Remove columns from a CSV using CMD/Batch












0















I currently have a CSV I download from a vendor which has data in it I do not require before importing it into our system.



I need to removed some columns from the CSV using a batch file as we are on citrix we have not got powershell as an option.



There are no column headers and below is what the CSV looks like:



"9826XXXXXXXXXX217","60007834        ","    10.00","D","22/11/2018"
"9826XXXXXXXXXX324","60008504 "," 12.00","D","22/11/2018"
"9826XXXXXXXXXX414","60010605 "," 12.00","D","22/11/2018"
"9826XXXXXXXXXX013","60011385 "," 12.00","D","22/11/2018"


I need the output to be 2nd, 3rd, 5th column.



"60007834        ","    10.00","22/11/2018"
"60008504 "," 12.00","22/11/2018"
"60010605 "," 12.00","22/11/2018"
"60011385 "," 12.00","22/11/2018"









share|improve this question


















  • 1





    for more complex csv manipulation with batch files you can take a look at this

    – npocmaka
    Nov 23 '18 at 9:52











  • What have you tried, where are you stuck? Please share your efforts!

    – aschipfl
    Nov 23 '18 at 10:15













  • @Echo off For %%F in (.txt) do ( (for /f "delims=" %%A in (%%F) do Call :Split %%A ) > "C:Run Folder%%~nF_New.txt" ) goto :Eof :Split if "%~2"=="" goto :eof Echo(%2,%3,%5 move ".txt" "C:Archive" >nul The above is what I ended with which takes columns 2,3,5 and pushes it to an output file which mirrors the old file name with "_New" and it removes the blank line at the end of the file.

    – ryall579
    Nov 28 '18 at 14:40


















0















I currently have a CSV I download from a vendor which has data in it I do not require before importing it into our system.



I need to removed some columns from the CSV using a batch file as we are on citrix we have not got powershell as an option.



There are no column headers and below is what the CSV looks like:



"9826XXXXXXXXXX217","60007834        ","    10.00","D","22/11/2018"
"9826XXXXXXXXXX324","60008504 "," 12.00","D","22/11/2018"
"9826XXXXXXXXXX414","60010605 "," 12.00","D","22/11/2018"
"9826XXXXXXXXXX013","60011385 "," 12.00","D","22/11/2018"


I need the output to be 2nd, 3rd, 5th column.



"60007834        ","    10.00","22/11/2018"
"60008504 "," 12.00","22/11/2018"
"60010605 "," 12.00","22/11/2018"
"60011385 "," 12.00","22/11/2018"









share|improve this question


















  • 1





    for more complex csv manipulation with batch files you can take a look at this

    – npocmaka
    Nov 23 '18 at 9:52











  • What have you tried, where are you stuck? Please share your efforts!

    – aschipfl
    Nov 23 '18 at 10:15













  • @Echo off For %%F in (.txt) do ( (for /f "delims=" %%A in (%%F) do Call :Split %%A ) > "C:Run Folder%%~nF_New.txt" ) goto :Eof :Split if "%~2"=="" goto :eof Echo(%2,%3,%5 move ".txt" "C:Archive" >nul The above is what I ended with which takes columns 2,3,5 and pushes it to an output file which mirrors the old file name with "_New" and it removes the blank line at the end of the file.

    – ryall579
    Nov 28 '18 at 14:40
















0












0








0








I currently have a CSV I download from a vendor which has data in it I do not require before importing it into our system.



I need to removed some columns from the CSV using a batch file as we are on citrix we have not got powershell as an option.



There are no column headers and below is what the CSV looks like:



"9826XXXXXXXXXX217","60007834        ","    10.00","D","22/11/2018"
"9826XXXXXXXXXX324","60008504 "," 12.00","D","22/11/2018"
"9826XXXXXXXXXX414","60010605 "," 12.00","D","22/11/2018"
"9826XXXXXXXXXX013","60011385 "," 12.00","D","22/11/2018"


I need the output to be 2nd, 3rd, 5th column.



