Determining the name of a single file
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- Posts: 10
- Joined: Tue Dec 09, 2014 10:28 am
I am trying to query to find out the file name that has been unzipped. We do it now from a file list, but I would like to query a directory that I know only has one file in it, for the file name.
In a list, we use the function Set Variable and assign the variable the results.
variable = ${ScanFile:name}
Where ScanFile is the current item variable.
I can only see how that is done in a LIST.
How can I assign a variable a file name if it is not being processed through a list?
Thank you,
B Smith
In a list, we use the function Set Variable and assign the variable the results.
variable = ${ScanFile:name}
Where ScanFile is the current item variable.
I can only see how that is done in a LIST.
How can I assign a variable a file name if it is not being processed through a list?
Thank you,
B Smith
- Support Specialist
- Posts: 590
- Joined: Tue Jul 17, 2012 2:12 pm
- Location: Phoenix, AZ
-
Barb,
You can get access to file information outside of a ForEach list (like you're indicating) by utilizing the "FileInfo" function.
Go into the Wizard for the assignment of a variable, then in the middle column, select "FileInfo" then utilize that to retrieve any information about a file provided as a "string" variable a input to the function.
This should give you what you're looking for.
You can get access to file information outside of a ForEach list (like you're indicating) by utilizing the "FileInfo" function.
Go into the Wizard for the assignment of a variable, then in the middle column, select "FileInfo" then utilize that to retrieve any information about a file provided as a "string" variable a input to the function.
This should give you what you're looking for.
Rick Elliott
Lead Solutions Consultant
(402) 944.4242
(800) 949-4696
Lead Solutions Consultant
(402) 944.4242
(800) 949-4696
- Posts: 10
- Joined: Tue Dec 09, 2014 10:28 am
Thank you Rick. I will try this.
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