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

试试我们新的Bash库产品realpath-lib吧,我们已经把它放在GitHub上了,可以免费、不受阻碍地使用。它有完整的文档,是一个很好的学习工具。

它可以解析本地、相对和绝对路径,除了Bash 4+,它没有任何依赖关系;所以它应该适用于任何地方。它是免费的,干净的,简单的和有教育意义的。

你可以:

get_realpath <absolute|relative|symlink|local file path>

这个函数是库的核心:

function get_realpath() {

if [[ -f "$1" ]]
then 
    # file *must* exist
    if cd "$(echo "${1%/*}")" &>/dev/null
    then 
        # file *may* not be local
        # exception is ./file.ext
        # try 'cd .; cd -;' *works!*
        local tmppwd="$PWD"
        cd - &>/dev/null
    else 
        # file *must* be local
        local tmppwd="$PWD"
    fi
else 
    # file *cannot* exist
    return 1 # failure
fi

# reassemble realpath
echo "$tmppwd"/"${1##*/}"
return 0 # success

}

它还包含get_dirname, get_filename, get_ stemname和validate_path函数。在不同平台上尝试它,并帮助改进它。

其他回答

我知道这是一个古老的问题。我仍在提供另一种选择。最近我遇到了同样的问题,并且发现没有现有的可移植命令来执行此操作。因此,我编写了下面的shell脚本,其中包括一个可以实现此功能的函数。

#! /bin/sh                                                                                                                                                

function normalize {
  local rc=0
  local ret

  if [ $# -gt 0 ] ; then
    # invalid
    if [ "x`echo $1 | grep -E '^/\.\.'`" != "x" ] ; then
      echo $1
      return -1
    fi

    # convert to absolute path
    if [ "x`echo $1 | grep -E '^\/'`" == "x" ] ; then
      normalize "`pwd`/$1"
      return $?
    fi

    ret=`echo $1 | sed 's;/\.\($\|/\);/;g' | sed 's;/[^/]*[^/.]\+[^/]*/\.\.\($\|/\);/;g'`
  else
    read line
    normalize "$line"
    return $?
  fi

  if [ "x`echo $ret | grep -E '/\.\.?(/|$)'`" != "x" ] ; then
    ret=`normalize "$ret"`
    rc=$?
  fi

  echo "$ret"
  return $rc
}

https://gist.github.com/bestofsong/8830bdf3e5eb9461d27313c3c282868c

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`

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...

realpath试试。以下是全文来源,特此捐赠给公众领域。

// realpath.c: display the absolute path to a file or directory.
// Adam Liss, August, 2007
// This program is provided "as-is" to the public domain, without express or
// implied warranty, for any non-profit use, provided this notice is maintained.

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <libgen.h>   
#include <limits.h>

static char *s_pMyName;
void usage(void);

int main(int argc, char *argv[])
{
    char
        sPath[PATH_MAX];


    s_pMyName = strdup(basename(argv[0]));

    if (argc < 2)
        usage();

    printf("%s\n", realpath(argv[1], sPath));
    return 0;
}    

void usage(void)
{
    fprintf(stderr, "usage: %s PATH\n", s_pMyName);
    exit(1);
}

不完全是一个答案,但可能是一个后续问题(最初的问题不明确):

如果你真的想遵循符号链接,Readlink是很好的。但也有一个用例仅仅是正常化。/和../和//序列,这可以纯语法地完成,而不需要规范化符号链接。Readlink做不到这一点,realpath也不行。

for f in $paths; do (cd $f; pwd); done

适用于现有路径,但不适用于其他路径。

sed脚本似乎是一个不错的选择,除了不能迭代地替换序列(/foo/bar/baz/../..)- > / foo / bar / . .-> /foo)没有使用像Perl这样的东西,这在所有系统上都是不安全的,或者使用一些丑陋的循环来比较sed的输出和它的输入。

FWIW,使用Java (JDK 6+)的一行程序:

jrunscript -e 'for (var i = 0; i < arguments.length; i++) {println(new java.io.File(new java.io.File(arguments[i]).toURI().normalize()))}' $paths