scripting question

Jurrian · 10-21-2008, 04:49 AM

Hi all,

I want to execute a small script and echo the commands that are inside.

Right now I am doing it as following:

echo cd /home/jurian/test
echo ./run

cd /home/jurian/test
./run

This is working fine, however, everytime I edit the code I need to replace it in 2 places. Is there any better way to solve this?

Thanx in advance,

Jurrian

Jurrian · 10-21-2008, 04:55 AM

hmm, found out that it is working, but that it first executes the cript and then echo's what it have done, is there a way to make it first echo wat it does and then execute it? That way any eventual errors will be more logic to read

Thanx in advance,

Jurrian

ZAMO · 10-21-2008, 04:58 AM

Jurrian,

If you want to execute the commands

Code:

 cd /home/jurian/test
./run

Then echo is not needed. In case, if you want to display them only, using echo,

Code:

echo -e "cd /home/jurian/test"
echo "./run"

Cheers

ZAMO · 10-21-2008, 05:01 AM

You can try the following :

Code:

echo -e "cd /home/jurian/test"
cd /home/jurian/test
echo -e "./run"
./run

To have a time difference , in between display and execute add sleep

Code:

echo -e "cd /home/jurian/test"
sleep 5
cd /home/jurian/test
echo -e "./run"
sleep 5
./run

Jurrian · 10-21-2008, 05:01 AM

Actually, I want to run and display them, while the output should first contain the command and after that the output of the command..

Thanx for your reply,

Jurrian

Jurrian · 10-21-2008, 05:15 AM

Quote:

Originally Posted by ZAMO

You can try the following :

Code:

echo -e "cd /home/jurian/test"
cd /home/jurian/test
echo -e "./run"
./run

To have a time difference , in between display and execute add sleep

Code:

echo -e "cd /home/jurian/test"
sleep 5
cd /home/jurian/test
echo -e "./run"
sleep 5
./run

Well, the point is, I still need to type the commands twice, so if I change the directory, I need to change it at 2 places, at the echo command and at the execute command. I wondered if there was any way to prevent this.

Thanx, Jurrian