Hi,
In a csv file, I want to select records where first column has zero or multiple spaces.
Eg: abc.csv
,123,a
,22,b
,11,c
a,11,d
So output should be:
,123,a
,22,b
,11,c
Please advise
Hi,
In a csv file, I want to select records where first column has zero or multiple spaces.
Eg: abc.csv
,123,a
,22,b
,11,c
a,11,d
So output should be:
,123,a
,22,b
,11,c
Please advise
What have you tried so far?
Tried below options of awk.. but it works only for zero spaces.
awk -F',' '! $1 ~ /./'
awk -F',' '$1=="" {print $0}'
awk -F"," ' ! $1 ' file
Or
sed -n "/^ *,/p" file
Hello vegasluxor,
Following may help you in same.
awk -F, '($1 == "" || $1 ~ /[[:space:]]/)' Input_file
But if your first field has space and some value then that input will be catch also, as you didn't mention this condition in your input so I am giving you above solution. If you have first field only empty NOT a space with values following may help you in same too.
awk '{match($1,/^ *,/);A=substr($1,RSTART,RLENGTH);if(A){print $0}}' Input_file
EDIT: Adding one more solution on same.
grep '^ *,' Input_file
Thanks,
R. Singh
Thanks !!!
second awk and grep worked!!!
---------- Post updated at 04:07 AM ---------- Previous update was at 04:01 AM ----------
@anbu
sed solution worked well!!! Thanks a lot