I have a input file which has user names and subject.
input
sk7865 /opt/apps/login
sk4888 /opt/apps/info
I am writing a shell script to take inputs from the above file and send a mail.
shell script
#!/bin/bash
while read a b
echo echo ""$a"" | mail -s ""$b"" "$a"#example.com
done < input
In the above script the actual command I wanted to use is:-
echo "hello world" | mail -s "a subject" someone#example.com
I want it to take arguments a,b at hello world, a subject to send the email to someone like I used it in the script. But it is not taking the arguments. I think it is something to do with double quotes. Please provide me with proper script.
You seem to be missing a do after your while:
#!/bin/bash
while read -r a b
do
echo "$a" | mail -s "$b" "$a"#somewhere.com
done < input
Remember to always run shellcheck on your bash scripts before posting here.
Also, you had an echo echo in there and double-double-quotes all over the place ""xyz"".
I have a Linux script that generates an HTML file with various outputs for various Linux commands.
Here is one of the outputs that creates an intrf.txt. I generated this file using this command
ip link show|sed '=;G'>intrf.txt
I did that as I want the lines of file to have line space between.
If I do the cat intrf.txt command on my shell I can see indeed the line spacing. If I run the script with the function below I see all lines of this file on my browser, but concatenated with no space between.
Maybe this is something simple, but I cannot figure it out.
function net_ifconfig
{
echo -e "GET http://google.com HTTP/1.0\n\n" | nc google.com 80 > /dev/null 2>&1
if [ $? -eq 0 ]; then
echo "<h2 style="background-color:#00FF00"><font size="5"> CHECK 2. LIST OF AVAILABLE INTERFACE</h2>"
ip link show|sed '=;G'>intrf.txt
cat intrf.txt
else
echo "CHECK 1. INTERNET IF OFFLINE"
echo "<h2 style="background-color:#FF0000"><font size="5"> INTERNET IS NOT CONNECTED</h2>"
fi
}
Seems if I put the command between pre tags it works.
echo "<pre>"
ip link show|sed '=;G'>intrf.txt
cat intrf.txt
echo "</pre>"
please bear with me if my terminology or syntax is less than stellar (still learning). I currently have a simple bash script that checks the arguments of the command and outputs files names with matching text. This part of my script works correctly via a grep command and piped to xargs for proper formatting.
When running the script, I run through a simple loop to check if the value is null and then move to running my variable/search if not.
My question is: Is it possible to have this script output via stdout AND also save a new file each time it is run with the user input and date/time? (but not overwrite) EX: report-bob-0729161500.rpt
I saw same other suggestions to use tee with the command, but I was trying to get it to work within the script. Similarly, another suggestion stated to utilize exec > >(tee -i logfile.txt), but I am unsure how to properly format this to include the date/time and $1 input into new files each time the script is executed.
Any help or suggested resources?
Thank you.
SEARCH=`[search_variable]`
if [ -z "$SEARCH" ]
then
echo "$1 not found."
else
echo -e "REPORT LISTING\n\n"
echo "$SEARCH"
fi
EDIT: I did try simply piping the echo statements to the tee command, which does work. However, I am still curious if anyone has other suggestions to accomplish this same task via alternative methods. Thank you.
With echo statements piped to tee:
SEARCH=`[search_variable]`
DATE=`date +"%m%d%y%k%M"`
if [ -z "$SEARCH" ]
then
echo "$1 not found."
else
echo -e "REPORT LISTING\n\n" | tee tps-list-$1-$DATE.rpt
echo "$SEARCH" | tee tps-list-$1-$DATE.rpt
fi
If you want to do it within the script, why then not just write to
both standard output and the file (using append where appropriate?).
Maybe a bit more writing, but it gives complete control.
Leon
one of the questions that I have been given to do for my Computer Science GCSE was:
Write a shell script that takes a string input from a user, asks for a file name and reports whether that string is present in the file.
However way I try to do it, I cannot create a shell script.
I don't need you to tell me the whole number, however, I have no idea where to start. I input the variable and the file name, however, I have no idea how to search for the chosen word in the chosen file. Any ideas?
