Better Way Of Writing This Command Of Leaving Multiple Blank Lines

Hello,

They gave me this:
Code:

#!/bin/bash

# Write error message on stderr and die
function die() {
  echo "$@" >&2
  exit 1
}

# Load var from properties files
function load() {
  for propertie in "$@" ; do
    [[ -f "${propertie}" ]] || die "load() : \"${propertie}\" don't exist !"
    while read ; do
echo "${REPLY%=*}"
echo "${REPLY#*=}"
      [[ -n "${REPLY%%=*}" && -z "${REPLY#*=}" ]] || die "load() : Variable \"${REPLY%%=*}\" is empty!"
      eval "${REPLY%%=*}"=\'"${REPLY#*=}"\'
    done < "${propertie}"
  done
}

load "manage_srcds.conf"
echo "SRCDS_TICKRATE=$SRCDS_TICKRATE"
echo "SRCDS_SCREEN=$SRCDS_SCREEN"
echo "SRCDS_SRCDS_GAME_NAME=$SRCDS_GAME_NAME"
echo "SRCDS_MAXPLAYERS=$SRCDS_MAXPLAYERS"

but, return:
Quote:
###################################################
###################################################
load() : Variable "###################################################" is empty!

cat manage_srcds.conf

I just wish that there was no empty variable excluding those who do not have integer.

Best regards,

i need to complete this exercise but my code has some issues
HERE is the PRoblem:
Create a script that can accept ANY amount of numbers from the command line. Process the numbers one at a time, where numbers greater than 10 print “large”, numbers less than or equal to 10 print “small”
E.g. process 5 10 15 would print
small
small
large

and here is my code so far
if [ $@ -le "10" ]
then
echo "smaller"
else
echo "bigger"
shift
fi
if [ $@ -le "10" ]
then
echo "smaller"
else
echo "bigger"
shift
fi
if [ $@ -le "10" ]
then
echo "smaller"
else
echo "bigger"
shift
fi
if [ $@ -le "10" ]
then
echo "smaller"
else
echo "bigger"
shift
fi

any help would be greatly appreciated

Hi guys,

i'm beginner with Unix, I tried my best, but I Really don't know how to finish it .

here's my problem : I have to do something like Student system - student's name | subjects | mark |credits.

after starting script it will ask you for Student's name, then which subject, then the mark will appear.

It's a homework, and I have my own idea how to solve it, but cannot finish it, i don't want complete solution, just a help!

Code:

#!/bin/bash
echo "Students name"
echo -n "Enter student's name: "
read name
echo
case $name in
Example1|Math) echo "C" ;;
Example2|Programming) echo "D" ;;
Example3|WWW) echo "A" ;;
Example4|Economie) echo "-" ;;
esac

I don't now how to bring together case with second things which is "enter subject".

I appreciate any help

Hello everyone,

I recently started using "function" and "case" in linux but I came across a slight problem

My code:

Code:

function Return
{
echo $1
}

case "$1" in 
 ""       ) echo "give parameters please";;
 [0-9]*   ) echo "No numbers please";;
 *        ) Return;;

So whenever the first parameter is a simple word I would like it returned, but it seems like there is no returned string (empty space)

Also I know I can just write "echo $1" instead of Return, but I'd like to use a function for this.

Can anyone tell me why the function doesn't return $1? Thanks!

I want to And search grep shell

but It's hard to grep variable


---------------------------------------------------------------
#!/bin/bash


if [ $# -eq 0 ]
then
echo "Ussage: phone searchfor [...searchfor]"
echo "(You didn't tell me what you want to search for )"

else

pass=0
find=""

for idx in $*
do
if [ -n "$idx" ]
then
if [ $pass -eq 0 ]
then
find=$(egrep "$idx" mydata)
pass=1

else

find=$("$find" | grep "$idx")

echo $find
fi
fi

done

if [ -z "$find" ]
then
echo "There is no such thing"
else

echo $find | awk -f display.awk

fi

fi

-----------------------------------------------------

there is one error : command not found

in find=$("$find" | grep "$idx")

how can I grep variable and store it into variable?

