ksh

Using the same sort command with the same input produces different results on different machines. How do I fix that? The man-page on OS X says: ******* WARNING ******* The locale specified by the environment affects sort order. Set LC_ALL=C to get the traditional sort order that uses native byte values. which might explain things. If some of your systems have no locale support, they would default to that locale (C), so you wouldn't have to set it on those. If you have some that supports locales and want the same behavior, set LC_ALL=C on those systems. That would be the way to have as many

I always believed that a sub-shell was not a child process, but another shell environment in the same process. I use a basic set of built-ins: (echo "Hello";read) On another terminal: ps -t pts/0 PID TTY TIME CMD 20104 pts/0 00:00:00 ksh So, no child process in kornShell (ksh). Enter bash, it appears to behave differently, given the same command: PID TTY TIME CMD 3458 pts/0 00:00:00 bash 20067 pts/0 00:00:00 bash So, a child process in bash. From reading the man pages for bash, it is obvious that another process is created for a sub-shell, however it fakes $$, which is sneeky. Is this

I want similar option like getche() in C. How can I read just a single character input from command line? Using read command can we do it? In ksh you can basically do: stty raw REPLY=$(dd bs=1 count=1 2> /dev/null) stty -raw SiegeX In bash, read can do it: read -n1 ans fess . read -n1 works for bash The stty raw mode prevents ctrl-c from working and can get you stuck in an input loop with no way out. Also the man page says stty -raw is not guaranteed to return your terminal to the same state. So, building on dtmilano's answer using stty -icanon -echo avoids those issues. #/bin/ksh ## /bin/{ksh

I have some shell command. I would like to write the output to the standard output and save it into variable as well. I'd like to solve it with one command. I have tried these things. ls > $VAR # redirects the output to file which name is stored in $VAR ls | tee -a $VAR # writes to standard output as well as in file which name is stored in $VAR VAR=`ls` # output into $VAR, but it is not sent to standard output VAR=`ls`;echo $VAR # ok, it works but these are two commands Any idea? How about: VAR=$(ls | tee /dev/tty) great answer from @Gary_Barker, but this isn't possible on all systems. On our

I've got a shell script which does the following to store the current day's date in a variable 'dt': date "+%a %d/%m/%Y" | read dt echo ${dt} How would i go about getting yesterdays date into a variable? Basically what i'm trying to achieve is to use grep to pull all of yesterday's lines from a log file, since each line in the log contains the date in "Mon 01/02/2010" format. Thanks a lot If you have Perl available (and your date doesn't have nice features like yesterday ), you can use: pax> date Thu Aug 18 19:29:49 XYZ 2010 pax> dt=$(perl -e 'use POSIX;print strftime "%d/%m/%Y%",localtime

My command's output is something like: 1540 "A B" 6 "C" 119 "D" The first column is always a number, followed by a space, then a double-quoted string. My purpose is to get the second column only, like: "A B" "C" "D" I intended to use <some_command> | awk '{print $2}' to accomplish this. But the question is, some values in the second column contain space(s), which happens to be the default delimiter for awk to separate the fields. Therefore, the output is messed up: "A "C" "D" How do I get the second column's value (with paired quotes) cleanly? Or use sed & regex. <some_command> | sed 's/^.* \(