Question

我是mule esb的新手，但我知道其他ESB模式我的问题是我已经做了一个示例，它将数据插入DB.its工作正常，但它没有给客户端任何响应，所以客户端从服务器什么都没有我的示例代码是

<mule xmlns:json="http://www.mulesoft.org/schema/mule/json"
    xmlns:http="http://www.mulesoft.org/schema/mule/http" xmlns:jdbc-ee="http://www.mulesoft.org/schema/mule/ee/jdbc"
    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-3.4.1" 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/ee/jdbc http://www.mulesoft.org/schema/mule/ee/jdbc/current/mule-jdbc-ee.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
http://www.mulesoft.org/schema/mule/http http://www.mulesoft.org/schema/mule/http/current/mule-http.xsd">

<jdbc-ee:postgresql-data-source name="PostgreSQL_Data_Source" user="youtilitydba" password="45782dfff1" url="jdbc:postgresql://localhost:5432/sample" transactionIsolation="UNSPECIFIED" doc:name="PostgreSQL Data Source"/>
    <jdbc-ee:connector name="Database" dataSource-ref="PostgreSQL_Data_Source" validateConnections="true" queryTimeout="-1" pollingFrequency="0" doc:name="Database"/>
    <flow name="insertintoDBFlow1" doc:name="insertintoDBFlow1">
        <http:inbound-endpoint exchange-pattern="request-response" host="localhost" port="8081" doc:name="HTTP" path="httpPost"/>
        <logger message="log about input message: #[payload]" level="INFO" doc:name="Logger"/>

                <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" queryTimeout="-1" connector-ref="Database" doc:name="Database">
            <jdbc-ee:query key="INSERT_TOKEN" value="insert into users(FirstName,lastname) values(#[message.payload.name],#[message.payload.id]);"/>
        </jdbc-ee:outbound-endpoint>
    </flow>

</mule>

我正在使用像这样的样本JSON客户端来调用它

curl -H "Content-Type: application/json" -d '{"name":"kk","id":"anil"}' http://localhost:8081/httpPost

当我正在运行这个json时，它没有给出任何回应，但我希望给出这样的回复{"ResponseJSON":{"Body":{"Datalist":{"Data":"Successfully Rows inserted"}},"Status":"200"}}

我如何格式化以上格式我使用 http响应构建器但是没有任何事情可以这样做..

Answer 1

您需要添加Echo组件，将有效负载设置为您要查找的值，并添加带有status="200"和contentType="application/json"的HTTP响应生成器。

在</jdbc-ee:outbound-endpoint>之后和</flow>

之前添加以下行

<response>
        <http:response-builder status="200" contentType="application/json" doc:name="HTTP Response Builder"/>
    </response>
    <response>
        <set-payload value="{&quot;Body&quot;:{&quot;Datalist&quot;:{&quot;Data&quot;:&quot;Successfully Rows inserted&quot;}},&quot;Status&quot;:&quot;200&quot;}}" doc:name="Set Payload"/>
    </response>
    <echo-component doc:name="Echo"/>

干杯，

-Marco。

如何使用Mule ESB返回客户端的custmise响应

1 个答案: