Running multiple unix commands in a single script


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# 1  
Running multiple unix commands in a single script

Hi,

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

my script like below:
Code:
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")

Last edited by Scott; 02-09-2010 at 08:47 AM.. Reason: Code tags
# 2  
Code:
#! /bin/bash

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

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

exit 0

# 3  
Quote:
Originally Posted by koti_rama
Hi,

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

my script like below:
Code:
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")


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


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


Last edited by Scott; 02-09-2010 at 10:39 AM.. Reason: Code tags
# 4  
Hi,
one simple way is to use the read command, ie
Code:
...
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
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