how to print the percentage of task completed on the same line

Shell Programming and Scripting
1

Hi

I have written a utility in shell program for which i want to add a code to display percentage of completion dynamically

My scripts runs approx about 30 to 45min , It appends exactly 2000 lines to one of the log file.

How to calculate percentage ?
I will note the total number of lines in the log file before running scripts.
Lets say Initial=8000.

Final is the total number of lines after execution
Final=8000+2000

In a loop for every 5 seconds i will count the lines in log file and store in
Current=8500

Percentage % = 8500-8000/2000*100 = 25%

I want to display Percentage as increasing value in the same line rather than printing a new line for every 5 seconds.

Thanks

2

You might want to look here.
Also here.

1 Like
3

If you want to calculate percentage by using one expression, use -

echo "Percentage% =" `expr \( \( 8500 - 8000 \) \* 100 \) / 2000`"%"
1 Like
4

This little example should show you output that stays on one line:

$ cat x
#!/bin/ksh

integer i=0

while [[ $i -lt 10 ]]; do
##
##  \r = carriage return
##  \c = suppress linefeed
##
  echo "\r$i\c"
  (( i=i+1 ))
  sleep 1
done

echo

exit 0
$
1 Like
5

Thanks gary_w

By changing echo it worked in bash shell.

here is the code for bash

#!/bin/bash

i=0

while [[ $i -lt 10 ]]; do
##
##  \r = carriage return
##  \c = suppress linefeed
##
  echo -en "\r$i\c\b"
  (( i=i+1 ))
  sleep 1
done

echo

exit 0
6

Actually, this works for me on my version of bash:

echo -en "\r$i"

The "n" argument means the "\c" is not needed, and neither is the "\b".
Alternatively, this works too:

echo -e "\r$i\c"