如何在Apache骆驼路线中提及动态目录路径 [英] How to mention dynamic directory path in Apache camel route

问题描述

我正在尝试从不同目录中获取文件.例如我有以下目录结构

I am trying to take files from different directories. For example I have following directory Structure

vendors/dir1/files/heelo.txt
vendors/dir2/files/hello2.txt

这里有3个目录:

1.供应商
2.dir1和dir2
3.文件

1.vendors
2.dir1 and dir2
3.files

由于2.dir1和dir2不同，所以我得动态取.

since 2. dir1 and dir2 is different, so I have to take it dynamically.

我编写了以下代码:

<routes xmlns="http://camel.apache.org/schema/spring"> 
<route id="com.performancebikes.Inventory1" autoStartup="false"> 
<from uri="b2bmbFileSystem://com.a/vendors/${file:name}/files"/>
<to uri="b2bmbMailBox://com.b/Files"/>
</route> 
</routes>

由于目录 ${file:name} 不起作用，请帮我解决这个问题

since it is directory ${file:name} is not working, please help me to resolve this

推荐答案

如果您想使用 vendors 下的每个文件，您可以递归地使用文件强>:

<from uri="b2bmbFileSystem://com.a/vendors/?recursive=true"/>
<to uri="b2bmbMailBox://com.b/Files"/>

如果消耗了vendors/dir2/files/hello2.txt文件，输出的文件会保存在com.b/Files/dir2/files/hello2.txt下code>，因此它重新创建与源文件系统中相同的相对路径.

If it consumes the file vendors/dir2/files/hello2.txt, the output file will be saved under com.b/Files/dir2/files/hello2.txt, so it recreates the same relative path as in the source file system.

如果不想重新创建相同的结构，可以将输出结构展平:

If you don't want to recreate the same structure, you can flatten the output structure:

<from uri="b2bmbFileSystem://com.a/vendors/?recursive=true"/>
<to uri="b2bmbMailBox://com.b/Files?flatten=true"/>

这会带来相同文件名出现在多个子目录中的风险，因此您在目标文件夹中出现冲突.

This comes with the risk that the same filename appears in multiple subdirectories and therefore you got a conflict in the target folder.

如果您想仅从两个特定目录中使用，您可以简单地创建两个路由:

If you want to consume only from the two specific directories you can simply create two routes:

<from uri="b2bmbFileSystem://com.a/vendors/dir1/files/"/>
<to uri="b2bmbMailBox://com.b/Files"/>

<from uri="b2bmbFileSystem://com.a/vendors/dir2/files/"/>
<to uri="b2bmbMailBox://com.b/Files"/>

只要路由不包含也被乘法的处理逻辑，有多个是没有问题的.

As long as the routes do not contain processing logic that is also multiplied, it is no problem to have multiple of them.

即使你有处理逻辑，你也可以像上面那样编写简单的文件收集器路由"，然后构建一个使用所有文件收集目录的路由，并在该路由中实现逻辑.

And even if you got processing logic, you could write simple "file collector routes" as the ones above and then build a route that consumes the directory where all files are collected and implement the logic in that route.

如果您想从大量特定目录中使用，您可以在您的应用程序中注入一个路由配置列表.例如，YAML 格式的路由配置可能如下所示:

If you want to consume from lots of specific directories, you could inject a List of route-configurations in your application. A route config in YAML format could for example look like this:

fileConsumer:
  routes:
    - routeId: "consumer1"
      source: "/path/to/source/directory"
      target: "/path/to/target/directory"
    - routeId: "consumer2"
      source: "/path/to/other/source/directory"
      target: "/path/to/other/target/directory"

如果您将其作为 List 注入，您可以在 RouteBuilder 类中迭代它以创建所有配置的路由:

If you inject this as List<RouteConfiguration> you can iterate over it in a RouteBuilder class to create all configured routes:

@Override
public void configure() {
    configuration.getRoutes().forEach(this::addRouteToContext);
}

private void addRouteToContext(final RouteConfiguration routeConfiguration) throws Exception {
    String fileReaderSourceUri = [build complete endpoint URI from directory];
    String fileReaderTargetUri = [build complete endpoint URI from directory];

    this.camelContext.addRoutes(new RouteBuilder() {
        @Override
        public void configure() throws Exception {
            from(fileReaderSourceUri)
            .routeId(routeConfiguration.getRouteId())
            .to(fileReaderTargetUri);
        }
    }
}

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