"60007834        ","    10.00","22/11/2018"
"60008504 "," 12.00","22/11/2018"
"60010605 "," 12.00","22/11/2018"
"60011385 "," 12.00","22/11/2018"









share|improve this question














I currently have a CSV I download from a vendor which has data in it I do not require before importing it into our system.



I need to removed some columns from the CSV using a batch file as we are on citrix we have not got powershell as an option.



There are no column headers and below is what the CSV looks like:



"9826XXXXXXXXXX217","60007834        ","    10.00","D","22/11/2018"
"9826XXXXXXXXXX324","60008504 "," 12.00","D","22/11/2018"
"9826XXXXXXXXXX414","60010605 "," 12.00","D","22/11/2018"
"9826XXXXXXXXXX013","60011385 "," 12.00","D","22/11/2018"


I need the output to be 2nd, 3rd, 5th column.



"60007834        ","    10.00","22/11/2018"
"60008504 "," 12.00","22/11/2018"
"60010605 "," 12.00","22/11/2018"
"60011385 "," 12.00","22/11/2018"






csv batch-file cmd






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 23 '18 at 9:37









ryall579ryall579

205




205








  • 1





    for more complex csv manipulation with batch files you can take a look at this

    – npocmaka
    Nov 23 '18 at 9:52











  • What have you tried, where are you stuck? Please share your efforts!

    – aschipfl
    Nov 23 '18 at 10:15













  • @Echo off For %%F in (.txt) do ( (for /f "delims=" %%A in (%%F) do Call :Split %%A ) > "C:Run Folder%%~nF_New.txt" ) goto :Eof :Split if "%~2"=="" goto :eof Echo(%2,%3,%5 move ".txt" "C:Archive" >nul The above is what I ended with which takes columns 2,3,5 and pushes it to an output file which mirrors the old file name with "_New" and it removes the blank line at the end of the file.

    – ryall579
    Nov 28 '18 at 14:40
















  • 1





    for more complex csv manipulation with batch files you can take a look at this

    – npocmaka
    Nov 23 '18 at 9:52











  • What have you tried, where are you stuck? Please share your efforts!

    – aschipfl
    Nov 23 '18 at 10:15













  • @Echo off For %%F in (.txt) do ( (for /f "delims=" %%A in (%%F) do Call :Split %%A ) > "C:Run Folder%%~nF_New.txt" ) goto :Eof :Split if "%~2"=="" goto :eof Echo(%2,%3,%5 move ".txt" "C:Archive" >nul The above is what I ended with which takes columns 2,3,5 and pushes it to an output file which mirrors the old file name with "_New" and it removes the blank line at the end of the file.

    – ryall579
    Nov 28 '18 at 14:40










1




1





for more complex csv manipulation with batch files you can take a look at this

– npocmaka
Nov 23 '18 at 9:52





for more complex csv manipulation with batch files you can take a look at this

– npocmaka
Nov 23 '18 at 9:52













What have you tried, where are you stuck? Please share your efforts!

– aschipfl
Nov 23 '18 at 10:15







What have you tried, where are you stuck? Please share your efforts!

– aschipfl
Nov 23 '18 at 10:15















@Echo off For %%F in (.txt) do ( (for /f "delims=" %%A in (%%F) do Call :Split %%A ) > "C:Run Folder%%~nF_New.txt" ) goto :Eof :Split if "%~2"=="" goto :eof Echo(%2,%3,%5 move ".txt" "C:Archive" >nul The above is what I ended with which takes columns 2,3,5 and pushes it to an output file which mirrors the old file name with "_New" and it removes the blank line at the end of the file.

– ryall579
Nov 28 '18 at 14:40







@Echo off For %%F in (.txt) do ( (for /f "delims=" %%A in (%%F) do Call :Split %%A ) > "C:Run Folder%%~nF_New.txt" ) goto :Eof :Split if "%~2"=="" goto :eof Echo(%2,%3,%5 move ".txt" "C:Archive" >nul The above is what I ended with which takes columns 2,3,5 and pushes it to an output file which mirrors the old file name with "_New" and it removes the blank line at the end of the file.

