CASE 在调试和正常模式下的行为不同[重复]

Question

我编写了一个脚本，其中“CASE”部分识别输入的字节/字符是 Lower、Upper、digit 还是某些特殊字符。该程序运行良好，但奇怪的是；当我输入字母“A”，然后运行 ​​

sh -x ./scrip "A"

输出

+ [ 1 -ne 1 ]
+ char=A
+ wc -c
+ echo A
+ NumOfChars=2
+ [ 2 -gt 2 ]
+ echo  Arguments are correct in numbers 
 Arguments are correct in numbers 
+ echo Upper case alphabet
Upper case alphabet

输出正确（即执行[A-Z]的情况

但是当我执行程序时是正常的方式，即

./script "A"

输出

Lower case alphabet

CASE [a-z] 被执行了，为什么？

脚本如下

if [ "$#" -ne 1 ]; then
    echo "Number of arguments are wrong"
    exit 1;

else
    char="$1"
    NumOfChars=$(echo "$char" | wc -c)
    if [ "$NumOfChars" -gt 2 ]; then
        echo "Number of characters are greater then one"
        exit 2;
    else
        echo " Arguments are correct in numbers ";
    fi
fi
case "$char"
in

[a-z] ) echo "Lower case alphabet";;
[A-Z] ) echo "Upper case alphabet";;
[0-9] ) echo "Digit";;
* )     echo "Non AlphaNumeric characters/byte";;
esac

Answer 1

所以你被两个不同的问题所困扰：

• 由于缺少 shebang，当您使用 sh ./script.sh resp 运行时，您的脚本将/可以由 不同的 shell 实现执行。 ./script.sh

• bash（以./script.sh 运行时恰好执行脚本）可以将A 匹配到特定语言环境的[a-z]。

解决方法是同时指定 shebang 和整理顺序：

#!/bin/sh
export LC_COLLATE="C"

# the rest of the script is unchanged.
if [ "$#" -ne 1 ]; then
    echo "Number of arguments are wrong"
    exit 1;

else
    char="$1"
    NumOfChars=$(echo "$char" | wc -c)
    if [ "$NumOfChars" -gt 2 ]; then
        echo "Number of characters are greater then one"
        exit 2;
    else
        echo " Arguments are correct in numbers ";
    fi
fi
case "$char"
in

[a-z] ) echo "Lower case alphabet";;
[A-Z] ) echo "Upper case alphabet";;
[0-9] ) echo "Digit";;
* )     echo "Non AlphaNumeric characters/byte";;
esac