我通常使用:

dirname $(which $BASH_SOURCE)

我会用这样的东西:

# Retrieve the full pathname of the called script
scriptPath=$(which $0)

# Check whether the path is a link or not
if [ -L $scriptPath ]; then

    # It is a link then retrieve the target path and get the directory name
    sourceDir=$(dirname $(readlink -f $scriptPath))

else

    # Otherwise just get the directory name of the script path
    sourceDir=$(dirname $scriptPath)

fi

$(dirname "$(readlink -f "$BASH_SOURCE")")

#!/bin/sh
PRG="$0"

# need this for relative symlinks
while [ -h "$PRG" ] ; do
   PRG=`readlink "$PRG"`
done

scriptdir=`dirname "$PRG"`

关键部分是,我正在减少问题的范围:我禁止通过路径间接执行脚本(如 /bin/sh [脚本路径与路径组件有关])。

这可以被检测到,因为0美元将是一个相对的路径,不解决与当前文件夹有关的任何文件。我相信使用#!机制的直接执行总是导致绝对0美元,包括当脚本在路径上找到时。

我也要求在象征性链接链接链接的任何字符和字符只包含一个合理的字符子,特别是不是 \n, >, * 或?. 这对于字符逻辑来说是必要的。

#!/bin/sh
(
    path="${0}"
    while test -n "${path}"; do
        # Make sure we have at least one slash and no leading dash.
        expr "${path}" : / > /dev/null || path="./${path}"
        # Filter out bad characters in the path name.
        expr "${path}" : ".*[*?<>\\]" > /dev/null && exit 1
        # Catch embedded new-lines and non-existing (or path-relative) files.
        # $0 should always be absolute when scripts are invoked through "#!".
        test "`ls -l -d "${path}" 2> /dev/null | wc -l`" -eq 1 || exit 1
        # Change to the folder containing the file to resolve relative links.
        folder=`expr "${path}" : "\(.*/\)[^/][^/]*/*$"` || exit 1
        path=`expr "x\`ls -l -d "${path}"\`" : "[^>]* -> \(.*\)"`
        cd "${folder}"
        # If the last path was not a link then we are in the target folder.
        test -n "${path}" || pwd
    done
)

这应该是这样做的:

DIR="$(dirname "$(realpath "$0")")"

这与路线上的交线和空间一起工作。

请参见男人的页面为 dirname 和 realpath。

请添加关于如何支持MacOS的评论,对不起,我可以验证。