– ryall579
Nov 28 '18 at 14:40














2 Answers
2






active

oldest

votes


















1














In case there are possibly commas inside the fields,

use a call to a sub passing the quoted arguments:



:: Q:Test20181123SO_53444017.cmd
@Echo off
(for /f "delims=" %%A in (test.csv) do Call :Split %%A
) > test_New.csv
goto :Eof
:Split
Echo(%2,%3,%5




> type test_New.csv
"60007834 "," 10.00","22/11/2018"
"60008504 "," 12.00","22/11/2018"
"60010605 "," 12.00","22/11/2018"
"60011385 "," 12.00","22/11/2018"


Edit: Variant processing all *.csv files (appending _New to the name)



:: Q:Test20181123SO_53444017.cmd
@Echo off
For %%F in (*.csv) do (
(for /f "delims=" %%A in (%%F) do Call :Split %%A
) > "X:Path%%~nF_New.csv"
)
goto :Eof
:Split
Echo(%2,%3,%5





share|improve this answer


























  • Thanks, this works a treat, anyway I could have the file name as a wild character like: (*.csv), when I try do that it returns a blank file

    – ryall579
    Nov 23 '18 at 10:54











  • You'll need an outer for loop to enumerate the files. See edit to the answer.

    – LotPings
    Nov 23 '18 at 11:00








  • 1





    just following the label :split insert an if "%~2"=="" goto :eof or a similar check on other columns

    – LotPings
    Nov 28 '18 at 11:58








  • 1





    There is only one redirection to change to directly save in the folder you'd like so ) > "%~dpnF_New.csv" --> ) > "X:Yourpath%~nF_New.csv" But this should be a new question and my answer here deserves an upvote ;-)

    – LotPings
    Nov 28 '18 at 12:29






  • 1





    Sorry missed one % sign, see changed answer. How to vote up

    – LotPings
    Nov 28 '18 at 12:55



















0














Not too difficult really.



from batch file, assuming file name is test.csv



(for /f "tokens=2,3,5 delims=," %%i in (test.csv) do echo %%i,%%j,%%k) > newcsv.csv


This will simply write values 2,3,5 to the new csv file..



To run it from cmdline instead, simply remove one of each %



(for /f "tokens=2,3,5 delims=," %i in (test.csv) do echo %i,%j,%k) > newcsv.csv


Please note, this assumes the data within your values do not contain , if they do, we need to make some changes.






share|improve this answer





















  • 1





    much faster: (for ... ...,%%k)>newcsv.csv (very noticeable with larger files)

    – Stephan
    Nov 23 '18 at 11:21











  • Thanks @Stephan , updated.

    – Gerhard Barnard
    Nov 23 '18 at 11:27













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














In case there are possibly commas inside the fields,

use a call to a sub passing the quoted arguments:



:: Q:Test20181123SO_53444017.cmd
@Echo off
(for /f "delims=" %%A in (test.csv) do Call :Split %%A
) > test_New.csv
goto :Eof
:Split
Echo(%2,%3,%5




> type test_New.csv
"60007834 "," 10.00","22/11/2018"
"60008504 "," 12.00","22/11/2018"
"60010605 "," 12.00","22/11/2018"
"60011385 "," 12.00","22/11/2018"


Edit: Variant processing all *.csv files (appending _New to the name)



:: Q:Test20181123SO_53444017.cmd
@Echo off
For %%F in (*.csv) do (
(for /f "delims=" %%A in (%%F) do Call :Split %%A
) > "X:Path%%~nF_New.csv"
)
goto :Eof
:Split
Echo(%2,%3,%5





share|improve this answer


























  • Thanks, this works a treat, anyway I could have the file name as a wild character like: (*.csv), when I try do that it returns a blank file

    – ryall579
    Nov 23 '18 at 10:54











  • You'll need an outer for loop to enumerate the files. See edit to the answer.

