#!/bin/bash
# both $1 and $2 are absolute paths
# returns $2 relative to $1

source=$1
target=$2

common_part=$source
back=
while [ "${target#$common_part}" = "${target}" ]; do
  common_part=$(dirname $common_part)
  back="../${back}"
done

echo ${back}${target#$common_part/}

test.sh:

#!/bin/bash                                                                 

cd /home/ubuntu
touch blah
TEST=/home/ubuntu/.//blah
echo TEST=$TEST
TMP=$(readlink -e "$TEST")
echo TMP=$TMP
REL=${TMP#$(pwd)/}
echo REL=$REL

测试:

$ ./test.sh 
TEST=/home/ubuntu/.//blah
TMP=/home/ubuntu/blah
REL=blah

假设您已经安装了:bash、pwd、dirname、echo;relpath是

#!/bin/bash
s=$(cd ${1%%/};pwd); d=$(cd $2;pwd); b=; while [ "${d#$s/}" == "${d}" ]
do s=$(dirname $s);b="../${b}"; done; echo ${b}${d#$s/}

我从pini和其他一些想法中得到了答案

注意:这要求两个路径都是现有文件夹。文件将无法工作。

我需要这样的东西，但它也解决了符号链接。我发现pwd有一个-P标志用于此目的。附加了我的脚本的一个片段。它在shell脚本的函数中，因此是$1和$2。结果值是从START_ABS到END_ABS的相对路径，位于UPDIRS变量中。为了执行pwd -P，将脚本cd放入每个参数目录，这也意味着将处理相对路径参数。干杯,吉姆

SAVE_DIR="$PWD"
cd "$1"
START_ABS=`pwd -P`
cd "$SAVE_DIR"
cd "$2"
END_ABS=`pwd -P`

START_WORK="$START_ABS"
UPDIRS=""

while test -n "${START_WORK}" -a "${END_ABS/#${START_WORK}}" '==' "$END_ABS";
do
    START_WORK=`dirname "$START_WORK"`"/"
    UPDIRS=${UPDIRS}"../"
done
UPDIRS="$UPDIRS${END_ABS/#${START_WORK}}"
cd "$SAVE_DIR"

这是对@pini目前评分最高的解决方案(遗憾的是，它只处理少数情况)的更正，全功能改进

提醒:'-z'测试如果字符串是零长度(=空)，'-n'测试如果字符串不是空。

# both $1 and $2 are absolute paths beginning with /
# returns relative path to $2/$target from $1/$source
source=$1
target=$2

common_part=$source # for now
result="" # for now

while [[ "${target#$common_part}" == "${target}" ]]; do
    # no match, means that candidate common part is not correct
    # go up one level (reduce common part)
    common_part="$(dirname $common_part)"
    # and record that we went back, with correct / handling
    if [[ -z $result ]]; then
        result=".."
    else
        result="../$result"
    fi
done

if [[ $common_part == "/" ]]; then
    # special case for root (no common path)
    result="$result/"
fi

# since we now have identified the common part,
# compute the non-common part
forward_part="${target#$common_part}"

# and now stick all parts together
if [[ -n $result ]] && [[ -n $forward_part ]]; then
    result="$result$forward_part"
elif [[ -n $forward_part ]]; then
    # extra slash removal
    result="${forward_part:1}"
fi

echo $result

测试用例:

compute_relative.sh "/A/B/C" "/A"           -->  "../.."
compute_relative.sh "/A/B/C" "/A/B"         -->  ".."
compute_relative.sh "/A/B/C" "/A/B/C"       -->  ""
compute_relative.sh "/A/B/C" "/A/B/C/D"     -->  "D"
compute_relative.sh "/A/B/C" "/A/B/C/D/E"   -->  "D/E"
compute_relative.sh "/A/B/C" "/A/B/D"       -->  "../D"
compute_relative.sh "/A/B/C" "/A/B/D/E"     -->  "../D/E"
compute_relative.sh "/A/B/C" "/A/D"         -->  "../../D"
compute_relative.sh "/A/B/C" "/A/D/E"       -->  "../../D/E"
compute_relative.sh "/A/B/C" "/D/E/F"       -->  "../../../D/E/F"

#!/bin/bash
# both $1 and $2 are absolute paths
# returns $2 relative to $1

source=$1
target=$2

common_part=$source
back=
while [ "${target#$common_part}" = "${target}" ]; do
  common_part=$(dirname $common_part)
  back="../${back}"
done

echo ${back}${target#$common_part/}