检测输入 用来检测输入的选项$1是否在PATH中的shell脚本
人气:0想了解用来检测输入的选项$1是否在PATH中的shell脚本的相关内容吗,在本文为您仔细讲解检测输入的相关知识和一些Code实例,欢迎阅读和指正,我们先划重点:检测输入,下面大家一起来学习吧。
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。
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