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=$1path=$2 retval=1 oldIFS=$IFSIFS=":" for directory in $path doif [ -x $directory/$cmd ] ; then retval=0 # if we"re here, we found $cmd in $directoryfi done IFS=$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" ！= "" ] ; thenif [ "${var%${var#?}}" = "/" ] ; then if [ ！ -x $var ] ; thenreturn 1 fielif ！ in_path $var $PATH ; then return 2fi fi} if [ $# -ne 1 ] ; then echo "Usage: $0 command" >&2 ; exit 1ficheckForCmdInPath "$1"case $? in 0 ） echo "$1 found in PATH" ;; 1 ） echo "$1 not found or not executable";; 2 ） echo "$1 not found in PATH" ;;esacexit 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。