selective printing awk

Shell Programming and Scripting
1

Hi there

my file looks like this

1 a b c d e f
2 a b b c d e f f g h e t t
3 a c b d e f
4 a b c

i want to print the line which has the fields containing ONLY a b c, in this case the line 4.

How can i awk it !!!?

Many Thanks in advance!

2

Assuming your 1 2 3 4 are not part of the file

awk '$0=="a b c" ' file
3

Thanks but it doesnt help...may be am wrong in the question.

I reframe the question:

John a b c d e f
Bob a b c
Fred b c a
Tom a c d f e

i want to print the lines Bob and Fred which contain the "a b c"

4 
awk '/a b c/' file

Regards

5

nawk -f saint.awk myFile

saint.awk:

BEGIN {
  str="a b c"
  n=split(str, strA, FS)
  for(i=1; i<=n; i++) {
    strA[strA]
    delete strA
  }
}
{
  for(i=2; i<=NF; i++)
    if (!($i in strA)) next

  print
}

YMMV depending on what you mean by 'a b c'.

6

It works but it prints only the line containing a b c in the given order and not if the line contains a b c in different order

7 
awk '/Bob/ || /Fred/ && /a/ && /b/ && /c/'  file
8 
# awk '/[a-c] [a-c] [a-c]$/' file
Bob a b c
Fred b c a
9

Fred x c b a