesoft.org/schema/mule/ee/data-mapper" xmlns:http="http://www.mulesoft.org/schema/mule/http"
    xmlns:json="http://www.mulesoft.org/schema/mule/json"
    xmlns="http://www.mulesoft.org/schema/mule/core" xmlns:doc="http://www.mulesoft.org/schema/mule/documentation" xmlns:spring="http://www.springframework.org/schema/beans" version="EE-3.3.2" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="
http://www.mulesoft.org/schema/mule/json http://www.mulesoft.org/schema/mule/json/current/mule-json.xsd 
http://www.mulesoft.org/schema/mule/http http://www.mulesoft.org/schema/mule/http/current/mule-http.xsd 
http://www.mulesoft.org/schema/mule/ee/jdbc http://www.mulesoft.org/schema/mule/ee/jdbc/current/mule-jdbc-ee.xsd 
http://www.mulesoft.org/schema/mule/ee/data-mapper http://www.mulesoft.org/schema/mule/ee/data-mapper/current/mule-data-mapper.xsd 
http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-current.xsd 
http://www.mulesoft.org/schema/mule/core http://www.mulesoft.org/schema/mule/core/current/mule.xsd ">
    <data-mapper:config name="Mapper" transformationGraphPath="mapper.grf" doc:name="DataMapper"></data-mapper:config>
    <jdbc-ee:mysql-data-source name="MySQL_Data_Source1" url="jdbc:mysql://localhost:3306/mule" user="root" password="1234" transactionIsolation="UNSPECIFIED" doc:name="MySQL Data Source"/>
    <jdbc-ee:connector name="Database" dataSource-ref="MySQL_Data_Source1" validateConnections="true" queryTimeout="-1" pollingFrequency="0" doc:name="Database"/>
    <flow name="httpPostTestFlow1" doc:name="httpPostTestFlow1"> 
        <http:inbound-endpoint exchange-pattern="request-response" host="localhost" port="8081" path="httpPost" doc:name="httpPost"></http:inbound-endpoint>

        <json:json-to-object-transformer doc:name="JSON to Object" returnClass="java.util.Map"></json:json-to-object-transformer>


                <jdbc-ee:outbound-endpoint exchange-pattern="one-way" queryKey="INSERT_TOKEN"  connector-ref="Database" doc:name="Database">
                    <jdbc-ee:query key="INSERT_TOKEN" value="insert into users(FirstName) values(#[message.payload.name]);"/>
                </jdbc-ee:outbound-endpoint>


              <!--  <http:outbound-endpoint exchange-pattern="request-response" host="localhost" port="80" path="post-debug.php" contentType="application/x-www-form-urlencoded" doc:name="post-debug.php"/> --> 


         <catch-exception-strategy doc:name="Catch Exception Strategy">
            <expression-component doc:name="Create error response">#[message.payload = "{\"status\":\"error\", \"message\":\"" + exception.cause.message + "\"}"]</expression-component>
            <jdbc-ee:outbound-endpoint exchange-pattern="one-way"   connector-ref="Database" doc:name="Database" queryTimeout="-1" queryKey="UPDATE_TOKEN">
                    <jdbc-ee:query key="UPDATE_TOKEN" value="update users SET fail_message = #[message.payload]"/>
                </jdbc-ee:outbound-endpoint>
        </catch-exception-strategy>


    </flow>
</mule>
4

1 回答 1

0

根据您希望拥有某种全局变量的标题，一种方法是您可以尝试使用 flowVars 作为快速修复。使用 Mule Studio 它与变量相同（在变形金刚下）。

您在 catch 异常策略中得到的消息就是异常。

于 2013-02-22T06:29:19.330 回答