用來檢測輸入的選項$1是否在PATH中的shell指令碼_linux shell

今天無意中發現一本挺有意思的shell編程的書，是e文的，內容是101個shell案例，堅持明天看一個，寫點心得。
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 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

這指令碼目的是用來檢測輸入的選項$1是否在PATH中。


這指令碼有幾個地方值得注意的：
1）它運用了函數嵌套，在checkForCmdInPath裡嵌套了in_path函數。
2）if [ "${var%${var#?}}" = "/" ] 這語句中的${var%${var#?}}是顯示變數的第一個字元，也可以用${varname:1:1} 或$(echo $var | cut -c1)替代。
3） elif ! in_path $var $PATH ; then 這意思是如果in_path $var $PATH 執行結果不為0的話則
問題：
發現輸入 echo ， echo_err， /etco_err 都返回正確結果，但輸入 /etc/echo_right (存在著執行檔案但不在PATH中)卻返回found in PATH。我想這指令碼還有需要完善的地方。