Search the File and Send to corresponding Email

Help Studio
1

Hello,

I am creating some .csv files of Persons name in Folder ‘Buyer’
EmailFiles

e.g
Rahul Buyer.csv
Mehul Buyer.csv
Pooja Buyer.csv
.
.
.
etc
I have another file Buyers Email.xlsx
2SendEmail

I want to send an Email to That Person’s Email ID associate with his name with his file names Attachment File
e.g

File Name : Mehul Buyer.csv
To : Mehul@gmail.com
Attachment : Mehul Buyer.csv

File Name : Rahul Buyer.csv
To : Rahul@gmail.com
Attachment : Rahul Buyer.csv

Any Idea or sample workflow would be appreciated.

Thanks

2

Too lazy to actually build the example, but this is your solution:

  • read your excel with the adresses and store this in a datatable (DT_EmailAddress)
  • loop through your input folder: for each ‘FileName’ in Directory.GetFiles(InputFolder)
  • while in the loop: extract the name from the filename: split(FileName, " "c).First → Recipient
  • do a lookupdatatable with your DT_EmailAddress, use the lookupvalue Recipient and return the email adress → RecipientEmail
  • send a mail using the FileName as attachment, Subject, and the RecipientEmail as the ‘To:’

And you’re done.

1 Like
3

Thanks for the advice to a beginner in the Uipath.

One more question is I need to take some action if any File is present in the Folder and It’s email ID is not listed in the Email Excel File.

Thanks

4

That’s not a qestion :slight_smile:

But yes I suppose you need to build in your validations.
Based on your ‘remark’, I’d suggest that after extracting the name from the filename, and doing the lookup in excel, you validate if the email extraction yielded a result. So if RecipientEmail = “” then throw an error, or whichever your cleanup action is.

5

Folder Path : “C:\RPA\Dev\Test\Buyer Files”

using split(FileName, " "c).First

O/p is : C:\RPA\Dev

Need the names of the Files.

Thanks :slight_smile:

6

You read it wrongly…

directory.getfiles(“C:\RPA\Dev\Test\Buyer Files”)

this reads all the files into an array of string.
With the for each filename in this array gives your each individual filepath of all files in that directory.
image

Edit: small mistake in this screenshot: split().last should be split().first