實現shell中的哪個函數

Question

我正試圖實現Unix的which函數，但不斷收到語法錯誤，我認爲是合法的？這是我的實現：

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 

我不斷收到以下錯誤

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

Answer 1

你的腳本有DOS換行符。使用dos2unix來轉換它，或者在可以爲你做轉換的編輯器中打開它（在vim中，你將運行:set fileformat=unix，然後使用: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在這些行的開頭？這些應該在該行的末端處。

發生了什麼事，但是，那是你do■找他們之後隱藏$'\r'性格，這將光標回行的開頭。因此，而不是看到do爲有效的令牌，或者正確打印

# 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' 

...我們得到了...

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

...因爲回車是do和坐在之間'。