从 Bash 的数组中选择随机项

Question

我正在用 Shell 脚本创建一个机器人：

# Array with expressions
expressions=("Ploink Poink" "I Need Oil" "Some Bytes are Missing!" "Poink Poink" "Piiiip Beeeep!!" "Hello" "Whoops! I'm out of memmory!")

# Seed random generator
RANDOM=$$$(date +%s)

# Loop loop loop loop loop loop ...
while [ 1 ]
do
    # Get random expression...
    selectedexpression=${expressions[$RANDOM % ${#RANDOM[*]}]}
    
    # Write to Shell
    echo $selectedexpression
    
    
    # Wait an half hour
    sleep 1 # It's one second for debugging, dear SOers
done

我希望它每秒从

expressions

打印一个随机项目。我尝试过这个，但它不起作用。它每次只打印第一个（Ploink Poink）。有人可以帮我吗？

Answer 1

将定义

selectedexpression

的行更改为

selectedexpression=${expressions[ $RANDOM % ${#expressions[@]} ]}

您希望

expression

的索引是从 0 到

expression

数组长度的随机数。这样就可以了。

Answer 2

arr[0]="Ploink Poink"
arr[1]="I Need Oil"
arr[2]="Some Bytes are Missing!"
arr[3]="Poink Poink"
arr[4]="Piiiip Beeeep!!"
arr[5]="Hello"
arr[6]="Whoops! I'm out of memmory!"
rand=$[$RANDOM % ${#arr[@]}]
echo $(date)
echo ${arr[$rand]}

Answer 3

使用shuf的解决方案：

expressions=("Ploink Poink" "I Need Oil" "Some Bytes are Missing!" "Poink Poink" "Piiiip Beeeep!!" "Hello" "Whoops! I'm out of memmory!")
selectedexpression=$(printf "%s\n" "${expressions[@]}" | shuf -n1)
echo $selectedexpression

或者可能更好：

select_random() {
    printf "%s\0" "$@" | shuf -z -n1 | tr -d '\0'
}

expressions=("Ploink Poink" "I Need Oil" "Some Bytes are Missing!" "Poink Poink" "Piiiip Beeeep!!" "Hello" "Whoops! I'm out of memmory!")
selectedexpression=$(select_random "${expressions[@]}")
echo "$selectedexpression"

Answer 4

这是另一个解决方案，可能比 Jacob Mattison 的解决方案更加随机（很难从

jot

手册页中看出）：

declare -a expressions=('Ploink' 'I Need Oil' 'Some Bytes are Missing' 'Poink Poink' 'Piiiip Beeeep' 'Hello' 'Whoops I am out of memory')
index=$( jot -r 1  0 $((${#expressions[@]} - 1)) )
selected_expression=${expressions[index]}

Answer 5

对于随机

1.

rand=("q" "w") 
r=$(shuf -i 0-${#rand[@]} -n 1)
echo ${rand[$r]}

echo `$((1 + $RAND % 5))` // **for number between 1..5**