how to print no. from n to 1 from the pattern in shell script-CodePudding

display the loop

7 6 5 4 3 2 1 
  5 4 3 2 1
    3 2 1
      1

how to print this pattern no idea of printing it from n to 1

read -p "Enter rows:" rows

for ((a=$rows;a>=1;a--))
do
for ((b=$rows;b>$a;b--))
do
    echo -n " "
done
for ((c=1;c<=$a;c  ))
do
    echo -n '* '
    if [ $c = $a ]
    then
        echo -e '\n'
    fi
done
done

please help to print this pattern loop not have enough spaces if i take n as 7 then printing it in 4 line will be difficult for me

CodePudding user response：

The immediate error is that your requirement calls for subtracting two items from the start in each iteration, but you only subtract one.

More generally, probably avoid echo -e entirely; learn to use printf if you need detailed control over newlines etc.

for ((a=$rows; a>=1; a-=2)); do
  printf -v sep "%$((rows - a))s" ""
  for ((b=$a; b>=1; b--)); do
    printf "%s%i" "$sep" "$b"
    sep=" "
  done
  printf '\n'
done

And, as demonstrated here, probably learn to indent your code to show how blocks are nested.

Tangentially, perhaps better to accept the number of rows as a command-line argument rather than prompt interactively for a value. Interactive I/O is pesky when you want to use your script as part of another larger script.