move to guid directory - VisualCron

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andyt
Free support Topic Starter

2009-04-16T08:10:43Z

Hi Guys,

I have a process which is watching for files of a certain type in a directory (XML, DOC and XLS). I then need to move these files into another directory. All of this I can do, however I need the directory I create to use a guid as its name.

Is there an easy way to do this?

Thanks,
Andy

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Support
Official support

2009-04-16T10:56:55Z

The source or destination directory?

Henrik
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andyt
Free support Topic Starter

2009-04-16T11:05:15Z

destination directory.

Support
Official support

2009-04-16T11:28:11Z

You need to find the Variable in the Variables browser.

VisualCron Variables->Jobs->[your job name]->Tasks->[Your task name]->Items->[Your copy file item]->Destination folder

Support attached the following image(s):

Henrik
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andyt
Free support Topic Starter

2009-04-17T07:46:45Z

Thanks! That's helpful, but I don't see the specific Item variables for my job. Am I setting up the job incorrectly?

This is what i'm trying to do:

1. Monitor a Folder
2. Once an .XML file is created in this folder, a subfolder should be created.
3. The Subfolder should be named with a GUID
4. The XML should then be moved into the subfolder

Thanks for your help!

andyt attached the following image(s):

Support
Official support

2009-04-17T07:56:44Z

The items property is part of the Copy files Task. So you need to look among the Variables for that Task.

Where you looking for Trigger Variables?

Henrik
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andyt
Free support Topic Starter

2009-04-19T06:06:07Z

Ok, thanks. I have now found the variable using the Copy Files task. However, what is the next step? How do I get Visual Cron to create a folder using the GUID name? Apologies, I'm new to Visualcron. Thanks.

Support
Official support

2009-04-19T08:40:20Z

What GUID do you want to use? Job guid, Task guid, random guid or other guid?

Henrik
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andyt
Free support Topic Starter

2009-04-19T08:41:53Z

Random GUID, the folder name needs to be unique each time.

Support
Official support

2009-04-19T08:49:21Z

#10

We don't have that Variable yet, but can create it for next version. You can always use the random file name Variable:

{FILE(GetRandomFileName)}

Henrik
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andyt
Free support Topic Starter

2009-04-19T08:56:40Z

#11

That's too bad. The applicatin I'm intergrating into requires a GUID folder name. Any other suggestions on how to get this done with the current version? Thanks.

Support
Official support

2009-04-19T08:58:13Z

#12

One way would be to create an application that creates a guid and use the output from that application. Not effective but it works. However, we have already added this functionality and will release that version today.

Henrik
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andyt
Free support Topic Starter

2009-04-19T09:08:05Z

#13

That's great news. Can you confirm what version will have the Random GUID variable feature? Will I be able to downloaded it soon? Thanks!

Support
Official support

2009-04-19T09:11:11Z

#14

Random GUID will be available in function {STRING(GUID)}. I don't know when but hopefully within some hours.

Henrik
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andyt
Free support Topic Starter

2009-04-19T09:19:07Z

#15

That's great news! I'll wait and download the latest release and see if it fixes the problem. Thanks very much for your help.

Support
Official support

2009-04-19T13:31:24Z

#16

You can now find this fix in version 5.1.1. Look here: http://www.visualcron.co...m.aspx?g=posts&t=531

Henrik
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andyt
Free support Topic Starter

2009-04-20T02:53:39Z

#17

Thanks Henrik! Works great!

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