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

realpath的问题是它在BSD(或OSX)上不可用。下面是一个简单的配方,摘自Linux Journal上一篇相当古老(2009年)的文章,它是相当可移植的:

function normpath() {
  # Remove all /./ sequences.
  local path=${1//\/.\//\/}

  # Remove dir/.. sequences.
  while [[ $path =~ ([^/][^/]*/\.\./) ]]; do
    path=${path/${BASH_REMATCH[0]}/}
  done
  echo $path
}

注意,这个变体也不需要路径存在。

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

一个使用node.js的简单解决方案:

#!/usr/bin/env node
process.stdout.write(require('path').resolve(process.argv[2]));

我做了一个内置函数来处理这个问题,并专注于最高的性能(为了好玩)。它不解析符号链接,因此基本上与realpath -sm相同。

## A bash-only mimic of `realpath -sm`. 
## Give it path[s] as argument[s] and it will convert them to clean absolute paths
abspath () { 
  ${*+false} && { >&2 echo $FUNCNAME: missing operand; return 1; };
  local c s p IFS='/';  ## path chunk, absolute path, input path, IFS for splitting paths into chunks
  local -i r=0;         ## return value

  for p in "$@"; do
    case "$p" in        ## Check for leading backslashes, identify relative/absolute path
    '') ((r|=1)); continue;;
    //[!/]*)  >&2 echo "paths =~ ^//[^/]* are impl-defined; not my problem"; ((r|=2)); continue;;
    /*) ;;
    *)  p="$PWD/$p";;   ## Prepend the current directory to form an absolute path
    esac

    s='';
    for c in $p; do     ## Let IFS split the path at '/'s
      case $c in        ### NOTE: IFS is '/'; so no quotes needed here
      ''|.) ;;          ## Skip duplicate '/'s and '/./'s
      ..) s="${s%/*}";; ## Trim the previous addition to the absolute path string
      *)  s+=/$c;;      ### NOTE: No quotes here intentionally. They make no difference, it seems
      esac;
    done;

    echo "${s:-/}";     ## If xpg_echo is set, use `echo -E` or `printf $'%s\n'` instead
  done
  return $r;
}

注意:这个函数不处理以//开头的路径,因为路径开头的两个双斜杠是实现定义的行为。但是,它可以很好地处理/、///等等。

这个函数似乎正确地处理了所有的边缘情况,但可能还有一些我没有处理的情况。

性能注意:当调用数千个参数时,abspath运行速度比realpath -sm慢10倍左右;当使用单个参数调用abspath时,在我的机器上,abspath的运行速度比realpath -sm快110倍,这主要是因为不需要每次都执行新程序。

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