hi,

i ve got the following variables defined,
Y=ls
Z="$Y"

can someone please tell me how to get 'the execution of ls' out of Z , like when i write
echo $Z i get ls

but how can i change or redefine the variables to get the execution of ls ??

Do you mean like this?

Always try commands in real-time before trying to put in a script.

try these:
echo `ls`
echo $(ls)
echo $ls
echo ls

When you understand how each of these works, then it will be easier to see what happens when they are assigned to variables.

Note the "`" (backtic)---not the same as the single quote (')

thanks for answering,

sadly i had tried the eval command , i typed eval ls, in my xterm and now all i get is the result of eval ls no matter what i type !!
do u know how to disable this command ?


maya

If I type "eval ls", I get the same answer as if typing "ls". After using eval, everything else works normally.

Can you give some more detail/examples of exactly what happens?

Also, why do you need to use eval?

I had been trying to get the answer by myself, so i tried
eval ls
to try to get the execution of ls
but then when i read the answer HOMEY posted, i tried it out, i copied it into a script called execute_ls
but everytime i type on the xterm :
chmod +x execute_ls
then
./execute_ls
all i get is
execute_ls execute_ls~
and when i typed set -x to trace whats happening
i had a line +echo <listoffilesinthedirectory>

its driving me nuts
i cant do anythg !

like this?

yes just like homey posted it

Can you get back to normal by simply closing the terminal?

Please post the exact content of the script file.

I tried the script and it worked fine. What happens if you simply run "ls" in that directory? The result should be the same as running the script.

my script execute_ls contains the following :

#!/bin/bash
Y=$(ls)
Z=$Y
echo $Z

then i go and write chmod +x execute_ls
then ./execute_ls
which gives me :
execute_ls execute_ls~

this doesnt work, i tried reopening a new terminal many times. i even restarted the computer !!
the only thg that works is when i add an echo line at the beginning like this:
echo "write>>"
read Y
echo "write>"
read Z
if [[ "Y" = "$(ls)" && "Z"="$Y"]] ; then
echo $Y
echo $Z
fi

this works, meaning that the script responds, by telling me to "write>>"
but it gives me no answer

dont blame me for being complicazed, everytime i call a script i get the result of an ls command !!

I am totally confused.

What happens if you open a new terminal and simply run "ls"?

This script:should do the same thing.

Maybe I'm not following, but this looks to me the correct behavior. Let me dissect your simple script:

1. #!/bin/bash
this is the so-called sha-bang: it tells to use /bin/bash as interpreter

2. Y=$(ls)
this assigns to Y the output of the command ls. This is called "command substitution" and the syntax $(command) is equivalent to `command` (with back-ticks).

3. Z=$Y
this assigns to Z the value of Y, which is still the output of the ls command

4. echo $Z
this prints to the standard output the value of Z, that is the output of ls: execute_ls execute_ls~. It looks like in the current directory you have only the script "execute_ls" and its backup file "execute_ls~" (that one created by text editors like gedit).

What am I missing here? Can you re-formulate your original question?

That's showing the scriptname and the backup scriptname file which is what it's supposed to do.
Maybe you wanted a more verbose listing as from ( ls -al )
Just a side note, I use script name which are less likely to get you into trouble executing something unwanted. For example: my_script