I am writing a script to check whether a "sysinfo" is installed in a system or not. sysinfo is a tool that gives information about the system. Its is installed in /usr/local/bin. so at CLI when I type sysinfo, it will display output.

#!/bin/bash
sysinfo=`/usr/local/bin/sysinfo`
echo "$sysinfo"
if [ "$?" = 0 ];then
echo "sysinfo installed"
else
echo "Please install sysinfo"
fi

Concept is, I run the command if it is present, it will show output and exit status of command is 0 else it is not installed. But it is not working as expected. Any thoughts , idea, suggestion.

Hi,

I am new to UNIX - i wrote this below script based on the requirement. But i am stuck at the concatenation (at the second last step of the code)

The below code is working fine till the concatenation(second last step) - I need to concatenate Hello to the "physId" e.g. - The filename is UM123456789.20150503 - i am extracting M123456789 and i need to append "HELO" to it at the end. But as per the below script - when i am using the concatenation, it is overwriting the M123456789 and the output thus becomes HELO456789. I am trying to get the output as - M123456789HELO - where am i going wrong?

on the terminal i checked - echo $0 and it gave -> /bin/sh. Hence i wrote the below code.

#!bin/sh
absolutePath=/abc/data/abc_unix/stg/decrypt/*.*
filepath=$(echo ${absolutePath%.*})
echo "$filepath"
filenameext=$(echo ${filepath#/abc*decrypt/})
echo "$filenameext"
file=$(echo ${filenameext#.*})
echo "$file"
extract_physId=$(echo ${file:1:9})
physId=$(echo ${extract_physId})
echo "$physId"
key="$physId"HELO
echo "$key"

Code:

PREFIX=192.168.1
OCTET=1
while [ $OCTET -lt "255" ]; do
    echo -en "Pinging ${PREFIX}.${OCTET}..."
    ping -c1 -w1 ${PREFIX}.${OCTET} > /dev/null 2>&1
    if [ "$?" -eq "0" ]; then
       echo "OK"
    else
       echo "Failed"
    fi
    let OCTET=$OCTET+1
done

My question is, in the penultimate line, why is it OCTET, and not $OCTET, with a dollar before it, meaning, the variable?

What if it were written:
Code:

let $OCTET=$OCTET+1

Like in C++ (I think) or Python or whatever?

Hi,

I've used unix before, but have just started to play with CYGWIN (1.1.3) on windows XP (32 bit). I'm having trouble getting 'echo' to behave as I would expect.

I've created a script called runstuff which contains these lines....
Code:

#!/bin/bash
# This line works as expected....
echo "the user is <$USER> and the hostname is <$HOSTNAME>"
MYVAR1=bill
MYVAR2=fred
# This lines doesn't.....
echo "my first variable is<$MYVAR1> and my second is <$MYVAR2>"

The output I get is this...
Quote:
$ runstuff
the user is <Mike> and the hostname is <MikesLaptop>
> and my second is <fredl

In the first echo statement it is just echoing a couple of external variables - this works as expected. In the 2nd echo example the two variables are internal, but the displayed output isn't what I'd expect, my guess is that the first part of the output "my first variable is <bill" has somehow been overwritten by the remainder of the string. Or am I missing something?

I have tried exporting the variables, and I've experimented with wrapping them in quotes, but the results are always the same.

Any help gratefully received.

i have been trying to get this peice of code to work but i cant seem to get it to work.
im just starting out coding and kinda poking around in the dark to get this resolved, any help would be much appriciated, thankyou


Quote:
#!/bin/bash

echo "Would you like to scan using nmap or hping?"

read scanner

if "$scanner" = "nmap" : then

echo "Enter the starting IP address :"

read FirstIP

echo "Enter the last IP address :"

read LastIP

echo "Enter the port number you want to scan for :"

read port

nmap -sT $FirstIP-$LastIP -p $port -oG web
cat web | grep open > web1
cat web1 | cut -f2 -d ":" | cut -f1 -d "(" > web2
cat web2


elit



echo "Which IP address would you like to scan?"

read IPaddress

echo echo "What port would you like to scan for ?"

read hpingport

echo "How many packets would you like to send?"

read packets

hping3 -c $packets $IPaddress -p $hpingport > hpingscan

cat hpingscan
thankyou for any help