Implementing which function in shell [duplicate] - bash

Question

This question already has answers here:
Bash script syntax error "do"?
(3 answers)
Closed 7 years ago.
I'm trying to implement Unix's which function, but keep getting syntax errors, on what I (think) is legal? This is my implementation:
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
I keep getting the following error
$ bash which.sh
: command not found:
'which.sh: line 5: syntax error near unexpected token `do
'which.sh: line 5: `do

Answer 1

Your script has DOS newlines. Use dos2unix to convert it, or open it in an editor that can do the conversion for you (in vim, you would run :set fileformat=unix and then save with :w).
$ bash which.sh
: command not found:
'which.sh: line 5: syntax error near unexpected token `do
'which.sh: line 5: `do
See the 's at the beginning of those lines? Those are supposed to be at the end of the line.
What's happening, however, is that your dos have a hidden $'\r' character after them, which sends the cursor back to the beginning of the line. Thus, instead of seeing do as a valid token, or correctly printing
# 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'
...we get...
# this is the error you get when your "do" is really a $'do\r'
'yntax error near unexpected token `do
...because a carriage return is sitting between the do and the '.