I'm trying to write a script where to run the script, the user will type something along the lines of
$./cpc -c test1.txt backup
into the terminal, where ./cpc is to run the script, -c is $option, test1.txt is $source and backup is $destination.
How would I assign the values typed in to the terminal to use them in my script, for example in
if [[ -z $option || -z $source || -z $destination ]]; then
echo "Error: Incorrect number of arguments." (etc)
as when checking the script online the following errors return: 'option/source/destination is referenced but not assigned.'
Sorry in advance if any of this doesn't make sense, I'm trying to be as clear as possible
The arguments are stored in the numbered parameters $1, $2, etc. So, just assign them
option=$1
source=$2
destination=$3
See also man getopt or getopts in man bash.
Related
Is it possible to have the output of commands separate_file.sh be visible to the caller of caller.sh automatically with the following setup?
I'm aware of adding >&2 to each of the commands in separate_file.sh, but I'm looking for a more automatic solution.
I'm also aware that the output is visible if I call separate_file.sh like so: separate_file.sh instead of $(separate_file.sh), but I'd like to preserve the -e option in caller.sh (it's used for other calls).
caller.sh:
#!/bin/bash -e
if [[ ! $(separate_file.sh) ]]; then
echo 'separate_file.sh failed'
fi
separate_file.sh:
#!/bin/bash
echo '1'
ls -la
when I read the book linux shell scripting cookbook
they say when you wanna print !,you shouldn't put it in double quote，or you can add \ before ! to escape it.
e.g.
$echo "Hello,world!"
bash: !:event not found error
$echo "Hello,world\\!"
Hello,world!
but in my situation(ubuntu14.04), I get the answer like that:
$echo "Hello,world!"
Hello,world!
$echo "Hello,world\\!"
Hello,world\!
So, why in my machine can't get the same answer?
Why the escape symbol \ was printed as a normal symbol?
When you're typing interactively to the shell, ! has special meaning, it's the history expansion character. To prevent this special meaning, you need to put it in single quotes or escape it.
echo 'Hello, world!'
echo "Hello, world\!'
The reason it's not happening on Ubuntu may be because it's running a newer version of bash, which is apparently more selective about when history expansion occurs. It seems to require ! to be followed by alphanumerics, not punctuation.
You don't need to do this in scripts, because history is not normally enabled there. It's just for interactive shells.
Create a shell script called file.sh:
#!/bin/bash
# file.sh: a sample shell script to demonstrate the concept of Bash shell functions
# define usage function
usage(){
echo "Usage: $0 filename"
exit 1
}
# define is_file_exits function
# $f -> store argument passed to the script
is_file_exits(){
local f="$1"
[[ -f "$f" ]] && return 0 || return 1
}
# invoke usage
# call usage() function if filename not supplied
[[ $# -eq 0 ]] && usage
# Invoke is_file_exits
if ( is_file_exits "$1" )
then
echo "File found"
else
echo "File not found"
fi
Run it as follows:
chmod +x file.sh
./file.sh
./file.sh /etc/resolv.conf
I wrote a zsh function to help me do some grepping at my job.
function rgrep (){
if [ -n "$1" ] && [ -n "$2" ]
then
exec grep -rnw $1 -r $2
elif [ -n "$1" ]
then
exec grep -rnw $1 -r "./"
else
echo "please enter one or two args"
fi
}
Works great, however, grep finishes executing I don't get thrown back into the shell. it just hangs at [process complete] any ideas?
I have the function in my .zshrc
In addition to getting rid of the unnecessary exec, you can remove the if statement as well.
function rgrep (){
grep -rwn "${1:?please enter one or two args}" -r "${2:-./}"
}
If $1 is not set (or null valued), an error will be raised and the given message displayed. If $2 is not set, a default value of ./ will be used in its place.
Do not use exec as it replace the existing shell.
exec [-cl] [-a name] [command [arguments]]
If command is supplied, it replaces the shell without creating a new process. If the -l option is supplied, the shell places a dash at the beginning of the zeroth argument passed to command. This is what the login program does. The -c option causes command to be executed with an empty environment. If -a is supplied, the shell passes name as the zeroth argument to command. If no command is specified, redirections may be used to affect the current shell environment. If there are no redirection errors, the return status is zero; otherwise the return status is non-zero.
Try this instead:
rgrep ()
{
if [ -n "$1" ] && [ -n "$2" ]
then
grep -rnw "$1" -r "$2"
elif [ -n "$1" ]
then
grep -rnw "$1" -r "./"
else
echo "please enter one or two args"
fi
}
As a completely different approach, I like to build command shortcuts like this as minimal shell scripts, rather than functions (or aliases):
% echo 'grep -rwn "$#"' >rgrep
% chmod +x rgrep
% ./rgrep
Usage: grep [OPTION]... PATTERN [FILE]...
Try `grep --help' for more information.
%
(This relies on a traditional behavior of Unix: executable text files without #! lines are considered shell scripts and are executed by /bin/sh. If that doesn't work on your system, or you need to run specifically under zsh, use an appropriate #! line.)
One of the main benefits of this approach is that shell scripts in a directory in your PATH are full citizens of the environment, not local to the current shell like functions and aliases. This means they can be used in situations where only executable files are viable commands, such as xargs, sudo, or remote invocation via ssh.
