bat file to copy column data to another column in csv

Question

My csv file contain 6 columns and I want to copy 6th column value inside first column path value.

Path,Firstname,Lastname,Name,Description,Images  
C:\Users\magoo\abc.jpg,alex,ross,pvs,data,123.jpg  
C:\Users\magoo\pqr.jpg,kris,ward,poi,data,784.jpg

Output I am looking for

Path,Firstname,Lastname,Name,Description,Images  
C:\Users\magoo\123.jpg,alex,ross,pvs,data,123.jpg  
C:\Users\magoo\784.jpg,kris,ward,poi,data,784.jpg  

I only want to only replace image name from first path column with 6th column value.
Kindly help as I am stuck here since long.I am using copy commands but it is replacing entire text from column one.

Code I tried:

@ECHO OFF
SETLOCAL 
( FOR /f "tokens=1-6delims=," %%a IN (ImportFile.csv) DO ( 
   ECHO(%%a%%f,%%b,%%c,%%d,%%e,%%f ) 
)>new.csv
GOTO :EOF

Answer 1

I just wrote this and it stripped my comments. .. then I fixed a bug :(

@echo off

Set Input_file=test.csv
Set Output_file=test_modified.csv

for /f "tokens=1-6 delims=," %%l in (%Input_file%) do call :process_tokens "%%l" "%%m" "%%n" "%%o" "%%p" "%%q"
goto :EOF

:process_tokens 
Set Col2=%~2
Set Col3=%~3
Set Col4=%~4
Set Col5=%~5
Set Col6=%~6

Set Col1=%~dp1%Col6%
echo %Col1%,%Col2%,%Col3%,%Col4%,%Col5%,%Col6% >>%Output_file%

goto :EOF

I am using label/functions to accomplish this.

In the ":process_tokens" function, look at the %~dp2%Col6% line.

%~dp1 means:

• %~ = get the passed parameter and strip quotes from it.
• dp = drive and path only (aka strip the filename)
• 1 = passed param #1

%Col6% is where I tack on the filename you are looking for.