CASE在調試和正常模式下表現不同

Question

我寫了一個腳本，其中「CASE」部分標識輸入的字節/字符是Lower，Upper，digit還是一些特殊字符。程序運行良好，但奇怪的是;當我輸入字母「A」，並運行

sh -x ./scrip "A" 

輸出

+ [ 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 

是正確的（即，殼體[AZ]執行

輸出但是，當我執行該程序是正常的方式即

./script "A" 

輸出

Lower case alphabet 

而是執行CASE [a-z]，爲什麼？

的腳本如下

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

所以你是兩個不同的問題咬傷：

• 因爲缺少家當你的腳本將/可以通過不同 shell的實現被執行當與sh ./script.sh分別運行時。 ./script.sh

• 的bash（恰好在作爲./script.sh運行時執行該腳本）可以匹配A到[a-z]特定區域設置。

的解決辦法是指定一個認領和整理序列：

#!/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