如何将默认值传递给Shell脚本中的变量？

Question

I have written this piece of code to move one directory to another. What I want with this code is as follows:

1. User will give a file name or a directory name.
2. User may give a destination folder. If user doesn't want a destination folder he/she will just press enter.
3. Then source directory will be copied to destination directory(if given) or a default directory(if destination not given)

Here is the code.

$source_folder
$destination_folder
$destination
read -r directory_source
read -r destination_folder
if [ "$destination_folder" = ""]
then
    $destination = "/var/www/html/"
else
    $destination = "$destination_folder"
fi
cp -a "$source_folder" "$destination"

Here is the input to this program:

Myfiles.sh (called the shell script in terminal)
Max Sum (This is the source directory)
(Pressed enter i.e. No destination given)

It gives following error:

/usr/local/myfiles/copyDirToHTML.sh: line 6: [: : unary operator expected
/usr/local/myfiles/copyDirToHTML.sh: line 10: =: command not found
cp: cannot stat ‘’: No such file or directory

Answer 1

To solve your problem

Change

if [ "$destination_folder" = ""]

to

if [ "$destination_folder" = "" ]

and change read -r directory_source to

read -r source_folder

Also you can use the below script. Passing arguments from cmd line

#!/bin/sh
source_folder=$1

if [ $# -eq 2 ]; then
    destination=$2
else
    destination="/var/www/html/"
fi

cp -a "$source_folder" "$destination"

Where $# is the number of arguments to the script.
$1 - > first argument...Similarly $2..

To run

./script.sh source_folder [destination]

destination is optional

Answer 2

I have got a solution to this problem. The working code is like this:

$source_folder
$destination_folder
read -r source_folder
read -r destination_folder
if [ "$destination_folder" = "" ]; then
   sudo cp -a "$source_folder" "/var/www/html/"
else
   sudo cp -a "$source_folder" "$destination_folder"
fi