How to get File Name in File Copy Task - VisualCron - Forum

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ishelp
2015-08-18T19:42:54Z
I need to get the File name in the File Copy task in order to rename it. I thought there would be a variable for this but can't seem to find anything. Maybe I have been staring at this too long. The files coming in to our system have a date and time in the name and that needs to be stripped off to go to it's destination.

Thanks,

Bill Huber
ErikC
2015-08-19T08:52:46Z
Hi Bill, welcome to the forum!

You could use the option 'Post process mask with Variable - use {NEWNAME()} in Variable'.

I do not know how you source filename is, but lets say it's: YYYYMMDDblabla.txt
So the 1st 8 characters are the date.

You need to trip off the 1st 8, so use the substring funtion:
Use this: {STRING(Substring|{NEWNAME()}|8|???)}

We don't know the length of the remaining part, so we need to get the length of the filename and subtract 8 from it.
Use this: {MATH(Add|Integer|{STRING(Length|{NEWNAME()})}|-8|0)}

Now you can combine both by replacing the ??? and you get:
{STRING(Substring||{NEWNAME()}||8|{MATH(Add|Integer|{STRING(Length||{NEWNAME()}|)}|-8|0)})}

This will copy the file to a new filename named: blabla.txt
It might be different in your situation, but this is a way to go.

Regards
Erik
Uses Visualcron since 2006.
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