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编写 bash 脚本未获得预期的输出

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编写 bash 脚本未获得预期的输出

这是我正在编写的 bash 脚本。您可以看到预期的输出。但我得到了其他的东西。我可能在哪里犯了错误？

#!/bin/bash
    #Demostrate how read actually works
    echo What cars do you like?

    read car1 car2 car3

    echo Your first car was: $car1
    echo Your second car was: $car2
    echo Your third car was: $car3

预期输出

./cars.sh
What cars do you like?
Jaguar Maserati Bentley Lotus
Your first car was: Jaguar
Your second car was: Maserati
Your third car was: Bentley Lotus

实际产量

[root@localhost ~]# ./cars.sh
What cars do you like?
Jaguar
Your first car was: Jaguar
Your second car was:
Your third car was:
[root@localhost ~]#

答案1

第二种情况是，您没有提供所有变量值。您的脚本没有任何问题。

# bash -x cars.sh
# What cars do you like?
# Jaguar Maserati Bentley Lotus

如果仍然遇到问题，请尝试使用 -x 进行调试。

答案2

read -a variable_name如果您有可变数量的输入参数，则可以像本例中那样使用数组变量。

例如：cars.sh

#!/bin/bash

read -a cars -p "What cars do you like? "

echo "You have entered ${#cars[@]} cars"

declare -i count=0
for car in "${cars[@]}"; do
        echo "Car number $(( ++count )) was: $car"
done

输出：

$ ./cars.sh
What cars do you like? bmw audi mercedes
You have entered 3 cars
Car number 1 was: bmw
Car number 2 was: audi
Car number 3 was: mercedes

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