Using grep can get this working, for example
viewEntry()
{
echo "Entering view entry"
echo -n "Enter Name: "
read input
if grep -q "$input" datafile
then
echo ""
echo -n "Information -> "
grep -w "$input" datafile
echo ""
else
echo "/!\Name Not Found/!\\"
fi
echo "Exiting view entry"
echo ""
}
dataFile is the file you would be reading from. Then making use of -q and -w arguments of grep, you should be able to navigate your chosen file.
This site does a great job explaining grep and your exact problem: http://www.cyberciti.biz/faq/howto-use-grep-command-in-linux-unix/
The following shell-script is a very quick approach to do what you suggested:
#!/bin/sh # Tell your shell with what program this script should be exectued
echo "Please enter the filename: "
read filename # read user input into variable filename
count=`grep -c $1 $filename` # store result of grep into variable count
if [ $count -gt 0 ] # check if count is greater than 0
then
echo "String is present:" $1
else
echo "String not found:" $1
fi
You should look at some tutorials to get the basics of shell-scripting. Your task isn't very complex, so after some reading you should be able understand what the script does and modify it according your needs.
I'm converting some Windows batch files to Unix scripts using sh. I have problems because some behavior is dependent on the %~dp0 macro available in batch files.
Is there any sh equivalent to this? Any way to obtain the directory where the executing script lives?
The problem (for you) with $0 is that it is set to whatever command line was use to invoke the script, not the location of the script itself. This can make it difficult to get the full path of the directory containing the script which is what you get from %~dp0 in a Windows batch file.
For example, consider the following script, dollar.sh:
#!/bin/bash
echo $0
If you'd run it you'll get the following output:
# ./dollar.sh
./dollar.sh
# /tmp/dollar.sh
/tmp/dollar.sh
So to get the fully qualified directory name of a script I do the following:
cd `dirname $0`
SCRIPTDIR=`pwd`
cd -
This works as follows:
cd to the directory of the script, using either the relative or absolute path from the command line.
Gets the absolute path of this directory and stores it in SCRIPTDIR.
Goes back to the previous working directory using "cd -".
Yes, you can! It's in the arguments. :)
look at
${0}
combining that with
{$var%Pattern}
Remove from $var the shortest part of $Pattern that matches the back end of $var.
what you want is just
${0%/*}
I recommend the Advanced Bash Scripting Guide
(that is also where the above information is from).
Especiall the part on Converting DOS Batch Files to Shell Scripts
might be useful for you. :)
If I have misunderstood you, you may have to combine that with the output of "pwd". Since it only contains the path the script was called with!
Try the following script:
#!/bin/bash
called_path=${0%/*}
stripped=${called_path#[^/]*}
real_path=`pwd`$stripped
echo "called path: $called_path"
echo "stripped: $stripped"
echo "pwd: `pwd`"
echo "real path: $real_path
This needs some work though.
I recommend using Dave Webb's approach unless that is impossible.
In bash under linux you can get the full path to the command with:
readlink /proc/$$/fd/255
and to get the directory:
dir=$(dirname $(readlink /proc/$$/fd/255))
It's ugly, but I have yet to find another way.
I was trying to find the path for a script that was being sourced from another script. And that was my problem, when sourcing the text just gets copied into the calling script, so $0 always returns information about the calling script.
I found a workaround, that only works in bash, $BASH_SOURCE always has the info about the script in which it is referred to. Even if the script is sourced it is correctly resolved to the original (sourced) script.
The correct answer is this one:
How do I determine the location of my script? I want to read some config files from the same place.
It is important to realize that in the general case, this problem has no solution. Any approach you might have heard of, and any approach that will be detailed below, has flaws and will only work in specific cases. First and foremost, try to avoid the problem entirely by not depending on the location of your script!
Before we dive into solutions, let's clear up some misunderstandings. It is important to understand that:
Your script does not actually have a location! Wherever the bytes end up coming from, there is no "one canonical path" for it. Never.
$0 is NOT the answer to your problem. If you think it is, you can either stop reading and write more bugs, or you can accept this and read on.
