sh

I have defined the same path in the $PATH variable 6 times. I wasn't logging out to check whether it worked. How can I remove the duplicates? The $PATH variable looks like this: echo $PATH /usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/games:/home/flacs/Programmes/USFOS/bin:/home/flacs/Programmes/USFOS/bin:/home/flacs/Programmes/USFOS/bin:/home/flacs/Programmes/USFOS/bin:/home/flacs/Programmes/USFOS/bin:/home/flacs/Programmes/USFOS/bin How would I reset it to just /usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/games You just execute: export PATH=/usr/local

This scripts will sort the files by date then move the first 2500 files to another directory. When I run below scripts, system prompt out Argument list too long msg. Anyone can help me enhance the scripts ? Thanks NUM_OF_FILES=2500 FROM_DIRECTORY=/apps/data01/RAID/RC/MD/IN_MSC/ERC/in DESTINATION_DIRECTORY=/apps/data01/RAID/RC/MD/IN_MSC/ERC/in_load if [ ! -d $DESTINATION_DIRECTORY ] then echo "unused_file directory does not exist!" mkdir $DESTINATION_DIRECTORY echo "$DESTINATION_DIRECTORY directory created!" else echo "$DESTINATION_DIRECTORY exist!" fi echo "Moving $NUM_OF_FILES oldest files to

The scala documentation shows that the way to create a scala script is like this: #!/bin/sh exec scala "$0" "$@" !# /* Script here */ I know that this executes scala with the name of the script file and the arguments passed to it, and that the scala command apparently knows to read a file that starts like this and ignore everything up to the reversed shebang !# My question is: is there any reason why I should use this (rather verbose) format for a scala script, rather than just: #!/bin/env scala /* Script here */ This, as far a I can tell from a quick test, does exactly the same thing, but is

This question already has an answer here: When to wrap quotes around a shell variable? 5 answers I want to create some scripts for filling some templates and inserting them into my project folder. I want to use a shell script for this, and the templates are very small so I want to embed them in the shell script. The problem is that echo seems to ignore the line breaks in my string. Either that, or the string doesn't contain line breaks to begin with. Here is an example: MY_STRING=" Hello, world! This Is A Multi lined String." echo -e $MY_STRING This outputs: Hello, world! This Is A Multi lined

While replacing external commands in a shell script, I used an array to get rid of awk's NF . Now, since I moved from bash to POSIX sh, I cannot get the array marked right: #!/bin/bash export RANGE="0 1 4 6 8 16 24 46 53" RANGE=($RANGE) echo arrayelements: $((${#RANGE[@]})) LAST=$((${#RANGE[@]}-1)) echo "Last element(replace NF): ${RANGE[$LAST]}" # ./foo arrayelements: 9 Last element(replace NF): 53 I'm using OpenBSD's, sh and it has exactly the same size as the ksh. Changing above to /bin/sh , it seems that the following doesn't work: set -A "$RANGE" set -- "$RANGE" How could I realise the

I'm writing a Bourne shell script and have a password input like this: echo -n 'Password: ' read password Obviously, I don't want the password being echoed to the terminal, so I want to turn off echo for the duration of the read. I know there's way to do this with stty , but I'll ask the question for the benefit of the community whilst I go read the manpage. ;) stty_orig=`stty -g` stty -echo echo 'hidden section' stty $stty_orig read -s password works on my linux box. You can use '-s' option of read command to hide user input. echo -n "Password:" read -s password if [ $password != "..." ] then