如何使用awk（或剪切）从管道分隔的文件中每行打印最后一个字段？

Question

I have a file like that

14618   | 184.72.251.121   | 184.72.128.0/17     | US | arin     | 2010-01-26 | AMAZON-AES - Amazon.com, Inc., US
14618   | 107.21.223.240   | 107.21.128.0/17     | US | arin     | 2011-05-03 | AMAZON-AES - Amazon.com, Inc., US
16509   | 52.85.180.156    | 52.85.180.0/23      | US | arin     |            | AMAZON-02 - Amazon.com, Inc., US
61337   | 85.199.214.99    | 85.199.212.0/22     | GB | ripencc  | 2005-02-08 | ECOM-AS ============================================, GB
31034   | 94.177.187.22    | 94.177.160.0/19     | IT | ripencc  | 2008-08-14 | ARUBA-ASN, IT
8816    | 212.45.144.206   | 212.45.128.0/19     | IT | ripencc  |            | IT-STCOM, IT

I must extract the last field and all its content. Notice that sometimes the penultimate column is empty.

I tried

awk 'BEGIN { ORS = " "}; {for(i=13;i<NF;++i) print $i}'

but sometimes it doesn't work or it misses some words.

Can you help me please?

Answer 1

Try this -

 awk -F'|' '{print $NF}' file
 AMAZON-AES - Amazon.com, Inc., US
 AMAZON-AES - Amazon.com, Inc., US
 AMAZON-02 - Amazon.com, Inc., US
 ECOM-AS ============================================, GB
 ARUBA-ASN, IT
 IT-STCOM, IT

If this is not your desired output then please post your expected output.

Answer 2

With sed: substitute the longest match up to | by nothing.

$ sed 's/.*|//' infile
 AMAZON-AES - Amazon.com, Inc., US
 AMAZON-AES - Amazon.com, Inc., US
 AMAZON-02 - Amazon.com, Inc., US
 ECOM-AS ============================================, GB
 ARUBA-ASN, IT
 IT-STCOM, IT

If you don't want the leading blanks, remove them as well:

$ sed 's/.*|[[:blank:]]*//' infile
AMAZON-AES - Amazon.com, Inc., US
AMAZON-AES - Amazon.com, Inc., US
AMAZON-02 - Amazon.com, Inc., US
ECOM-AS ============================================, GB
ARUBA-ASN, IT
IT-STCOM, IT

This matches up to the last | and includes as many blanks as possible after the | , then removes the match.

Answer 3

还原每一行，以剪切方式读取第一列，然后还原：

rev input | cut -f1 -d\| | rev

Answer 4

It looks like you would be better off using awk with the | character as the field separator ( awk -F'|' ) but, if they're always going to be nicely lined up like that, cut may well be a better option:

cut -c81- inputFile

That would give you all the characters from position 81 onwards in each line, which is roughly right just from me eyeballing the data. A bit of trial and error would give a more accurate value to use in place of 81 since I may be off by a couple - my eyeballs aren't as young as they once were :-)

Answer 5

It appears that you just want the 7th | -delimited field of your input:

cut -d\| -f7

 AMAZON-AES - Amazon.com, Inc., US
 AMAZON-AES - Amazon.com, Inc., US
 AMAZON-02 - Amazon.com, Inc., US
 ECOM-AS ============================================, GB
 ARUBA-ASN, IT
 IT-STCOM, IT

If you want to get rid of the leading space, you can remove that with cut too:

cut -d\| -f7- | cut -c2-

Answer 6

I'm assuming that by the last field you mean the last column. This is what you need:

awk -F'\|' '{print $7}' MyFile

It will print the last column of your file in this case.

Answer 7

You can use grep :

$ grep -o '[^|]*$' file
 AMAZON-AES - Amazon.com, Inc., US
 AMAZON-AES - Amazon.com, Inc., US
 AMAZON-02 - Amazon.com, Inc., US
 ECOM-AS ============================================, GB
 ARUBA-ASN, IT
 IT-STCOM, IT

If you don't want leading spaces:

grep -o '[^|]*$' file | cut -d' ' -f2-
AMAZON-AES - Amazon.com, Inc., US
AMAZON-AES - Amazon.com, Inc., US
AMAZON-02 - Amazon.com, Inc., US
ECOM-AS ============================================, GB
ARUBA-ASN, IT
IT-STCOM, IT

You can also use Bash parameter expansion:

$ while read -r line; do echo ${line##*|}; done <file
AMAZON-AES - Amazon.com, Inc., US
AMAZON-AES - Amazon.com, Inc., US
AMAZON-02 - Amazon.com, Inc., US
ECOM-AS ============================================, GB
ARUBA-ASN, IT
IT-STCOM, IT