我正在使用Apache camel和jboss fuse,我已经创建了一个示例路由蓝图,下面列出了
<?xml version="1.0" encoding="UTF-8"?>
<blueprint xmlns="http://www.osgi.org/xmlns/blueprint/v1.0.0"
xmlns:camel="http://camel.apache.org/schema/blueprint"
xmlns:cxf="http://camel.apache.org/schema/blueprint/cxf"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.osgi.org/xmlns/blueprint/v1.0.0 http://www.osgi.org/xmlns/blueprint/v1.0.0/blueprint.xsd http://camel.apache.org/schema/blueprint http://camel.apache.org/schema/blueprint/camel-blueprint.xsd">
<cxf:rsServer address="/testservice" id="testserver" serviceClass="com.company.HelloBean">
<camelContext id="testContext" trace="false" xmlns="http://camel.apache.org/schema/blueprint">
<route id="testRoute" >
<from id="_from1" uri="cxfrs:bean:testserver"/>
<bean beanType="com.company.HelloBean"
id="_bean1" method="hello"/>
</route>
</camelContext>
</blueprint>
和实现它的java类。
@Path("/testservicenew")
public class HelloBean {
@POST
@Path("/test")
@Produces(MediaType.APPLICATION_JSON)
@Consumes(MediaType.APPLICATION_JSON)
public String hello(Person name ) {
return "Hello:"+name.getName();
}
}
但是,当我发送错误的JSON时,它会返回错误的请求,我会使用一些自定义的拦截器,这样我就可以使用自定义的主体和头部来控制返回的响应
https://stackoverflow.com/questions/44379693
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