用来检测输入的选项$1是否在PATH中的shell脚本

今天无意中发现一本挺有意思的shell编程的书，是e文的，内容是101个shell案例，坚持明天看一个，写点心得。
下面是例子001：

#!/bin/sh
# inpath - verifies that a specified program is either valid as is,
#  or that it can be found in the path directory list.

in_path()
{
 # given a command and the path, try to find the command. returns
 # 0 if found and executable, 1 if not. note that this temporarily modifies
 # the ifs (input field separator) but restores it upon completion.

 cmd=$1    path=$2     retval=1
 oldifs=$ifs  ifs=":"

 for directory in $path
 do
  if [ -x $directory/$cmd ] ; then
   retval=0   # if we're here, we found $cmd in $directory
  fi
 done
 ifs=$oldifs
 return $retval
}

checkforcmdinpath()
{
 var=$1

 # the variable slicing notation in the following conditional
 # needs some explanation: ${var#expr} returns everything after
 # the match for 'expr' in the variable value (if any), and
 # ${var%expr} returns everything that doesn't match (in this
 # case, just the very first character. you can also do this in
 # bash with ${var:0:1}, and you could use cut too: cut -c1.

 if [ "$var" != "" ] ; then
  if [ "${var%${var#?}}" = "/" ] ; then
   if [ ! -x $var ] ; then
    return 1
   fi
  elif ! in_path $var $path ; then
   return 2
  fi
 fi
}

 
if [ $# -ne 1 ] ; then
 echo "usage: $0 command" >&2 ; exit 1
fi

checkforcmdinpath "$1"
case $? in
 0 ) echo "$1 found in path"         ;;
 1 ) echo "$1 not found or not executable"  ;;
 2 ) echo "$1 not found in path"       ;;
esac

exit 0

这脚本目的是用来检测输入的选项$1是否在path中。


这脚本有几个地方值得注意的：
1）它运用了函数嵌套，在checkforcmdinpath里嵌套了in_path函数。
2）if [ "${var%${var#?}}" = "/" ] 这语句中的${var%${var#?}}是显示变量的第一个字符，也可以用${varname:1:1} 或$(echo $var | cut -c1)替代。
3） elif ! in_path $var $path ; then 这意思是如果in_path $var $path 执行结果不为0的话则
问题：
发现输入 echo ， echo_err， /etco_err 都返回正确结果，但输入 /etc/echo_right (存在着执行文件但不在path中)却返回found in path。我想这脚本还有需要完善的地方。