IBM Content Collector FS到P8归档(IBM Content Collector FS to P8 Archiving)

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IBM Content Collector FS到P8归档(IBM Content Collector FS to P8 Archiving)

在我当前使用IBM Content Collector 4.0.1 SP5和IBM Filenet P8 Content Engine 5.2.1的项目中,我需要从文件系统中收集文件并将它们添加到某个P8的对象库中。

报告文件被添加到八个固定文件夹的文件夹中,在它们下面随着时间的推移,结构随着yyyy / mm形式的进一步嵌套文件夹增长:

我从FS to P8 Archiving (Replicate File System and Detect Duplicates).ctms开始FS to P8 Archiving (Replicate File System and Detect Duplicates).ctms任务路由示例,以便在我的对象存储中拥有复制结构。

让我们关注P8 File Document in folder task的P8 File Document in folder task ,特别是File in Folder Options 。

问题是我的存储库中创建的路径还包含驱动器号(例如: E:\Report\AMM_000001_00001\2017\05 ),而我希望只从Report文件夹开始复制文件夹结构。

我怎样才能做到这一点? 我应该使用正则表达式吗?

In my current project using IBM Content Collector 4.0.1 SP5 with IBM Filenet P8 Content Engine 5.2.1 I need to collect files from file system and add them to a certain P8's object store.

Report files are added into folders eight fixed folders and under them the structure grows dinamically over time with further nested folders following the form yyyy/mm:

I started from the FS to P8 Archiving (Replicate File System and Detect Duplicates).ctms task route example in order to have the replicated structure in my object store.

Let's focus on the P8 File Document in folder task and particularly on the File in Folder Options.

The problem is that the path created in my repository contains also the drive letter (e.g.: E:\Report\AMM_000001_00001\2017\05) whereas I would like to have only the folder structure replicated starting from the Report folder.

How can I achieve this? Should I use regular expressions for this?

最满意答案

我设法使用“P8 File Document In Folder”任务中的正则表达式来实现所需的结果,如下所示:

Purpose: Get a folder path without a drive letter. Regular expression: ^[^\\]* Replacement string: $1 Sample text: C:\folder 1\folder 2 Sample result: \folder 1\folder 2

如此处所述。

I managed to achieve the desired result using regular expressions in the "P8 File Document In Folder" task, like this:

Purpose: Get a folder path without a drive letter. Regular expression: ^[^\\]* Replacement string: $1 Sample text: C:\folder 1\folder 2 Sample result: \folder 1\folder 2

as described here.

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