我有并行读取员工、地址和联系人文件并将其转换为 beanIO 对象并合并 beanIO 对象以生成完整的 employeeDetails 对象的用例。
Emp 文件:
1 Foo Engineer
2 Bar AssistantEngineer
员工联系文件:
1 8912345678 foo@org.com
2 7812345678 bar@org.com
员工地址文件:
1 city1 1234
2 city2 2345
Exchange 中 EmployeeDetailsBeanIODataFormat 对象的预期输出:
1 Foo Engineer foo@org.com city1 1234
2 Bar AssistantEngineer bar@org.com city2 2345
我有以下路线
from("file://C:/cameltest/employee.txt").to("seda:beanIO");
from("file://C:/cameltest/employeeContact.txt").to("seda:beanIOContact");
from("file://C:/cameltest/employeeAddress.txt").to("seda:beanIOAddress");
每个文件都转换为beanio对象
BeanIODataFormat empFormat = new BeanIODataFormat("beanIO.xml","emp");
BeanIODataFormat empContactFormat = new BeanIODataFormat("beanIO.xml", "empContact");
BeanIODataFormat empAddressFormat = new BeanIODataFormat("beanIO.xml", "empAddress");
from("seda:beanIO").unmarshal(empFormat).log("body - ${body}");
from("seda:beanIOContact").unmarshal(empContactFormat).log("Contact body ${body}");
from("seda:beanIO").unmarshal(empAddressFormat).log("Address body - ${body}");
输出正确记录 Bean 对象。
现在我需要合并对象以形成 EmployeeDetails 对象。有人可以让我知道该怎么做吗?我已阅读过,似乎可以使用聚合器来完成这项工作,但不确定该方法。
任何关于此示例的想法都会有所帮助。 欢迎提出建议,是否建议首先根据员工 ID 合并文件并从中创建一个对象?在这种情况下,我不想将合并的文件写入磁盘,因为 IO 会降低性能。
提前致谢。
最佳答案
在解码后使用拆分器拆分每条消息
from("seda:beanIO").unmarshal(empFormat).split(body()).to("seda:aggregate");
from("seda:beanIOContact").unmarshal(empContactFormat).split(body()).to("seda:aggregate");
from("seda:beanIOAddress").unmarshal(empAddressFormat).split(body()).to("seda:aggregate");
下面是聚合器的样子。 详细信息对象作为 header 存储在 olddExchange 中。 最重要的参数如下
- correlationExpression: simple("${body.id}") 关联具有相同 id(1 或 2)的所有消息
- 完成大小=3。每个文件一个。
from("seda:aggregate").aggregate(simple("${body.id}"), (oldExchange,newExchange) -> {
if (oldExchange == null) {
EmployeeDetails details = buildDetails(new EmployeeDetails(), newExchange);
newExchange.getIn().setHeader("details", details);
return newExchange;
}
EmployeeDetails details = oldExchange.getIn().getHeader("details", EmployeeDetails.class);
buildDetails(details, newExchange);
oldExchange.getIn().setHeader("details", details);
return oldExchange;
}).completionSize(3).log("Details - ${header.details}")
和
private EmployeeDetails buildDetails(EmployeeDetails details, Exchange newExchange) {
Object newBody = newExchange.getIn().getBody();
if (newBody instanceof Employee) {
details.setId(((Employee) newBody).getId());
details.setName(((Employee) newBody).getName());
details.setJob(((Employee) newBody).getJob());
} else if (newBody instanceof EmployeeContact) {
details.setEmail(((EmployeeContact) newBody).getEmail());
} else if (newBody instanceof EmployeeAddress) {
details.setCity(((EmployeeAddress) newBody).getCity());
details.setCode(((EmployeeAddress) newBody).getCode());
}
return details;
}
那么结果将是 2 个细节对象
Details - EmployeeDetails(id=1, name=Foo, job=Engineer, email=foo@org.com, city=city1, code=1234)
Details - EmployeeDetails(id=2, name=Bar, job=AssistantEnginee, email=bar@org.com, city=city2, code=2345)
关于java - Apache Camel : File to BeanIO and merge beanIO objects based on id,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46860535/