所以我尝试使用apache camel访问简单队列服务。
Java DSL方法工作正常,但我尝试使用xml配置。
private AmazonSQS sqs;
sqs = new AmazonSQSClient(credentials);
Region sqsRegion = Region.getRegion(Regions.US_WEST_2);
sqs.setRegion(sqsRegion);
上面的代码工作正常,但我决定构建bean。
<context:property-placeholder location="classpath:/default.properties" />
<bean name="sqsClient" class="com.amazonaws.services.sqs.AmazonSQSClient">
<constructor-arg>
<bean class="com.amazonaws.auth.BasicAWSCredentials">
<constructor-arg value="${access.key}"/>
<constructor-arg value="${secret.key}"/>
</bean>
</constructor-arg>
<property name="region" value="com.amazonaws.regions.Region"/>
</bean>
我收到了错误
无法将[java.lang.String]类型的属性值转换为 属性&#39; region&#39;所需的类型[com.amazonaws.regions.Region]; 嵌套异常是java.lang.IllegalStateException:无法转换 类型[java.lang.String]的值为必需的类型 [com.amazonaws.regions.Region]对于属性&#39; region&#39;:没有匹配 编辑或转换策略发现
我还没有通过Spring xml找到有关配置sqs的任何信息。有时我认为apache camel已经过时或者没有人使用sqs。 此外,下一步是连接可在Java DSL实现中正常工作的Extended SQS库,但我不知道如何通过xml配置队列。
UPD:
感谢@jbird,我用这种方式解决了问题:
<context:property-placeholder location="classpath:/default.properties" />
<bean id="awsRegion" class="org.springframework.beans.factory.config.MethodInvokingFactoryBean">
<property name="targetClass" value="com.amazonaws.regions.RegionUtils"/>
<property name="targetMethod" value="getRegion"/>
<property name="arguments">
<list>
<value>${aws.region}</value>
</list>
</property>
</bean>
<bean name="sqsClient" class="com.amazonaws.services.sqs.AmazonSQSClient">
<constructor-arg>
<bean class="com.amazonaws.auth.BasicAWSCredentials">
<constructor-arg value="${access.key}"/>
<constructor-arg value="${secret.key}"/>
</bean>
</constructor-arg>
<property name="region" ref="awsRegion"/>
</bean>
所以,我刚刚解析了包含aws.key,aws.secret和区域设置的default.properties
文件。
然后我遇到了下一个问题。 Apache camel在加载路由器后停止运行,依此类推。
[ main] SpringCamelContext INFO Route: route1 started and consuming from: Endpoint[aws-sqs://queue?amazonSQSClient=%23sqsClient]
[ main] SpringCamelContext INFO Total 1 routes, of which 1 are started.
[ main] SpringCamelContext INFO Apache Camel 2.17.2 (CamelContext: camel-1) started in 6.105 seconds
Process finished with exit code 0
路由器:
import org.apache.camel.builder.RouteBuilder;
import org.springframework.stereotype.Component;
@Component
public class MyRouteBuilder extends RouteBuilder {
public void configure() {
from("aws-sqs://queue?amazonSQSClient=#sqsClient")
.log("We have a message! ${body}")
.to("file:target/output?fileName=login-message-${date:now:MMDDyy-HHmmss}.json");
}
}
和camel-context.xml
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:context="http://www.springframework.org/schema/context"
xsi:schemaLocation="
http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans.xsd
http://camel.apache.org/schema/spring http://camel.apache.org/schema/spring/camel-spring.xsd
http://www.springframework.org/schema/context http://www.springframework.org/schema/context/spring-context-3.2.xsd
">
<context:property-placeholder location="classpath:/default.properties" />
<bean id="awsRegion" class="org.springframework.beans.factory.config.MethodInvokingFactoryBean">
<property name="targetClass" value="com.amazonaws.regions.RegionUtils"/>
<property name="targetMethod" value="getRegion"/>
<property name="arguments">
<list>
<value>${aws.region}</value>
</list>
</property>
</bean>
<bean name="sqsClient" class="com.amazonaws.services.sqs.AmazonSQSClient">
<constructor-arg>
<bean class="com.amazonaws.auth.BasicAWSCredentials">
<constructor-arg value="${access.key}"/>
<constructor-arg value="${secret.key}"/>
</bean>
</constructor-arg>
<property name="region" ref="awsRegion"/>
</bean>
<!-- enable Spring @Component scan -->
<context:component-scan base-package="com.test.router"/>
<camelContext xmlns="http://camel.apache.org/schema/spring">
<contextScan/>
</camelContext>
</beans>
答案 0 :(得分:1)
region
属性的值是字符串值"com.amazonaws.regions.Region"
。期待类型为com.amazonaws.regions.Region
的对象。因此,您需要引用类型为Region
的对象,而不是提供String值。
答案 1 :(得分:0)
假设这个问题。
第一:
要向对象添加属性,我只需创建新bean
<bean id="awsRegion" class="org.springframework.beans.factory.config.MethodInvokingFactoryBean">
<property name="targetClass" value="com.amazonaws.regions.RegionUtils"/>
<property name="targetMethod" value="getRegion"/>
<property name="arguments">
<list>
<value>${aws.region}</value>
</list>
</property>
</bean>
并创建参考
<property name="region" ref="awsRegion"/>
第二
运行Apache Camel
Main main = new Main();
main.setApplicationContextUri("/META-INF/spring/camel-contex.xml");
main.run(args);
多数民众赞成!