如何在MacOS上获取shell脚本的绝对路径名？

Question

readlink -f在MacOS上不存在.我设法在网上找到的唯一可用的Mac OS解决方案是这样的:

if [[ $(echo $0 | awk '/^\//') == $0 ]]; then
    ABSPATH=$(dirname $0)
else
    ABSPATH=$PWD/$(dirname $0)
fi

任何人都可以向这个看似微不足道的任务提出更优雅的建议吗？

Answer 1

另一个(也相当丑陋)的选择:

ABSPATH=$(cd "$(dirname "$0")"; pwd)

Answer 2

获取shell脚本的绝对路径

从我的.bashrc中挖出一些旧脚本,并稍微更新了语法,添加了一个测试套件.

支持

• source ./script(由.点运算符调用时)
• 绝对路径/路径/到/脚本
• 像./script这样的相对路径
• /path/dir1/../dir2/dir3/../script
• 从符号链接调用时
• 当符号链接嵌套时,例如) foo->dir1/dir2/bar bar->./../doe doe->script
• 调用者更改脚本名称时

它已在成功的实际项目中进行了测试和使用,但可能存在我不知道的极端情况.
如果您能够找到这种情况,请告诉我.
(首先,我知道这不会在sh shell上运行)

码

pushd . > /dev/null
SCRIPT_PATH="${BASH_SOURCE[0]}";
  while([ -h "${SCRIPT_PATH}" ]) do 
    cd "`dirname "${SCRIPT_PATH}"`"
    SCRIPT_PATH="$(readlink "`basename "${SCRIPT_PATH}"`")"; 
  done
cd "`dirname "${SCRIPT_PATH}"`" > /dev/null
SCRIPT_PATH="`pwd`";
popd  > /dev/null
echo "srcipt=[${SCRIPT_PATH}]"
echo "pwd   =[`pwd`]"

已知的问题

脚本必须位于某个磁盘上,让它通过网络. 如果您尝试从PIPE运行此脚本,它将无法正常工作

wget -o /dev/null -O - http://host.domain/dir/script.sh |bash

从技术上讲,它是未定义的.
实际上,没有理智的方法来检测这一点.

使用的测试用例

以及检查它是否有效的当前测试用例.

#!/bin/bash
# setup test enviroment
mkdir -p dir1/dir2
mkdir -p dir3/dir4
ln -s ./dir1/dir2/foo bar
ln -s ./../../dir3/dir4/test.sh dir1/dir2/foo
ln -s ./dir1/dir2/foo2 bar2
ln -s ./../../dir3/dir4/doe dir1/dir2/foo2
cp test.sh ./dir1/dir2/
cp test.sh ./dir3/dir4/
cp test.sh ./dir3/dir4/doe
P="`pwd`"
echo "--- 01"
echo "base  =[${P}]" && ./test.sh
echo "--- 02"
echo "base  =[${P}]" && `pwd`/test.sh
echo "--- 03"
echo "base  =[${P}]" && ./dir1/dir2/../../test.sh
echo "--- 04"
echo "base  =[${P}/dir3/dir4]" && ./bar
echo "--- 05"
echo "base  =[${P}/dir3/dir4]" && ./bar2
echo "--- 06"
echo "base  =[${P}/dir3/dir4]" && `pwd`/bar
echo "--- 07"
echo "base  =[${P}/dir3/dir4]" && `pwd`/bar2
echo "--- 08"
echo "base  =[${P}/dir1/dir2]" && `pwd`/dir3/dir4/../../dir1/dir2/test.sh
echo "--- 09"
echo "base  =[${P}/dir1/dir2]" && ./dir1/dir2/test.sh
echo "--- 10"
echo "base  =[${P}/dir3/dir4]" && ./dir3/dir4/doe
echo "--- 11"
echo "base  =[${P}/dir3/dir4]" && ./dir3/dir4/test.sh
echo "--- 12"
echo "base  =[${P}/dir3/dir4]" && `pwd`/dir3/dir4/doe
echo "--- 13"
echo "base  =[${P}/dir3/dir4]" && `pwd`/dir3/dir4/test.sh
echo "--- 14"
echo "base  =[${P}/dir3/dir4]" && `pwd`/dir1/dir2/../../dir3/dir4/doe
echo "--- 15"
echo "base  =[${P}/dir3/dir4]" && `pwd`/dir1/dir2/../../dir3/dir4/test.sh
echo "--- 16"
echo "base s=[${P}]" && source test.sh
echo "--- 17"
echo "base s=[${P}]" && source `pwd`/test.sh
echo "--- 18"
echo "base s=[${P}/dir1/dir2]" && source ./dir1/dir2/test.sh
echo "--- 19"
echo "base s=[${P}/dir3/dir4]" && source ./dir1/dir2/../../dir3/dir4/test.sh
echo "--- 20"
echo "base s=[${P}/dir3/dir4]" && source `pwd`/dir1/dir2/../../dir3/dir4/test.sh
echo "--- 21"
pushd . >/dev/null
cd ..
echo "base x=[${P}/dir3/dir4]"
./`basename "${P}"`/bar
popd  >/dev/null

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