如何检查BASH列表中是否存在变量 [英] How do I check if a variable exists in a list in BASH

问题描述

我正在尝试用bash编写脚本来检查用户输入的有效性.
我想将输入(例如变量x)与有效值列表匹配.

I am trying to write a script in bash that check the validity of a user input.
I want to match the input (say variable x) to a list of valid values.

我现在想出的是:

for item in $list
do
    if [ "$x" == "$item" ]; then
        echo "In the list"
        exit
    fi
done

我的问题是，是否有更简单的方法来做到这一点，
对于大多数编程语言来说，它类似于list.contains(x).

My question is if there is a simpler way to do this,
something like a list.contains(x) for most programming languages.

添加:
说的是:

Addition:
Say list is:

list="11 22 33"

我的代码将仅针对这些值回显消息，因为list被视为数组而不是字符串， 所有的字符串操作都将验证1，而我希望它失败.

my code will echo the message only for those values since list is treated as an array and not a string, all the string manipulations will validate 1 while I would want it to fail.

推荐答案

[[ $list =~ (^|[[:space:]])$x($|[[:space:]]) ]] && echo 'yes' || echo 'no'

或创建一个函数:

contains() {
    [[ $1 =~ (^|[[:space:]])$2($|[[:space:]]) ]] && exit(0) || exit(1)
}

使用它:

contains aList anItem
echo $? # 0： match, 1: failed

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