给定一个绝对路径或相对路径(在类unix系统中),我希望在解析任何中间符号链接之后确定目标的完整路径。同时还可以解析~用户名符号。

如果目标是一个目录,则可以将chdir()放入该目录,然后调用getcwd(),但我确实希望从shell脚本而不是编写C helper来完成此操作。不幸的是,shell倾向于试图向用户隐藏符号链接的存在(这是OS X上的bash):

$ ls -ld foo bar
drwxr-xr-x   2 greg  greg  68 Aug 11 22:36 bar
lrwxr-xr-x   1 greg  greg   3 Aug 11 22:36 foo -> bar
$ cd foo
$ pwd
/Users/greg/tmp/foo
$

我想要的是一个函数resolve(),这样当从上面的例子中的tmp目录执行时,resolve("foo") == "/Users/greg/tmp/bar"。


当前回答

readlink -e [filepath]

看来这正是你想要的 -它接受任意路径,解析所有符号链接,并返回“真实”路径 -可能所有系统都已经有了“标准*nix”

其他回答

普通shell脚本通常必须找到它们的“主”目录,即使它们是作为符号链接调用的。因此脚本必须从$0开始找到它们的“真实”位置。

cat `mvn`

在我的系统上输出一个包含以下内容的脚本,这应该能很好地提示您需要什么。

if [ -z "$M2_HOME" ] ; then
  ## resolve links - $0 may be a link to maven's home
  PRG="$0"

  # need this for relative symlinks
  while [ -h "$PRG" ] ; do
    ls=`ls -ld "$PRG"`
    link=`expr "$ls" : '.*-> \(.*\)$'`
    if expr "$link" : '/.*' > /dev/null; then
      PRG="$link"
    else
      PRG="`dirname "$PRG"`/$link"
    fi
  done

  saveddir=`pwd`

  M2_HOME=`dirname "$PRG"`/..

  # make it fully qualified
  M2_HOME=`cd "$M2_HOME" && pwd`

为了解决Mac不兼容的问题,我想到了

echo `php -r "echo realpath('foo');"`

不是很好,但是跨越了操作系统

这是在Bash 3.2.57中测试过的最佳解决方案:

# Read a path (similar to `readlink`) recursively, until the physical path without any links (like `cd -P`) is found.
# Accepts any existing path, prints its physical path and exits `0`, exits `1` if some contained links don't exist.
# Motivation: `${BASH_SOURCE[0]}` often contains links; using it directly to extract your project's path may fail.
#
# Example: Safely `source` a file located relative to the current script
#
#     source "$(dirname "$(rreadlink "${BASH_SOURCE[0]}")")/relative/script.sh"
#Inspiration: https://stackoverflow.com/a/51089005/6307827
rreadlink () {
    declare p="$1" d l
    while :; do
        d="$(cd -P "$(dirname "$p")" && pwd)" || return $? #absolute path without symlinks
        p="$d/$(basename "$p")"
        if [ -h "$p" ]; then
            l="$(readlink "$p")" || break

            #A link must be resolved from its fully resolved parent dir.
            d="$(cd "$d" && cd -P "$(dirname "$l")" && pwd)" || return $?
            p="$d/$(basename "$l")"
        else
            break
        fi
    done
    printf '%s\n' "$p"
}
readlink -e [filepath]

看来这正是你想要的 -它接受任意路径,解析所有符号链接,并返回“真实”路径 -可能所有系统都已经有了“标准*nix”

在这里,我提出了一个我认为是跨平台(至少Linux和macOS)的解决方案,目前对我来说效果很好。

crosspath()
{
    local ref="$1"
    if [ -x "$(which realpath)" ]; then
        path="$(realpath "$ref")"
    else
        path="$(readlink -f "$ref" 2> /dev/null)"
        if [ $? -gt 0 ]; then
            if [ -x "$(which readlink)" ]; then
                if [ ! -z "$(readlink "$ref")" ]; then
                    ref="$(readlink "$ref")"
                fi
            else
                echo "realpath and readlink not available. The following may not be the final path." 1>&2
            fi
            if [ -d "$ref" ]; then
                path="$(cd "$ref"; pwd -P)"
            else
                path="$(cd $(dirname "$ref"); pwd -P)/$(basename "$ref")"
            fi
        fi
    fi
    echo "$path"
}

这里有一个macOS(仅?)解决方案。也许更适合最初的问题。

mac_realpath()
{
    local ref="$1"
    if [[ ! -z "$(readlink "$ref")" ]]; then
        ref="$(readlink "$1")"
    fi
    if [[ -d "$ref" ]]; then
        echo "$(cd "$ref"; pwd -P)"
    else
        echo "$(cd $(dirname "$ref"); pwd -P)/$(basename "$ref")"
    fi
}