奇怪的awk行为，打印了$ 1，但没有打印$ 0？

Question

This is printed fine ls | xargs -I var sh -c 'echo var | awk "{print $1}"' ls | xargs -I var sh -c 'echo var | awk "{print $1}"'

while this is not ls | xargs -I var sh -c 'echo var | awk "{print $0}"' ls | xargs -I var sh -c 'echo var | awk "{print $0}"'

Obviously this is not my use case and is just to reproduce the problem.

So while $0 stands for complete line, it is weird for $0 to not print while $1 gets printed.

Reference OS - Ubuntu 14.04

Answer 1

When you use:

ls | xargs -I var sh -c 'echo var | awk "{print $0}"'

That has $0 in double quotes hence $0 gets expanded by shell which is in this case has value sh without quotes. awk treats it as a variable and prints blank.

To make it work escape the $ :

ls | xargs -I var sh -c 'echo var | awk "{print \$0}"'

Why $1 works:

ls | xargs -I var sh -c 'echo var | awk "{print $1}"'

$1 also gets expanded by shell and it is empty. Hence awk just executes an empty print which is equivalent of print $0 and you get filename in output.

However it is not a good idea to use awk command in double quotes and parsing ls output is error prone.

Answer 2

For others landing here: If your syntax was perfect but you are potentially using windows files, remember the \\r will effectively overwrite your text and make it seem like $0 is blank....

First run dos2unix <file> , then try your awk .

Note: removing \\r with sed or tr isn't enough