我想转换/foo/bar/../ foo

是否有bash命令可以做到这一点?


编辑:在我的实际案例中,目录确实存在。


当前回答

如果你只想规范化一个路径,不管是否存在,不涉及文件系统,不解析任何链接,也不使用外部utils,这里有一个从Python的posixpath.normpath转换而来的纯Bash函数。

#!/usr/bin/env bash

# Normalize path, eliminating double slashes, etc.
# Usage: new_path="$(normpath "${old_path}")"
# Translated from Python's posixpath.normpath:
# https://github.com/python/cpython/blob/master/Lib/posixpath.py#L337
normpath() {
  local IFS=/ initial_slashes='' comp comps=()
  if [[ $1 == /* ]]; then
    initial_slashes='/'
    [[ $1 == //* && $1 != ///* ]] && initial_slashes='//'
  fi
  for comp in $1; do
    [[ -z ${comp} || ${comp} == '.' ]] && continue
    if [[ ${comp} != '..' || (-z ${initial_slashes} && ${#comps[@]} -eq 0) || (\
      ${#comps[@]} -gt 0 && ${comps[-1]} == '..') ]]; then
      comps+=("${comp}")
    elif ((${#comps[@]})); then
      unset 'comps[-1]'
    fi
  done
  comp="${initial_slashes}${comps[*]}"
  printf '%s\n' "${comp:-.}"
}

例子:

new_path="$(normpath '/foo/bar/..')"
echo "${new_path}"
# /foo

normpath "relative/path/with trailing slashs////"
# relative/path/with trailing slashs

normpath "////a/../lot/././/mess////./here/./../"
# /lot/mess

normpath ""
# .
# (empty path resolved to dot)

Personally, I cannot understand why Shell, a language often used for manipulating files, doesn't offer basic functions to deal with paths. In python, we have nice libraries like os.path or pathlib, which offers a whole bunch of tools to extract filename, extension, basename, path segments, split or join paths, to get absolute or normalized paths, to determine relations between paths, to do everything without much brain. And they take care of edge cases, and they're reliable. In Shell, to do any of these, either we call external executables, or we have to reinvent wheels with these extremely rudimentary and arcane syntaxes...

其他回答

话多,回答有点晚。我需要写一个,因为我卡住了旧的RHEL4/5。 I处理绝对和相对链接,并简化//,/。/和somedir/../条目。

test -x /usr/bin/readlink || readlink () {
        echo $(/bin/ls -l $1 | /bin/cut -d'>' -f 2)
    }


test -x /usr/bin/realpath || realpath () {
    local PATH=/bin:/usr/bin
    local inputpath=$1
    local changemade=1
    while [ $changemade -ne 0 ]
    do
        changemade=0
        local realpath=""
        local token=
        for token in ${inputpath//\// }
        do 
            case $token in
            ""|".") # noop
                ;;
            "..") # up one directory
                changemade=1
                realpath=$(dirname $realpath)
                ;;
            *)
                if [ -h $realpath/$token ] 
                then
                    changemade=1
                    target=`readlink $realpath/$token`
                    if [ "${target:0:1}" = '/' ]
                    then
                        realpath=$target
                    else
                        realpath="$realpath/$target"
                    fi
                else
                    realpath="$realpath/$token"
                fi
                ;;
            esac
        done
        inputpath=$realpath
    done
    echo $realpath
}

mkdir -p /tmp/bar
(cd /tmp ; ln -s /tmp/bar foo; ln -s ../.././usr /tmp/bar/link2usr)
echo `realpath /tmp/foo`

今天我发现可以使用stat命令来解析路径。

对于"~/Documents"这样的目录:

你可以运行这个:

stat -f %N ~/文档

获取完整路径:

/用户/我/文档

对于符号链接,你可以使用%Y格式选项:

stat -f %Y example_symlink

这可能会返回如下结果:

/usr/local/sbin/example_symlink

格式选项可能在*NIX的其他版本上有所不同,但这些在OSX上对我有用。

基于loveborg出色的python代码片段,我这样写:

#!/bin/sh

# Version of readlink that follows links to the end; good for Mac OS X

for file in "$@"; do
  while [ -h "$file" ]; do
    l=`readlink $file`
    case "$l" in
      /*) file="$l";;
      *) file=`dirname "$file"`/"$l"
    esac
  done
  #echo $file
  python -c "import os,sys; print os.path.abspath(sys.argv[1])" "$file"
done

我不知道是否有直接的bash命令可以做到这一点,但我通常会这样做

normalDir="`cd "${dirToNormalize}";pwd`"
echo "${normalDir}"

而且效果很好。

使用coreutils包中的readlink实用程序。

MY_PATH=$(readlink -f "$0")