    – LotPings
    Nov 23 '18 at 11:00








  • 1





    just following the label :split insert an if "%~2"=="" goto :eof or a similar check on other columns

    – LotPings
    Nov 28 '18 at 11:58








  • 1





    There is only one redirection to change to directly save in the folder you'd like so ) > "%~dpnF_New.csv" --> ) > "X:Yourpath%~nF_New.csv" But this should be a new question and my answer here deserves an upvote ;-)

    – LotPings
    Nov 28 '18 at 12:29






  • 1





    Sorry missed one % sign, see changed answer. How to vote up

    – LotPings
    Nov 28 '18 at 12:55
















1














In case there are possibly commas inside the fields,

use a call to a sub passing the quoted arguments:



:: Q:Test20181123SO_53444017.cmd
@Echo off
(for /f "delims=" %%A in (test.csv) do Call :Split %%A
) > test_New.csv
goto :Eof
:Split
Echo(%2,%3,%5




> type test_New.csv
"60007834 "," 10.00","22/11/2018"
"60008504 "," 12.00","22/11/2018"
"60010605 "," 12.00","22/11/2018"
"60011385 "," 12.00","22/11/2018"


Edit: Variant processing all *.csv files (appending _New to the name)



:: Q:Test20181123SO_53444017.cmd
@Echo off
For %%F in (*.csv) do (
(for /f "delims=" %%A in (%%F) do Call :Split %%A
) > "X:Path%%~nF_New.csv"
)
goto :Eof
:Split
Echo(%2,%3,%5





share|improve this answer


























  • Thanks, this works a treat, anyway I could have the file name as a wild character like: (*.csv), when I try do that it returns a blank file

    – ryall579
    Nov 23 '18 at 10:54











  • You'll need an outer for loop to enumerate the files. See edit to the answer.

    – LotPings
    Nov 23 '18 at 11:00








  • 1





    just following the label :split insert an if "%~2"=="" goto :eof or a similar check on other columns

    – LotPings
    Nov 28 '18 at 11:58








  • 1





    There is only one redirection to change to directly save in the folder you'd like so ) > "%~dpnF_New.csv" --> ) > "X:Yourpath%~nF_New.csv" But this should be a new question and my answer here deserves an upvote ;-)

    – LotPings
    Nov 28 '18 at 12:29






  • 1





    Sorry missed one % sign, see changed answer. How to vote up

    – LotPings
    Nov 28 '18 at 12:55














1












1








1







In case there are possibly commas inside the fields,

use a call to a sub passing the quoted arguments:



:: Q:Test20181123SO_53444017.cmd
@Echo off
(for /f "delims=" %%A in (test.csv) do Call :Split %%A
) > test_New.csv
goto :Eof
:Split
Echo(%2,%3,%5




> type test_New.csv
"60007834 "," 10.00","22/11/2018"
"60008504 "," 12.00","22/11/2018"
"60010605 "," 12.00","22/11/2018"
"60011385 "," 12.00","22/11/2018"


Edit: Variant processing all *.csv files (appending _New to the name)



:: Q:Test20181123SO_53444017.cmd
@Echo off
For %%F in (*.csv) do (
(for /f "delims=" %%A in (%%F) do Call :Split %%A
) > "X:Path%%~nF_New.csv"
)
goto :Eof
:Split
Echo(%2,%3,%5





share|improve this answer















In case there are possibly commas inside the fields,

use a call to a sub passing the quoted arguments:



:: Q:Test20181123SO_53444017.cmd
@Echo off
(for /f "delims=" %%A in (test.csv) do Call :Split %%A
) > test_New.csv
goto :Eof
:Split
Echo(%2,%3,%5




> type test_New.csv
"60007834 "," 10.00","22/11/2018"
"60008504 "," 12.00","22/11/2018"
"60010605 "," 12.00","22/11/2018"
"60011385 "," 12.00","22/11/2018"


Edit: Variant processing all *.csv files (appending _New to the name)



:: Q:Test20181123SO_53444017.cmd
@Echo off
For %%F in (*.csv) do (
(for /f "delims=" %%A in (%%F) do Call :Split %%A
) > "X:Path%%~nF_New.csv"
)
goto :Eof
:Split
Echo(%2,%3,%5






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 28 '18 at 12:53

























answered Nov 23 '18 at 10:40









LotPingsLotPings

19k61532




19k61532













  • Thanks, this works a treat, anyway I could have the file name as a wild character like: (*.csv), when I try do that it returns a blank file

    – ryall579
    Nov 23 '18 at 10:54











  • You'll need an outer for loop to enumerate the files. See edit to the answer.

    – LotPings
    Nov 23 '18 at 11:00








  • 1





    just following the label :split insert an if "%~2"=="" goto :eof or a similar check on other columns

    – LotPings
    Nov 28 '18 at 11:58








  • 1





    There is only one redirection to change to directly save in the folder you'd like so ) > "%~dpnF_New.csv" --> ) > "X:Yourpath%~nF_New.csv" But this should be a new question and my answer here deserves an upvote ;-)

    – LotPings
    Nov 28 '18 at 12:29






  • 1





    Sorry missed one % sign, see changed answer. How to vote up

    – LotPings
    Nov 28 '18 at 12:55



















  • Thanks, this works a treat, anyway I could have the file name as a wild character like: (*.csv), when I try do that it returns a blank file

    – ryall579
    Nov 23 '18 at 10:54











  • You'll need an outer for loop to enumerate the files. See edit to the answer.

    – LotPings
    Nov 23 '18 at 11:00








  • 1





    just following the label :split insert an if "%~2"=="" goto :eof or a similar check on other columns

    – LotPings
    Nov 28 '18 at 11:58








  • 1





    There is only one redirection to change to directly save in the folder you'd like so ) > "%~dpnF_New.csv" --> ) > "X:Yourpath%~nF_New.csv" But this should be a new question and my answer here deserves an upvote ;-)

    – LotPings
    Nov 28 '18 at 12:29






  • 1





    Sorry missed one % sign, see changed answer. How to vote up

    – LotPings
    Nov 28 '18 at 12:55

















Thanks, this works a treat, anyway I could have the file name as a wild character like: (*.csv), when I try do that it returns a blank file

– ryall579
Nov 23 '18 at 10:54





Thanks, this works a treat, anyway I could have the file name as a wild character like: (*.csv), when I try do that it returns a blank file

– ryall579
Nov 23 '18 at 10:54













You'll need an outer for loop to enumerate the files. See edit to the answer.

– LotPings
Nov 23 '18 at 11:00







You'll need an outer for loop to enumerate the files. See edit to the answer.

– LotPings
Nov 23 '18 at 11:00






1




1





just following the label :split insert an if "%~2"=="" goto :eof or a similar check on other columns

– LotPings
Nov 28 '18 at 11:58







just following the label :split insert an if "%~2"=="" goto :eof or a similar check on other columns

– LotPings
Nov 28 '18 at 11:58






1




1





There is only one redirection to change to directly save in the folder you'd like so ) > "%~dpnF_New.csv" --> ) > "X:Yourpath%~nF_New.csv" But this should be a new question and my answer here deserves an upvote ;-)

– LotPings
Nov 28 '18 at 12:29





There is only one redirection to change to directly save in the folder you'd like so ) > "%~dpnF_New.csv" --> ) > "X:Yourpath%~nF_New.csv" But this should be a new question and my answer here deserves an upvote ;-)

– LotPings
Nov 28 '18 at 12:29




1




1





Sorry missed one % sign, see changed answer. How to vote up

– LotPings
Nov 28 '18 at 12:55





Sorry missed one % sign, see changed answer. How to vote up

– LotPings
Nov 28 '18 at 12:55













0














Not too difficult really.



from batch file, assuming file name is test.csv



(for /f "tokens=2,3,5 delims=," %%i in (test.csv) do echo %%i,%%j,%%k) > newcsv.csv


This will simply write values 2,3,5 to the new csv file..



