PATH中的shell脚本是否可以用来检测输入的选项$1，相信很多没有经验的人对此束手无策，为此本文总结了问题出现的原因和解决方法，通过这篇文章希望你能解决这个问题。
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。
看完上述内容，你们掌握PATH中的shell脚本是否可以用来检测输入的选项$1的方法了吗？如果还想学到更多技能或想了解更多相关内容，欢迎关注创新互联行业资讯频道，感谢各位的阅读！
