Command 'cut' doesn't show last column CSV

Question

I've created a CSV from shell. Then I need to filter the information by column. I used this command:

$cut -d ';' -f 12,22 big_file.csv

The input looks like:

ACT;XXXXXX;MCD;881XXXX;881017XXXXXX;ABCD;BMORRR;GEN;88XXXXXXXXXX;00000;01;2;000008608008602;AAAAAAAAAAA;0051;;;;;;093505;
ACT;XXXXXX;MCD;881XXXX;881017XXXXXX;ABCD;BMORRR;GEN;88XXXXXXXXXX;00000;01;3;000008608008602;AAAAAAAAAAA;0051;;;;;;085000;anl@mail.com

The output is:

ID CLIENT;email
00000xxxxxxxxx
00000000xxxxxx;anl@mail.com

As you can see, the last column does not appear (note, that the semicolon is missing in the first line). I want this:

ID CLIENT;email
00000xxxxxxxxx;
00000000xxxxxx;anl@mail.com

I have another CSV file with information and it works. I've reviewed the csv and the columns exist.

What does the input look like? Is the trouble that you want a trailing semicolon even if the input line has no column 22? — hmakholm left over Monica, Aug 02 '19 at 17:18
Please add sample input (no descriptions, no images, no links) and your desired output for that sample input to your question (no comment). — Cyrus, Aug 02 '19 at 17:21
Hmm for me `echo "a;b;" | cut -d';' -f1,3` does output `a;`. Are you sure the line with no email actually has enough semicolons? — hmakholm left over Monica, Aug 02 '19 at 17:31
Where from is the `ID CLIENT;email` in the output? Can't duplicate, there are 21 fields in the input file. Don't you want `-f12,21` ? That could be it. Maybe you have some `\r` characters in your input. So the `;` is there but behind the `0` — KamilCuk, Aug 02 '19 at 17:35
I think wich the problem is with the 'end of line', because I tested with a CSV created from excel, and then with the mine. The outputs are differents. — Eder Armando Anillo Lora, Aug 02 '19 at 17:37
@HenningMakholm I've reviewed, everything is fine. Bellow, your answer was excellent — Eder Armando Anillo Lora, Aug 05 '19 at 14:59

score 1 · Accepted Answer · answered Aug 02 '19 at 17:26

1

There doesn't seem to be a way to make cut do this. The next step up in expressivity is awk, which does it easily:

$ cat testfile
one;two;three;four
1;2;3
first;second
only
$ awk -F';' '{ OFS=FS; print $1, $3 }' < testfile
one;three
1;3
first;
only;
$

answered Aug 02 '19 at 17:26

hmakholm left over Monica

23,074
3
51
73

jottbe · Answer 2 · 2019-08-05T22:26:11.607

1

You don't get the semicolon in the output of your second line, because your second line contains just 21 fields (the first contains 23 fields). You can check that using:

(cat bigfile.csv | tr -d -c ";\n" ; echo "1234567890123456789012") | cat -n | grep -v -E ";{22}"

This will output all lines from bigfile.txt with less than 22 semicolons along with the corresponding line numbers.

To fix that, you can add a bunch of empty fields at the end of each line and pipe the result to cut like this:

sed -e's|^\(.*\)|\1;;;;;;;;;;;;;;;;;;;;;;;;|g' bigfile.csv | cut -d ';' -f 12,22 | cut -d ';' -f 12,22

The result is:

XXXXXXXXYYY;XXXNNN
XXXXYYYYXXXXX;

edited Aug 05 '19 at 22:26

answered Aug 02 '19 at 21:22

jottbe

4,228
1
15
31

No, I just want 12 and 22. – Eder Armando Anillo Lora Aug 05 '19 at 15:02
So specify it with a comma. `-f 12,22` – KamilCuk Aug 05 '19 at 21:14
@EderArmandoAnilloLora: ok, I see, you are missing the one semicolon. I update my answer. – jottbe Aug 05 '19 at 22:00
Thanks for your help. My copy from file was wrong. Sorry, I've modified the post. I've used the @Henning Makholm answer and It's works. – Eder Armando Anillo Lora Aug 20 '19 at 21:34

Command 'cut' doesn't show last column CSV

2 Answers2