I have the following lines
3, 3, 100
4, 2, 50
8, 5, 80
.
.
.
and I want the following output
line starts at 3 and ends at 3 with value 100
line starts at 4 and ends at 2 with value 50
line starts at 8 and ends at 5 with value 80
.
.
.
I tried the following: sed 's/^/line starts at /' then applying this command for the output: sed 's/, / and ends at /' then applying this command for the output sed 's/, / with value /'. Is there any way to do it in a single line?
awk is good for this kind of formatted input - formatted output:
awk -F, '{printf("line starts at %d and ends at %d with value %d\n", $1, $2, $3)}' file
line starts at 3 and ends at 3 with value 100
line starts at 4 and ends at 2 with value 50
line starts at 8 and ends at 5 with value 80
A shell while read loop with printf:
while IFS=', ' read c1 c2 c3; do
printf 'line starts at %s and ends at %s with value %s\n' \
"$c1" "$c2" "$c3"
done <file
By setting the IFS variable to a space and a comma, the read command will use those characters as field delimiters.
Output:
line starts at 3 and ends at 3 with value 100
line starts at 4 and ends at 2 with value 50
line starts at 8 and ends at 5 with value 80
It turns out that there is -e option in sed
sed -e 's/^/line starts at /g' -e 's/, / and ends at /' -e 's/, / with value at /'
There is a easy and fast way to do this in shell itself:-
# cat f
3, 3, 100
4, 2, 50
8, 5, 80
# cat f | while read line ; do IFS=", " array=($line) ; echo "line starts at ${array[0]} and ends at ${array[1]} with value ${array[2]}"; done
line starts at 3 and ends at 3 with value 100
line starts at 4 and ends at 2 with value 50
line starts at 8 and ends at 5 with value 80
