Bash:检查用户输入是否正确

Question

我写了一个bash脚本,它通过用户输入获取计算数字.我遇到的问题是如果用户错误地输入了字母或空格,整个脚本都会失败,用户必须重新开始.

必须有一种简单的方法来检查输入是否只是数字,如果意外输入任何其他内容,将再次请求输入？

Answer 1

使用while循环

number=""
while [[ ! $number =~ ^[0-9]+$ ]]; do
    echo Please enter your age
    read number
done
echo You are $number years old

Answer 2

要避免繁重的正则表达式引擎,请使用简单的glob:

if [[ ! $input || $input = *[^0-9]* ]]; then
    echo "Error: '$input' is not a number." >&2
fi

Answer 3

允许在数字_33或33___之前和之后留出空格,但不允许空间为3__3.而且没有信件.0或

- 从用户获取输入直到正确

    # -- get input until Correct
    unset get_num
    while [[ ! ${get_num} =~ ^[0-9]+$ ]]; do
        echo "Please enter in a number:"
        read get_num
    done
    echo This is a number :  ${get_num}

- 获取输入直到正确(在范围内)

    # -- get input until Correct (within range)
    unset get_num
    while [[ ! ${get_num} =~ ^[0-9]+$ ]]; do
        echo "Please enter in a number within range of (1-30):"
        read get_num
        ! [[ ${get_num} -ge 1 && ${get_num} -le 30  ]] && unset get_num
    done
    echo This is a number withn a range :  ${get_num}

- 获取输入直到正确(在范围内)(完全正则表达式方式)

"由于正则表达式处理文本而不是数字,因此匹配给定范围内的数字需要额外注意.您不能只写[0-255]来匹配0到255之间的数字.虽然是有效的正则表达式,它匹配完全不同的东西.[0-255]是一个有三个元素的字符类:字符范围0-2,字符5和字符5(再次).这个字符类匹配单个数字0,1,2或5,就像[0125]." ~ http://www.regular-expressions.info/numericranges.html

    # -- get input until Correct (within range) (another way)
    unset get_num
    while [[ ! ${get_num} =~ ^([1-9]|1[0-9]|2[0-9]|30)$ ]]; do
        echo "Please enter in a number within range of (1-30):"
        read get_num
    done
    echo This is a number withn a range :  ${get_num}

- 获取输入,并仅检查它(没有循环)

    # -- get input, and check it only (no while loop)
    unset get_num
    echo "Please enter in a number:"
    read get_num
    if [[ ! ${get_num} =~ ^[0-9]+$ ]] ;then 
        echo "${get_num} isn't a number" 
    else
        echo "${get_num} is a number"
    fi