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Scripting queries - Newbie

Hi,

I have a file containing a list of commands with argument, how to run the commands automatically?

Assuming the contents of file "command2run" are as follows:
ls -ls
ls -la
ls -L

I tried the following:
for i in `cat command2run`
do
$i
done

The arguments were ran seperately instead of "ls -ls". Please advise?

Thanks in advance.

Ok ... a few things:

1. Please use [code][/code] tags around code and data

2. Do not use a for loop to read a file as it will under go word splitting (which is why you see separate entries). Use a while loop

3. If you are looking to execute the items in the file you can source it, ie use source or . prior to file name on the command line

Code:

$ . command2run

Hi,

I used . command2run but it just not able to go back to the command prompt after it finished. Not even I do and control-c.

Then use "source" Or a little help for you to use while to read a file:

Code:

while read line; do <do something here with $line>; done< input_filename

Ok, not sure what to tell you as it works as described for me.

However, here are some assumptions I have made which may be incorrect:

1. Your standard shell is bash, to check:

Code:

$ echo $SHELL

/bin/bash

2. Your file contains the following data:

Code:

$ cat command2run

ls -ls

ls -la

ls -L

3. When run I receive:

Code:

$ . command2run

total 16

12 -rwxr-xr-x 1 grail users 10444 17.09.2014 18:07 bashrc.tar.gz*

 4 -rwx------ 1 grail users    46 17.09.2014 18:28 d.sh*

total 64

drwxr-xr-x 2 grail users  4096 17.09.2014 18:26 ./

drwxr-xr-x 4 grail users  4096 19.09.2014 14:22 ../

-rw------- 1 grail users 19679 14.09.2014 16:04 .bashrc

-rw------- 1 grail users 11025 14.09.2014 15:24 .bashrc_help

-rw------- 1 grail users  7828 14.09.2014 15:24 .bashrc_help.no_color

-rwxr-xr-x 1 grail users 10444 17.09.2014 18:07 bashrc.tar.gz*

-rwx------ 1 grail users    46 17.09.2014 18:28 d.sh*

bashrc.tar.gz*  d.sh*

Obviously your output will be different

run it with sh command directly

Quote:

sh command2run