如何在没有 readlink 或 realpath 的情况下递归解析符号链接？

Question

如果 readlink 和 realpath 不可用，最好的可移植（POSIX？）编写脚本查找 link 目标的方法是什么？

你会 ls -l，如果它以 l 开头，则将 -> 之后的文本替换为 sed 并重复直到它不再以 l 开头=] ?

Answer 1

限制：-printf 不是 POSIX 指定的选项

#!/bin/bash

tmp=<symlink-name>
tmp1=''
while tmp=$(find "$tmp" -prune -printf "%l" 2>/dev/null); do
    target="$tmp1" && tmp1="$tmp"
done;
echo "$target"

Answer 2

每 BashFAQ #29 (which also endorses the GNU find approach ):

一个广泛使用的（虽然不是纯POSIX）选项是使用perl:

target=/path/to/symlink-name perl -le 'print readlink $ENV{target}'

如果你的符号链接的名称保证不包含->，你可以解析ls的输出。

下面的代码结合了这两种方法：

# define the best readlink function available for this platform
if command -v readlink >/dev/null 2>/dev/null; then
  # first choice: Use the real readlink command
  readlink() {
    command readlink -- "$@"
  }
elif find . -maxdepth 0 -printf '%l' >/dev/null 2>/dev/null; then
  # second choice: use GNU find
  readlink() {
    local ll candidate >/dev/null 2>&1 ||:
    if candidate=$(find "" -maxdepth 0 -printf '%l') && [ "$candidate" ]; then
      printf '%s\n' "$candidate"
    else
      printf '%s\n' ""
    fi
  }
elif command -v perl >/dev/null 2>/dev/null; then
  # third choice: use perl
  readlink() {
    local candidate ||:
    candidate=$(target= perl -le 'print readlink $ENV{target}')
    if [ "$candidate" ]; then
      printf '%s\n' "$candidate"
    else
      printf '%s\n' ""
    fi
  }
else
  # fourth choice: parse ls -ld
  readlink() {
    local ll candidate >/dev/null 2>&1 ||:
    ll=$(LC_ALL=C ls -ld -- "" 2>/dev/null)
    candidate=${ll#* -> }
    if [ "$candidate" = "$ll" ]; then
      printf '%s\n' ""
    else
      printf '%s\n' "$candidate"
    fi
  }
fi

readlink_recursive() {
    local path prev_path oldwd found_recursion >/dev/null 2>&1 ||:
    oldwd=$PWD; path=; found_recursion=0

    while [ -L "$path" ] && [ "$found_recursion" = 0 ]; do
        if [ "$path" != "${path%/*}" ]; then
          cd -- "${path%/*}" || {
            cd -- "$oldwd" ||:
            echo "ERROR: Directory '${path%/*}' does not exist in '$PWD'" >&2
            return 1
          }
          path=${PWD}/${path##*/}
        fi
        path=$(readlink "$path")
        if [ -d "$path" ]; then
          cd -- "$path"
          path=$PWD
          break
        fi
        if [ "$path" != "${path%/*}" ]; then
          cd -- "${path%/*}" || {
            echo "ERROR: Could not traverse from $PWD to ${path%/*}" >&2
            return 1
          }
          path=${PWD}/${path##*/}
        elif [ "$PWD" != "$oldwd" ]; then
          path=${PWD}/$path
        fi
        for prev_path; do
          if [ "$path" = "$prev_path" ]; then
            found_recursion=1
            break
          fi
        done
        set -- "$path" "$@" # record path for recursion check
    done

    if [ "$path" != "${path%/../*}" ]; then
      cd "${path%/*}" || {
        echo "ERROR: Directory '${path%/*}' does not exist in $PWD" >&2
        return 1
      }
      printf '%s\n' "$PWD/${path##*/}"
    else
      printf '%s\n' "$path"
    fi
    cd -- "$oldwd" ||:
}

Answer 3

这是另一个与 Charles Duffy 的解决方案非常相似的解决方案。我经验不足，因此可能存在一些非 POSIX 或性能问题。在查看了 Charles 的解决方案并替换了我不明白的任何内容后，我得出了这个结论:-P 很可能在解决这里的任何问题后，您最终会再次使用 Charles 的解决方案。

resolve() {
    local arg path absolute ll dir prev_path oldwd found_recursion base >/dev/null 2>&1 ||:
    arg=""; path=""; oldwd=$PWD; found_recursion=0
    dir=$(dirname "$path")
    cd -- "$dir" || {
        cd -- "$oldwd" ||:
        echo "While resolving '$arg' could not go to '$dir'" >&2
        return 1
    }
    if [ $PWD = "/" ]; then
        absolute="/$(basename $path)"
    else
        absolute="$PWD/$(basename $path)"
    fi
    [ "$path" != "$absolute" ] && set -- "$absolute"
    while [ -L "$absolute" ] && [ "$found_recursion" = 0 ]; do
        ll=$(LC_ALL=C \ls -ld -- "$absolute" 2>/dev/null)
        path=${ll#* -> }
        dir=$(dirname "$path")
        cd -- "$dir" || {
            cd -- "$oldwd" ||:
            echo "While resolving '$arg' could not go to '$dir'" >&2
            return 1
        }
        base=$(basename "$path")
        absolute="$PWD/$base"
        for prev_path; do
            if [ "$absolute" = "$prev_path" ]; then
                found_recursion=1
                break
            fi
        done
        set -- "$absolute" "$@"
    done
    if [ -d "$absolute" ]; then
        cd -- "$absolute" || {
            cd -- "$oldwd" ||:
            echo "While resolving '$arg' could not go to '$absolute'" >&2
            return 1
        }
        printf '%s\n' "$PWD"
    else
        printf '%s\n' "$absolute"
    fi
}

编辑：现在使用 $PWD 和 printf，规范目录结果。

如何在没有 readlink 或 realpath 的情况下递归解析符号链接？

How to recursively resolve symlinks without readlink or realpath?

shell

posix

sh