I hope that's an example of how not to do something because it doesn't work.
The idea, I think, is to run the ls command to list the current directory, and feed its output into the more program so you can view it page by page. But you can't dump that text into a variable and expect it to work because the shell doesn't evaluate variables that way unless you force it to with eval.
I guess I just should have asked what the statement
$listpage
means -- does it mean to display the value of listpage, and if so, why didn't it use the echo command?
Think more literally. It's just a variable, and the shell does the same thing with variables for every statement whether they're at the beginning of the line or the end. For each statement, the shell
1) Splits apart on semicolons or pipes, and evaluates each part(turns variables into their contents, etc).
2) Executes each part.
will print 'asdf' because it
1) converts $CMD asdf into echo asdf
2) runs "echo asdf".
Now this:
does this:
1) Converts $CMD into ls | more
2) Executes ls | more
See something wrong? Pipes were supposed to be handled in step 1 but weren't, because it was in a variable. Step 2 doesn't do anything to pipes, so just feeds it into ls raw. ls complains there's no such files "|" and "more".
Now this:
1) Converts eval "$CMD" into eval "ls | more"
2) Executes eval "ls | more", shoving the string "ls | more" back into step 1.
1) Splits ls | more into two statements, ls and more
2) Executes ls, piping it into more.
eval's just a special keyword that bumps the shell back into step 1.
Last edited by Corona688; 03-10-2011 at 04:25 PM..
I booted into single user mode with
/usr/sbin/reboot -- -s
but after doing a control -d
my
who -r
shows
run-level 3 Nov 17 14:07 3 0 S
I was expecting it to show run-level S
why is this still in run level 3?
thanks (1 Reply)
hi All,
i have never used sed in Unix environment, but i have one script which is using this following command:
cat audit_session_rpt_MSP_20140331.lst|sed -n '/Apr 14/!p'| sed -n '/Page/!p'| sed -n '/UserName/!p' |\
egrep -v '^-|^=|^\*'|sed '/^$/d'|sed -e '1,7d'... (1 Reply)
I know $0 is the entire file's contents (at least I think that is what it is!), but what exactly is: $0!~
This was a snippet from a larger line
awk '$0!~/^$/ {print $0}'
This deletes blank lines, but I want to know specifically the $0!~ part... I am guessing /^$/ is regex for blank line...... (5 Replies)
Was wondering if someone could interpret this for me -- I'm not sure what everything means. It's a shell script from my bash book:
cd ()
{
builtin cd "$@"
es=$?
echo "$OLDPWD ->$PWD"
return $es
}
what I don't quite understand is the "$@". I think, if I understand... (6 Replies)
Hi,
So I am new to Unix, and I need to check the performance of some apps I am running. But I don't know how to interpret the output from TOP.
Could somebody please explain the difference between the different values. And also explain how I can have a process which has a %CPU > 100?
... (7 Replies)
Can someone help me out here. I can't get this piece of code to work. i.e. $ALL_EVENTS does not get interpreted in the if brackets. The first part is the code, the second part is the execution of the code. Note: $ALL_EVENTS does equal 2, but there is no value once passed to the if statement. ... (4 Replies)
Hi guys,
I have no idea on unix but suddenly, my cobol programs calls a unix script that i know nothing about.
can you guys interpret these lines for me?
i know its a print command but I want to actually know how many copies it prints.
qprt -da -P $1 -t '6' -i '6' -l '70' $2
qprt -da... (1 Reply)
Should I be concerned about the %runocc value be always 100. The CPU is 99% idle all the time and the paging is 0
MyMachine|10/23/2007 00:00:05|1.0|100||
MyMachine|10/23/2007 00:05:04|1.0|100||
MyMachine|10/23/2007 00:10:04|1.0|100||
MyMachine|10/23/2007 00:15:04|1.0|100||... (2 Replies)