KSH 脚本检查 case 语句中的字母数字

Question

下面是我要实现的简化模型：

#!bin/ksh
string=AUS00
case $string in
[[:alnum:]] ) echo "alphanumeric" ;;
*) echo "nope" ;;
esac

我无法验证字母数字代码。

约束条件：

验证需要在 case 语句中进行
不支持 alnum 函数
仅阳性检查。无法检查是否缺少字母数字。

非常感谢

Answer 1

模式 [[:alnum:]] 将匹配 单个字母数字字符 。您的字符串超过一个字符，因此不匹配。

如果你想检查你的字符串包含一个字母字符，你需要 *[[:alnum:]]*

如果你想检查你的字符串 只包含 个 alnum 字符，我会翻转检查以查看字符串是否包含非 alnum 字符：

for string in alnumOnly 'not all alnum'; do
    case "$string" in
        *[^[:alnum:]]*) echo "$string -> nope" ;;
        *)              echo "$string -> alphanumeric" ;;
    esac
done

alnumOnly -> alphanumeric
not all alnum -> nope

我意识到 ksh（甚至 ksh88）实现了 bash 所描述的 "extended patterns":

A pattern-list is a list of one or more patterns separated from each other with a |. Composite patterns can be formed with one or more of the following:

?(pattern-list) Optionally matches any one of the given patterns.

*(pattern-list) Matches zero or more occurrences of the given patterns.

+(pattern-list) Matches one or more occurrences of the given patterns.

@(pattern-list) Matches exactly one of the given patterns.

!(pattern-list) Matches anything, except one of the given patterns.

所以我们可以这样做：

case "$string" in
    +([[:alnum:]]) ) echo "$string -> alphanumeric" ;;
    *              ) echo "string -> nope" ;;
esac

KSH 脚本检查 case 语句中的字母数字

KSH script checks for alphanumeric in case statement

unix

ksh

case