...
Try this:
${0%/*}
This should work for bash shell:
dir=$(dirname $(readlink -m $BASH_SOURCE))
Test script:
#!/bin/bash
echo $(dirname $(readlink -m $BASH_SOURCE))
Run test:
$ ./somedir/test.sh
/tmp/somedir
$ source ./somedir/test.sh
/tmp/somedir
$ bash ./somedir/test.sh
/tmp/somedir
$ . ./somedir/test.sh
/tmp/somedir
This is a script can get the shell file real path when executed or sourced.
Tested in bash, zsh, ksh, dash.
BTW: you shall clean the verbose code by yourself.
#!/usr/bin/env bash
echo "---------------- GET SELF PATH ----------------"
echo "NOW \$(pwd) >>> $(pwd)"
ORIGINAL_PWD_GETSELFPATHVAR=$(pwd)
echo "NOW \$0 >>> $0"
echo "NOW \$_ >>> $_"
echo "NOW \${0##*/} >>> ${0##*/}"
if test -n "$BASH"; then
echo "RUNNING IN BASH..."
SH_FILE_RUN_PATH_GETSELFPATHVAR=${BASH_SOURCE[0]}
elif test -n "$ZSH_NAME"; then
echo "RUNNING IN ZSH..."
SH_FILE_RUN_PATH_GETSELFPATHVAR=${(%):-%x}
elif test -n "$KSH_VERSION"; then
echo "RUNNING IN KSH..."
SH_FILE_RUN_PATH_GETSELFPATHVAR=${.sh.file}
else
echo "RUNNING IN DASH OR OTHERS ELSE..."
SH_FILE_RUN_PATH_GETSELFPATHVAR=$(lsof -p $$ -Fn0 | tr -d '\0' | grep "${0##*/}" | tail -1 | sed 's/^[^\/]*//g')
fi
echo "EXECUTING FILE PATH: $SH_FILE_RUN_PATH_GETSELFPATHVAR"
cd "$(dirname "$SH_FILE_RUN_PATH_GETSELFPATHVAR")" || return 1
SH_FILE_RUN_BASENAME_GETSELFPATHVAR=$(basename "$SH_FILE_RUN_PATH_GETSELFPATHVAR")
# Iterate down a (possible) chain of symlinks as lsof of macOS doesn't have -f option.
while [ -L "$SH_FILE_RUN_BASENAME_GETSELFPATHVAR" ]; do
SH_FILE_REAL_PATH_GETSELFPATHVAR=$(readlink "$SH_FILE_RUN_BASENAME_GETSELFPATHVAR")
cd "$(dirname "$SH_FILE_REAL_PATH_GETSELFPATHVAR")" || return 1
SH_FILE_RUN_BASENAME_GETSELFPATHVAR=$(basename "$SH_FILE_REAL_PATH_GETSELFPATHVAR")
done
# Compute the canonicalized name by finding the physical path
# for the directory we're in and appending the target file.
SH_SELF_PATH_DIR_RESULT=$(pwd -P)
SH_FILE_REAL_PATH_GETSELFPATHVAR=$SH_SELF_PATH_DIR_RESULT/$SH_FILE_RUN_BASENAME_GETSELFPATHVAR
echo "EXECUTING REAL PATH: $SH_FILE_REAL_PATH_GETSELFPATHVAR"
echo "EXECUTING FILE DIR: $SH_SELF_PATH_DIR_RESULT"
cd "$ORIGINAL_PWD_GETSELFPATHVAR" || return 1
unset ORIGINAL_PWD_GETSELFPATHVAR
unset SH_FILE_RUN_PATH_GETSELFPATHVAR
unset SH_FILE_RUN_BASENAME_GETSELFPATHVAR
unset SH_FILE_REAL_PATH_GETSELFPATHVAR
echo "---------------- GET SELF PATH ----------------"
# USE $SH_SELF_PATH_DIR_RESULT BEBLOW
I have tried $0 before, namely:
dirname $0
and it just returns "." even when the script is being sourced by another script:
. ../somedir/somescript.sh