I want to get the name of the File, which is uploaded into the AARI Form, in order to further use it in my Task Bot.

Hi @MohammedReyanZedra ,

You can try using the AARI Web → Get Storage File in order to get the uploaded file details.

Hi @Padmakumar,
I have done the same to get the file and store it in my local,
But when I want to get the file name, I am unable to retrieve it.
 

When I am extracting $MeetingFile.File:getName$ i am getting error like 
 

Please let me know if you think I can do it in any other way.

Thank You

Hi @MohammedReyanZedra,

were you able to get the file path of the file that you are uploading to the AARI? May be you can follow the link below. Once you get the path then you can do string manipulation and get the desired value.

https://docs.automationanywhere.com/bundle/enterprise-v2019/page/enterprise-cloud/topics/hbc/aari-use-case-file.html

Hope this helps!

Hi @Abhay Naik ,
I tried that, but again we are naming the file here in Get Storage File like this 
 

I want to know the name of the file $FileVar$ which in my case is Barclays_bank_Statement, I want to get this name in my Task Bot
 

Thank you for your response.

Please check the process that you have created it.

Checkin the input value in the process task and use the same variable in message box.

Great @Tamil Arasu10 
Thank you so much,

We have to use the same $inputtFile0]$ but as string INPUT in our TaskBot and use it as file name.

Hi I am dealing with a similar issue, but when I try the above method, I get this as file name whereas the file that I uploaded was named “uae pvt inv 90911081”, how do i get only the filename part

Great @Tamil Arasu10 
Thank you so much,

We have to use the same $inputuFile0]$ but as string INPUT in our TaskBot and use it as file name.

Hi I am dealing with a similar issue, but when I try the above method, I get this as file name whereas the file that I uploaded was named “uae pvt inv 90911081”, how do i get only the filename part

 

You can make use of RegEx to extract the required part from the above string.

Great @Tamil Arasu10 
Thank you so much,

We have to use the same $inputuFile0]$ but as string INPUT in our TaskBot and use it as file name.

Hi I am dealing with a similar issue, but when I try the above method, I get this as file name whereas the file that I uploaded was named “uae pvt inv 90911081”, how do i get only the filename part

 

 

You can make use of RegEx to extract the required part from the above string.

 

Ok i will do that, thank you

Hi @khondoker,

You should be able to use String: Extract text command where the file name would always be located after the keyword ‘?fn=’ and before the file extension, also you can use a find and replace to replace all the ‘+’ with spaces.

Please check the process that you have created it.

 

 

Checkin the input value in the process task and use the same variable in message box.

 

 

 

For me using the Desktop file option with a string variable holding the value of the file location helped.