我正在尝试在Ubuntu中编写一个极其简单的脚本,它将允许我传递它一个文件名或目录,并且能够在它是一个文件时做一些特定的事情,当它是一个目录时做一些其他的事情。我遇到的问题是,当目录名,或可能文件,有空格或其他可转义字符在名称中。
下面是我的基本代码,还有几个测试。
#!/bin/bash
PASSED=$1
if [ -d "${PASSED}" ] ; then
echo "$PASSED is a directory";
else
if [ -f "${PASSED}" ]; then
echo "${PASSED} is a file";
else
echo "${PASSED} is not valid";
exit 1
fi
fi
这是输出:
andy@server~ $ ./scripts/testmove.sh /home/andy/
/home/andy/ is a directory
andy@server~ $ ./scripts/testmove.sh /home/andy/blah.txt
/home/andy/blah.txt is a file
andy@server~ $ ./scripts/testmove.sh /home/andy/blah\ with\ a\ space.txt
/home/andy/blah with a space.txt is not valid
andy@server~ $ ./scripts/testmove.sh /home/andy\ with\ a\ space/
/home/andy with a space/ is not valid
所有这些路径都是有效的,并且存在。
至少在编写代码时不要使用这棵浓密的树:
#!/bin/bash
PASSED=$1
if [ -d "${PASSED}" ]
then echo "${PASSED} is a directory";
elif [ -f "${PASSED}" ]
then echo "${PASSED} is a file";
else echo "${PASSED} is not valid";
exit 1
fi
当我把它放入一个文件“xx.sh”,并创建一个文件“xx sh”,并运行它,我得到:
$ cp /dev/null "xx sh"
$ for file in . xx*; do sh "$file"; done
. is a directory
xx sh is a file
xx.sh is a file
$
如果您遇到了问题,您应该通过添加以下内容来调试脚本:
ls -ld "${PASSED}"
这将显示ls对传递脚本的名称的想法。
使用"file"命令可能会有用:
#!/bin/bash
check_file(){
if [ -z "${1}" ] ;then
echo "Please input something"
return;
fi
f="${1}"
result="$(file $f)"
if [[ $result == *"cannot open"* ]] ;then
echo "NO FILE FOUND ($result) ";
elif [[ $result == *"directory"* ]] ;then
echo "DIRECTORY FOUND ($result) ";
else
echo "FILE FOUND ($result) ";
fi
}
check_file "${1}"
输出示例:
$ ./f.bash login
DIRECTORY FOUND (login: directory)
$ ./f.bash ldasdas
NO FILE FOUND (ldasdas: cannot open `ldasdas' (No such file or directory))
$ ./f.bash evil.php
FILE FOUND (evil.php: PHP script, ASCII text)
供你参考:上面的答案是有效的,但你可以使用-s来帮助在奇怪的情况下,首先检查一个有效的文件:
#!/bin/bash
check_file(){
local file="${1}"
[[ -s "${file}" ]] || { echo "is not valid"; return; }
[[ -d "${file}" ]] && { echo "is a directory"; return; }
[[ -f "${file}" ]] && { echo "is a file"; return; }
}
check_file ${1}