实现shell中的哪个函数

Question

我正试图实现Unix的which函数，但不断收到语法错误，我认为是合法的？这是我的实现：

IFS=":" 
x=false 

for i in $* 
do 
    for j in $PATH 
    do 
     if [ -x "${j}/$i" ];then 
      echo $j/$i 
      x=true 
      break 
     fi 
    done 
    if [ $x == false ]; then 
     echo my_which $i not found in --$PATH-- 

    fi 
    x=false 

done 

我不断收到以下错误

$ bash which.sh 
: command not found: 
'which.sh: line 5: syntax error near unexpected token `do 
'which.sh: line 5: `do 

Answer 1

你的脚本有DOS换行符。使用dos2unix来转换它，或者在可以为你做转换的编辑器中打开它（在vim中，你将运行:set fileformat=unix，然后使用:w保存）。

$ bash which.sh 
: command not found: 
'which.sh: line 5: syntax error near unexpected token `do 
'which.sh: line 5: `do 

查看' s在这些行的开头？这些应该在该行的末端处。

发生了什么事，但是，那是你do■找他们之后隐藏$'\r'性格，这将光标回行的开头。因此，而不是看到do为有效的令牌，或者正确打印

# this is the error you would get if your "do" were really a "do", but it were still 
# ...somehow bad syntax. 
syntax error near unexpected token `do' 

...我们得到了...

# this is the error you get when your "do" is really a $'do\r' 
'yntax error near unexpected token `do 

...因为回车是do和坐在之间'。