Extract File Name from File Directory

[SH_94]

Dear Community,

May i know how can i extract the file name from the file directory as below:

Directory examples

1.C:\Users\XXXX\Desktop\Alteryx\ABC\UDJD Input.csv

2.C:\Users\XXXX\Desktop\Alteryx\ABC\URKELC\XJX1828.csv

3.C:\Users\XXXX\Desktop\Alteryx\ABC\5324.csv

4.C:\Users\XXXX\Desktop\Alteryx\BD\92.csv

5.C:\Users\XXXX\Desktop\BD33\1nbfi.csv

Results wanted

1.UDJD Input

2. XJX1828

3.5324

4.92

5.1nbfi

Many thanks for your help

[fmvizcaino]

Hey @SH_94 ,

We can use regex to get that filename.

If you want to learn more about regex: https://community.alteryx.com/t5/Alteryx-Designer-Knowledge-Base/Tool-Mastery-RegEx/ta-p/37689

Best,

Fernando Vizcaino

[HomesickSurfer · Accepted solution]

Hi @SH_94 

Either of two simple formulas:

FileGetFileName([FullPath])

FileGetFileName([FileName])

[atcodedog05 · Accepted solution]

Hi @HomesickSurfer 

Mindblown 😮 where did you come across this. This works amazingly for this scenario.

Workflow:

Thank you for this knowledge share 🙂 @HomesickSurfer 

[HomesickSurfer]

Thank you @atcodedog05 

I make good use of the various 'GET' functions.  They're efficient.

@SH_94 , you may wish to additionally filter to exclude backups, temp files, etc as follows, or config your file specification in the Directory tool to only include filetypes you seek.

FileGetExt([FullPath])!=".bak" OR "~$"

[atcodedog05]

Definitely bookmarking this post. So much knowledgable 😀

Thank you @HomesickSurfer