在Bash中检查传递的参数是否为文件或目录

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。

当前回答

使用统计

function delete_dir () {
  type="$(stat --printf=%F "$1")"
  if [ $? -ne 0 ]; then
    echo "$1 directory does not exist. Nothing to delete."
  elif [ "$type" == "regular file" ]; then
    echo "$1 is a file, not a directory."
    exit 1
  elif [ "$type" == "directory" ]; then
    echo "Deleting $1 directory."
    rm -r "$1"
  fi
}

function delete_file () {
  type="$(stat --printf=%F "$1")"
  if [ $? -ne 0 ]; then
    echo "$1 file does not exist. Nothing to delete."
  elif [ "$type" == "directory" ]; then
    echo "$1 is a regular file, not a directory."
    exit 1
  elif [ "$type" == "regular file" ]; then
    echo "Deleting $1 regular file."
    rm "$1"
  fi
}

https://linux.die.net/man/2/stat https://en.m.wikipedia.org/wiki/Unix_file_types

2021-11-26 10:11:37

其他回答

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。

2011-01-12 03:31:42

这应该有用。我不知道为什么会失败。你正确地引用了变量。如果使用双[[]]，会发生什么?

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

双方括号是对[]的bash扩展。它不要求变量被引用，即使它们包含空格。

同样值得尝试:-e测试路径是否存在，而不测试它是什么类型的文件。

2011-01-12 03:27:31

使用统计

function delete_dir () {
  type="$(stat --printf=%F "$1")"
  if [ $? -ne 0 ]; then
    echo "$1 directory does not exist. Nothing to delete."
  elif [ "$type" == "regular file" ]; then
    echo "$1 is a file, not a directory."
    exit 1
  elif [ "$type" == "directory" ]; then
    echo "Deleting $1 directory."
    rm -r "$1"
  fi
}

function delete_file () {
  type="$(stat --printf=%F "$1")"
  if [ $? -ne 0 ]; then
    echo "$1 file does not exist. Nothing to delete."
  elif [ "$type" == "directory" ]; then
    echo "$1 is a regular file, not a directory."
    exit 1
  elif [ "$type" == "regular file" ]; then
    echo "Deleting $1 regular file."
    rm "$1"
  fi
}

https://linux.die.net/man/2/stat https://en.m.wikipedia.org/wiki/Unix_file_types

2021-11-26 10:11:37

在/bin/test上使用-f和-d开关:

F_NAME="${1}"

if test -f "${F_NAME}"
then                                   
   echo "${F_NAME} is a file"
elif test -d "${F_NAME}"
then
   echo "${F_NAME} is a directory"
else                                   
   echo "${F_NAME} is not valid"
fi

2017-08-26 20:44:29

更优雅的解决方案

echo "Enter the file name"
read x
if [ -f $x ]
then
    echo "This is a regular file"
else
    echo "This is a directory"
fi

2018-08-07 04:44:47

在Bash中检查传递的参数是否为文件或目录

推荐文章

最新文章

标签