mjhjmtu

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 %dn", $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 %sn' 
 "$c1" "$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

Hope it helps