AWK - 比较后打印完整的输入字符串

Question

I have a file a.text:

hello world
my world
hello universe

I want to print the complete string if the second word is "world":

[root@sc-rdops-vm18-dhcp-57-128:/var/log] cat a | awk -F " " '{if($2=="world") print $1}'
hello
my

But the output which I want is:

[root@sc-rdops-vm18-dhcp-57-128:/var/log] cat a | awk -F " " '{if($2=="world") print <Something here>}'
hello world
my world

Any pointers on how I can do this?

Thanks in advance.

Answer 1

awk '{if ($2=="world") {print}}' file

Output:

hello world
my world

Answer 2

First off, since you are writing a single if statement, you can use the awk 'filter{commands;}' pattern, like so

awk -F " " '$2=="world" { print <Something here> }'

To print the entire line you can use print $0

awk -F " " '$2=="world"{print $0}' file

which can be written as

awk -F " " '$2=="world"{print}' file

But {print} is the default action, so it can be omitted after the filter like this:

awk -F " " '$2=="world"' file

Or even without the -F option, since the space is the default FS value

awk '$2=="world"' file

Answer 3

If you want / have to use awk to solve your problem:

 awk '$0~/world/' file.txt

If a line (ie, $0 ) matches the string "world" (ie, ~/world/ ) the entire line is printed

If you only want to check the second column for world :

 awk '$2 == "world"' file.txt