Script Parameters in Bash

Script parameters in Bash

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.

Bash script passing parameter with spaces using variables

You should change CMD="curl -H $HEADER $URL" in CMD="curl -H \"$HEADER\" \"$URL\"".

How can I write a bash script that takes parameters in any given order?

Here's the script I wrote to achieve this:

#!/bin/bash

# The code below was written to replace the -print and -delete options for
# -p and -d respectively, because getopts can't handle long args, I suggest
# you only use arguments of single letters to make your code more simple, but if 
# you can't avoid it then this is a workaround

for ARGS in "$@"; do
shift
        case "$ARGS" in
                "-print") set -- "$@" "-p" ;;
                "-delete") set -- "$@" "-d" ;;
                *) set -- "$@" "$ARGS"
        esac
done
# getopts works like this: you put all your arguments between single quotes
# if you put a ':' character after the argument letter (just like I did with
# the 'f' letter in the example below), it means this argument NEEDS an extra
# parameter. If you just use letters without the ':' it means it doesn't need
# anything but the argument itself (it's what I did for the '-p' and '-d' options)

while getopts 'f:pd' flag; do
        case "${flag}" in
                f) FILE=${OPTARG} ;;
                p) COMMAND="print" ;;
                d) COMMAND="delete" ;;
        esac
done

echo "$COMMAND $FILE"

And below examples of it running:

$ ./script.sh -f filename -print
print filename
$ ./script.sh -print -f filename
print filename
$ ./script.sh -f filename -delete
delete filename
$ ./script.sh -delete -f filename
delete filename

Bash script with both parameter values and optional flags

You should replace getopts u:a:f:h: with getopts u:a:f:h.

Removing the : you tell getopts that h has no additional argument.

How to pass Named Parameter in Shell Script

You may want to read the getopts manual page

    while getopts "U:P:J:I:" flag
    do
             case "${flag}" in
                    U) TEST_USER=${OPTARG};;
                    P) TEST_PWD=${OPTARG};;
                    J) TEST_JOBID=${OPTARG};;
                    I) TEST_PROJECTID=${OPTARG};;
             esac
    done
    echo "USER: $TEST_USER";
    echo "PWD: $TEST_PWD";
    echo "JOBID: $TEST_JOBID";
    echo "PROJECTID: $TEST_PROJECTID";

./getopts.sh -U devops@gmail.com -P xxxxxx -J 8a809e2496 -I 80e2ea54b231f

USER: devops@gmail.com
PWD: xxxxxx
JOBID: 8a809e2496
PROJECTID: 80e2ea54b231f

With getopt, it's more complicated, but it allows long options and "=".

    #!/bin/bash
    TEMP=$(getopt -n "$0" -a -l "user:,password:,jobid:,projectid:" -- -- "$@")

    [ $? -eq 0 ] || exit

    eval set --  "$TEMP"

    while [ $# -gt 0 ]
    do
             case "$1" in
                    --user) TEST_USER="$2"; shift;;
                    --password) TEST_PWD="$2"; shift;;
                    --jobid) TEST_JOBID="$2"; shift;;
                    --projectid) TEST_PROJECTID="$2"; shift;;
                    --) shift;;
             esac
             shift;
    done
    echo "USER: $TEST_USER";
    echo "PWD: $TEST_PWD";
    echo "JOBID: $TEST_JOBID";
    echo "PROJECTID: $TEST_PROJECTID";

Some tests:

$ ./test.sh -user=jules -password=kabas -jobid 5555 -projectid 999 -c
./test.sh: unrecognized option '-c'

$ ./test.sh -user=jules -password=kabas -jobid 5555 -projectid 999   
USER: jules
PWD: kabas
JOBID: 5555
PROJECTID: 999

Shell how to make some of the script arguments mandatory

I made this work by checking that the variables req and out have some value after the for loop with the below code:

  if [[ -z $out  ||  -z $req ]]
  then
    echo "ERROR: -q  and -o are mandatory arguments. See usage: \n";
    exit 1;
  fi

wget bash script with parameters

The standard format for the bash (or sh or similar) command is bash scriptfilename arg1 arg2 .... If you leave off all the first argument (the name or path of the script to run), it reads the script from stdin. Unfortunately, there's no way to leave off the firs argument but pass the others. Fortunately, you can pass /dev/stdin as the first argument and get the same effect (at least on most unix systems):

wget -O - https://myserver/install/Setup.sh | bash /dev/stdin parameter1

If you're on a system that doesn't have /dev/stdin, you might have to look around for an alternative way to specify stdin explicitly (/dev/fd/0 or something like that).

Edit: Léa Gris suggestion of bash -s arg1 arg2 ... is probably a better way to do this.

How to pass two parameters or arguments in bash scripting

As a start you can do this:

#!/bin/bash
dayoffset=$1
processMode=$2
echo "Do something with $dayoffset and $processMode."

Usage:

./start.sh 1 true

Another:

#!/bin/bash

while [[ $# -gt 0 ]]; do
    case "$1" in
    -dayoffset)
        day_offset=$2
        shift
        ;;
    -processMode)
        if [[ $2 != true && $2 != false ]]; then
            echo "Option argument to '-processMode' can only be 'true' or 'false'."
            exit 1
        fi
        process_mode=$2
        shift
        ;;
    *)
        echo "Invalid argument: $1"
        exit 1
    esac
    shift
done

echo "Do something with $day_offset and $process_mode."

Usage:

./start.sh -dayoffset 1 -processMode true

Example argument parsing with day offset:

#!/bin/bash 
dayoffset=$1
date -d "now + $dayoffset days"

Test:

$ bash script.sh 0
Fri Aug 15 09:44:42 UTC 2014
$ bash script.sh 5
Wed Aug 20 09:44:43 UTC 2014

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