为何bash中read -p仅能正确读取第一个变量？

我通过教程了解到使用read -p可将用户输入保存为变量，于是编写了一段类似Madlibs的测试脚本，希望能多次将用户输入转为变量后输出。脚本示例如下：

read -p "What is your favorite fruit?" fruit
echo Oh, I see...
sleep 2s

read -p "And what is your mama's name?" mama
echo Amazing...
sleep2s

read -p "And how many years have you been a little bitch?" years
echo Understood...
sleep 2s

echo So let me get this straight...
echo Your favorite fruit is $fruit
echo Your mama's name is $mama
echo And you have been a little bitch for $years years.

当我输入Apples、Martha、3后，输出结果如下：

So let me get this straight...
Your favorite fruit is Apples.
Your mama's name is $mama
And you have been a little bitch for $years

只有第一个变量能正常输出，其余变量均显示变量名。我尝试过修改变量命名和变量引用方式，但问题依旧。请问我是否忽略了某些细节，或是对read命令的变量处理逻辑存在误解？

问题出在脚本里的sleep2s这一行——sleep命令和它的参数之间必须有空格，写成sleep2s会被系统识别成一个不存在的命令，脚本执行到这里就会报错终止，后面的read命令根本没机会运行，mama和years变量自然没有被赋值，输出时就会直接显示变量名而非输入内容。

修正后的脚本只需要把错误的sleep2s改成sleep 2s即可：

read -p "What is your favorite fruit?" fruit
echo Oh, I see...
sleep 2s

read -p "And what is your mama's name?" mama
echo Amazing...
sleep 2s

read -p "And how many years have you been a little bitch?" years
echo Understood...
sleep 2s

echo So let me get this straight...
echo Your favorite fruit is $fruit
echo Your mama's name is $mama
echo And you have been a little bitch for $years years.

内容的提问来源于stack exchange，提问作者mvd