This doesn't provide the ability to give default arguments (or not easily, anyway), but IMAO the benefits outweigh the drawbacks. (And in the specific case of defaulting grep to search PWD recursively, the real solution is to install ack.)
I'm trying to make a shell script which should be used like this:
ocrscript.sh -from /home/kristoffer/test.png -to /home/kristoffer/test.txt
The script will then ocr convert the image file to a text file. Here is what I have come up with so far:
#!/bin/bash
export HOME=/home/kristoffer
/usr/local/bin/abbyyocr9 -rl Swedish -if ???fromvalue??? -of ???tovalue??? 2>&1
But I don't know how to get the -from and -to values. Any ideas on how to do it?
The arguments that you provide to a bashscript will appear in the variables $1 and $2 and $3 where the number refers to the argument. $0 is the command itself.
The arguments are seperated by spaces, so if you would provide the -from and -to in the command, they will end up in these variables too, so for this:
./ocrscript.sh -from /home/kristoffer/test.png -to /home/kristoffer/test.txt
You'll get:
$0 # ocrscript.sh
$1 # -from
$2 # /home/kristoffer/test.png
$3 # -to
$4 # /home/kristoffer/test.txt
It might be easier to omit the -from and the -to, like:
ocrscript.sh /home/kristoffer/test.png /home/kristoffer/test.txt
Then you'll have:
$1 # /home/kristoffer/test.png
$2 # /home/kristoffer/test.txt
The downside is that you'll have to supply it in the right order. There are libraries that can make it easier to parse named arguments on the command line, but usually for simple shell scripts you should just use the easy way, if it's no problem.
Then you can do:
/usr/local/bin/abbyyocr9 -rl Swedish -if "$1" -of "$2" 2>&1
The double quotes around the $1 and the $2 are not always necessary but are adviced, because some strings won't work if you don't put them between double quotes.
If you're not completely attached to using "from" and "to" as your option names, it's fairly easy to implement this using getopts:
while getopts f:t: opts; do
case ${opts} in
f) FROM_VAL=${OPTARG} ;;
t) TO_VAL=${OPTARG} ;;
esac
done
getopts is a program that processes command line arguments and conveniently parses them for you.
f:t: specifies that you're expecting 2 parameters that contain values (indicated by the colon). Something like f:t:v says that -v will only be interpreted as a flag.
opts is where the current parameter is stored. The case statement is where you will process this.
${OPTARG} contains the value following the parameter. ${FROM_VAL} for example will get the value /home/kristoffer/test.png if you ran your script like:
ocrscript.sh -f /home/kristoffer/test.png -t /home/kristoffer/test.txt
As the others are suggesting, if this is your first time writing bash scripts you should really read up on some basics. This was just a quick tutorial on how getopts works.
Use the variables "$1", "$2", "$3" and so on to access arguments. To access all of them you can use "$#", or to get the count of arguments $# (might be useful to check for too few or too many arguments).
I needed to make sure that my scripts are entirely portable between various machines, shells and even cygwin versions. Further, my colleagues who were the ones I had to write the scripts for, are programmers, so I ended up using this:
for ((i=1;i<=$#;i++));
do
if [ ${!i} = "-s" ]
then ((i++))
var1=${!i};
elif [ ${!i} = "-log" ];
then ((i++))
logFile=${!i};
elif [ ${!i} = "-x" ];
then ((i++))
var2=${!i};
elif [ ${!i} = "-p" ];
then ((i++))
var3=${!i};
elif [ ${!i} = "-b" ];
then ((i++))
var4=${!i};
elif [ ${!i} = "-l" ];
then ((i++))
var5=${!i};
elif [ ${!i} = "-a" ];
then ((i++))
var6=${!i};
fi
done;
Rationale: I included a launcher.sh script as well, since the whole operation had several steps which were quasi independent on each other (I'm saying "quasi", because even though each script could be run on its own, they were usually all run together), and in two days I found out, that about half of my colleagues, being programmers and all, were too good to be using the launcher file, follow the "usage", or read the HELP which was displayed every time they did something wrong and they were making a mess of the whole thing, running scripts with arguments in the wrong order and complaining that the scripts didn't work properly. Being the choleric I am I decided to overhaul all my scripts to make sure that they are colleague-proof. The code segment above was the first thing.
In bash $1 is the first argument passed to the script, $2 second and so on
/usr/local/bin/abbyyocr9 -rl Swedish -if "$1" -of "$2" 2>&1
So you can use:
./your_script.sh some_source_file.png destination_file.txt
Explanation on double quotes;
consider three scripts:
# foo.sh
bash bar.sh $1
# cat foo2.sh
bash bar.sh "$1"
# bar.sh
echo "1-$1" "2-$2"
Now invoke:
$ bash foo.sh "a b"
1-a 2-b
$ bash foo2.sh "a b"
1-a b 2-
When you invoke foo.sh "a b" then it invokes bar.sh a b (two arguments), and with foo2.sh "a b" it invokes bar.sh "a b" (1 argument). Always have in mind how parameters are passed and expaned in bash, it will save you a lot of headache.
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