shell script to print rows if there is a value in column 2

Question

1 2 3 4
3 1 4 2
1 4 3 2
4 2 1 3

How can I print all rows, on which there is a 2 in the second column.

Solution

Using awk (or its cousin gawk):

gawk '$2==2 {print $0}' inputfile

In awk columns are indicated by a $, with $1 the first column, $2 the second, etc. The whole line is given by $0. So this example reads: if column 2 is equal to 2, print the whole line.

Edit: as devnull said below:

gawk '$2==2' inputfile 

is enough. To print the lines with 2 in the second column and 4 in the last use $NF, which stands for the Number of Fields (i.e. the last column):

gawk '$2==2 && $NF==4'  inputfile

Comments

[devnull]

{print $0} is redundant here. You could simply say awk '{$2==2}' inputfile.

[Stéphane Chazelas]

@devnull, ITYM awk '$2==2' or awk '$2=="2"' if you want string comparison (not consider 02 or 2.0 as the same as 2). gawk is more the grand-daughter of awk than its cousin ;-)

[devnull]

@StephaneChazelas Right, it seems that the OP would want to quote 2.

[Swish41]

is it possible to specify and ,if i want to print row with 2 in second column and 4 in last column.

[devnull]

@KevinParker Say awk '$2=="2" && $NF=="4"' inputfile

[Marek Zakrzewski]

or awk '$2 ~ /2/ && $NF ~ /'TEST'/ {print }' filename

[terdon]

@val0x00ff no, that will also print foobar2barfoo, ~// is a match operator and will be true for substrings. Also, you can't have nested ', you'd have to use ".