regex - awk lines not match -

using awk, need avoid print lines specific word in specific field... before call it's duplicate, did search, , found this: match string not containg phrase, think it's other context, because here need match in specific column.

given data:

abc      abc def      aaa ghi      abc 

i want filter print full lines text not containing abc in fist field(lines 2 , 3).

as first try, tried (?!...) , expected work way return def , ghi:

echo -e "abc\tabc\ndef\taaa\nghi\tabc" | awk '$1 ~ /(?!abc)/ {print}' 

but return nothing.

second try... using not before expression !/.../ :

~$ echo -e "abc\tabc\ndef\taaa\nghi\tabc" | awk '$1 ~ !/abc/ {print}' 

return nothing too.

i don't understand why...

so shot fails (as expected time):

~$ echo -e "abc\tabc\ndef\taaa\nghi\tabc" | awk '!/abc/ {print}' def     aaa 

as expected time, second field filtered too. so, seems work $0.

then did homework , got working way:

~$ echo -e "abc\tabc\ndef\taaa\nghi\tabc" | awk '{fl=$0;$0=$1}; !/abc/ {print fl}; { $0=fl }' def     aaa ghi     abc 

anyway, looks workaround me, , have bad feeling i'm not doing in right way.

someone know better way, or @ least explain why first , second tries not worked ?

you can use following idiom:

awk '$1 !~ /abc/' file 

if first field not contain abc line gets printed. (the matches not operator !~). note command contains condition , no print action. can work since default action in awk print , therefore not need specified explicitly.