awk Can not Select Column with empty value - awk

Question

i am trying to select a column with its missing value
here is my input file separated by tab
1 2 3
4 5
6
7 8
9
i am trying to select the first column in which output will look like
1
4
7
and the length of my column would be 5 in this case
I have tried
awk '$1!=""{print $1}' ./demo.txt
but it returns
1
4
6
7
9
can anybody help with this I am new in AWK

Answer 1

You can use cut:
$ cut -f 1 file # the default delimiter is a tab
Or with sed:
$ sed 's/[[:blank:]].*$//' file
Or awk:
$ awk '{sub(/[[:blank:]].*$/,"")}1' file
Or:
$ awk 'BEGIN{FS=OFS="\t"} {print $1}' file
All those print the first column and all five lines (blank or not)
Prints:
1
4
7

Answer 2

Tell awk to use a tab (\t) as the input field delimiter (-F):
$ awk -F'\t' '{ print $1 }' demo.txt
1
4
7
If you want to print multiple columns, maintaining the same delimiter for output, another approach using the FS and OFS variables:
$ awk 'BEGIN { FS=OFS="\t" } { print $1,$3 }' demo.txt
1 3
4 5
7
9

Answer 3

With sed something like:
sed 's/^\([^[:blank:]]*\).*/\1/' demo.txt

Answer 4

Using FIELDWIDTHS in gnu-awk you can do this for fixed width separated data:
awk 'BEGIN {FIELDWIDTHS = "4 4 *"} {print $1}' file
1
4
7
For demo purpose:
awk 'BEGIN {FIELDWIDTHS = "4 4 *"} {print NR ":", $1}' file
1: 1
2: 4
3:
4: 7
5:

Answer 5

if they're all single digits in 1st column :
echo \
'1 2 3
4 5
6
7 8
9' |
mawk NF=1 FS=
gcat -n
1 1
2 4
3
4 7
5
that's literally all you need. To play it safe, then do
nawk NF=1 FS='[[:space:]]' # overly-verbose so-called
# "proper" posix form
gawk NF=1 FS='[ \t]' # suffices unless the input
# happens to have uncommon bytes
# like \013 \v or \014 \f
or a very fringe way of fudging NF :
mawk 'NF ^= FS="[ \t]"'