To run it from cmdline instead, simply remove one of each %



(for /f "tokens=2,3,5 delims=," %i in (test.csv) do echo %i,%j,%k) > newcsv.csv


Please note, this assumes the data within your values do not contain , if they do, we need to make some changes.






share|improve this answer





















  • 1





    much faster: (for ... ...,%%k)>newcsv.csv (very noticeable with larger files)

    – Stephan
    Nov 23 '18 at 11:21











  • Thanks @Stephan , updated.

    – Gerhard Barnard
    Nov 23 '18 at 11:27


















0














Not too difficult really.



from batch file, assuming file name is test.csv



(for /f "tokens=2,3,5 delims=," %%i in (test.csv) do echo %%i,%%j,%%k) > newcsv.csv


This will simply write values 2,3,5 to the new csv file..



To run it from cmdline instead, simply remove one of each %



(for /f "tokens=2,3,5 delims=," %i in (test.csv) do echo %i,%j,%k) > newcsv.csv


Please note, this assumes the data within your values do not contain , if they do, we need to make some changes.






share|improve this answer





















  • 1





    much faster: (for ... ...,%%k)>newcsv.csv (very noticeable with larger files)

    – Stephan
    Nov 23 '18 at 11:21











  • Thanks @Stephan , updated.

    – Gerhard Barnard
    Nov 23 '18 at 11:27
















0












0








0







Not too difficult really.



from batch file, assuming file name is test.csv



(for /f "tokens=2,3,5 delims=," %%i in (test.csv) do echo %%i,%%j,%%k) > newcsv.csv


This will simply write values 2,3,5 to the new csv file..



To run it from cmdline instead, simply remove one of each %



(for /f "tokens=2,3,5 delims=," %i in (test.csv) do echo %i,%j,%k) > newcsv.csv


Please note, this assumes the data within your values do not contain , if they do, we need to make some changes.






share|improve this answer















Not too difficult really.



from batch file, assuming file name is test.csv



(for /f "tokens=2,3,5 delims=," %%i in (test.csv) do echo %%i,%%j,%%k) > newcsv.csv


This will simply write values 2,3,5 to the new csv file..



To run it from cmdline instead, simply remove one of each %



(for /f "tokens=2,3,5 delims=," %i in (test.csv) do echo %i,%j,%k) > newcsv.csv


Please note, this assumes the data within your values do not contain , if they do, we need to make some changes.







share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 23 '18 at 11:27

























answered Nov 23 '18 at 9:48









Gerhard BarnardGerhard Barnard

7,85131232




7,85131232








  • 1





    much faster: (for ... ...,%%k)>newcsv.csv (very noticeable with larger files)

    – Stephan
    Nov 23 '18 at 11:21











  • Thanks @Stephan , updated.

    – Gerhard Barnard
    Nov 23 '18 at 11:27
















  • 1





    much faster: (for ... ...,%%k)>newcsv.csv (very noticeable with larger files)

    – Stephan
    Nov 23 '18 at 11:21











  • Thanks @Stephan , updated.

    – Gerhard Barnard
    Nov 23 '18 at 11:27










1




1





much faster: (for ... ...,%%k)>newcsv.csv (very noticeable with larger files)

– Stephan
Nov 23 '18 at 11:21





much faster: (for ... ...,%%k)>newcsv.csv (very noticeable with larger files)

– Stephan
Nov 23 '18 at 11:21













Thanks @Stephan , updated.

– Gerhard Barnard
Nov 23 '18 at 11:27







Thanks @Stephan , updated.

– Gerhard Barnard
Nov 23 '18 at 11:27




















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