Running multiple unix commands in a single script

Shell Programming and Scripting
1

Hi,

I would like to write a script with include more than 6 unix commands.

my script like below:

echo " script started"

ls -ld 

bdf | grep "rama"

tail -10 log.txt
...
..
...

now, i want to run above unix commands one by one.
example:
first the ls -ld command will be executed, then the script will wait for user input( ENTER from KEYBOARD) to continue for execute next command(bdf | grep "rama")

2 
#! /bin/bash

function continue ()
{
  echo -n "Continue? "
  read KEY
}

echo "1st command"
continue
echo "2nd command"
continue
echo "3rd command"

exit 0
3

After each command, give one more command as "read".
So it will wait for ur enter.

eg.

echo " script started"
 read;
ls -ld 
 read;
bdf | grep "rama"
 read;
tail -10 log.txt
read;
4

Hi,
one simple way is to use the read command, ie

...
ls -ld 
read input
bdf | grep "rama"
...

whatever the user enter is put into the variable input, which You can use or discard as You please.

Best regards,
Lakris