Question

我有两个像这样的字符串

/home/user/Desktop/aaaa/Final/
/home/user/Desktop/aaaa/folder3333/IMAG0486.jpg

我的结果字符串应该是这样的

/home/user/Desktop/aaaa/Final/folder3333/IMAG0486.jpg

意味着当我在这种情况下找到最长的匹配时

“/ home / user中/桌面/ AAAA”

然后我添加第二个字符串的其余部分

“folder3333 / IMAG0486.jpg”

到第一个字符串

，结果字符串是

/home/user/Desktop/aaaa/Final/folder3333/IMAG0486.jpg

Answer 1

将/上的路径拆分为数组。迭代数组直到找到差异，将第二个数组的其余部分添加到输出中。我在代码中留下了调试打印件，通过删除它们可以大大缩短。

#! /bin/bash
s1=/home/user/Desktop/aaaa/Final/
s2=/home/user/Desktop/aaaa/folder3333/IMAG0486.jpg
expect=/home/user/Desktop/aaaa/Final/folder3333/IMAG0486.jpg
out=$s1

_IFS=$IFS
IFS=/
parts1=($s1)
parts2=($s2)
IFS=$_IFS

matching=1
for ((i=0;i<${#parts2[@]};i++)) ; do
    if [[ $matching && ${parts1[i]} == ${parts2[i]} ]] ; then
        echo same ${parts2[i]}
    else
        echo different ${parts1[i]} ${parts2[i]}
        matching=0
        out+=${parts2[i]}/
    fi
done
out=${out%/}

echo $expect
echo $out

Answer 2

This适用于我并且有效unit tests：

path_common()
{
    if [ -z "${2-}" ]
    then
        return 2
    fi

    # Remove repeated slashes
    for param
    do
        param=$(printf %s. "$1" | tr -s "/")
        set -- "$@" "${param%.}"
        shift
    done

    common_path=$1
    shift

    for param
    do
        while case ${param%/}/ in "${common_path%/}/"*) false;; esac; do
            new_common_path=${common_path%/*}
            if [ "$new_common_path" = "$common_path" ]
            then
                return 1 # Dead end
            fi
            common_path=$new_common_path
        done
    done
    printf %s "$common_path"
}

Answer 3

不如其他解决方案可靠，但在常规情况下工作正常：

#!/bin/bash

p1="/home/user/Desktop/aaaa/Final/"
p2="/home/user/Desktop/aaaa/folder3333/IMAG0486.jpg"

printf "%s|%s" "$p1" "$p2" | sed -e 's_^\(\(.*\)\/.*\)|\2/\(.*\)$_\1\3_'

找到shell中两个路径字符串的最长匹配

3 个答案: