PHP中查找文件的相对路径或者绝对路径的工具
以前面试题目中有一题为求两个文件的相对路径,当时觉得没有电脑,书写代码太麻烦,没有做那个题目,今日工作比较闲遐,就想起这事来,完成了两个函数,代码未经过实际生产环境考验,如引用至生产环境,还请慎重阅读,主要是参考学习
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1<?php2/**3 * 文件的相对路径或者绝对路径查找工具4 *5 * @author 清源教育<support@tsingyuan.cn >6 * @version $Id:v 1.0 ,2013/10/13 17:29:007 * @copyright (c) Copyright;tsingyuan,20138 */9 10/**11 * 返回文件的绝对路径12 *13 * @param string $filename14 * @return string15 */16function absoluteroute($filename)17{18 $split = '/\/|\\\/';19 $currentdir = preg_split($split, dirname(__FILE__));20 $dirarr = preg_split($split, $filename);21 $diracount = count($dirarr);22 $cda = count($currentdir);23 if(strpos(PHP_OS, 'WIN') !== false)24 $reg = '/\w\:/';25 else26 $reg = '/\//';27 if(!preg_match($reg, $dirarr[0]))28 {29 foreach($dirarr as $nk=>$name)30 {31 if($name == "." || $name == '..')32 {33 if($name == '..')34 $filenamearr = array_slice($currentdir, 0, -($nk+1));35 if($name == '.' && $nk == '0')36 $filenamearr = array_slice($currentdir, 0);37 }38 else39 {40 $filenamearr[] = $name;41 }42 }43 $filename = implode('/', $filenamearr);44 }45 return $filename;46}47 48/**49 * 返回两个文件的相对路径 (PS:^_^不错的php学习交流群:276167802,验证:csl,有兴趣的话可以加入进来一起讨论)50 * (为了保证输入的相对路径参数可用,故此函数依赖absoluteroute)51 *52 * @param string $filenamea53 * @param string $filenameb54 * @return string55 */56function relativeroute($filenamea, $filenameb)57{58 $split = '/\/|\\\/';59 $filenamea = absoluteroute($filenamea);60 $filenameb = absoluteroute($filenameb);61 $dira =preg_split($split, $filenamea);62 $dirb =preg_split($split, $filenameb);63 $flag = true;64 if(count($dira) >= count($dirb))65 {66 $tmp = array();67 $tmp = $dira;68 $dira = $dirb;69 $dirb = $tmp;70 $flag = false;71 }72 foreach($dira as $k=>$v)73 if($v != $dirb[$k])break;74 75 $dirr = array_slice($dirb, $k);76 77 $k == 1 ? $dtag = '/' : $dtag = './';78 $result = $dtag.implode('/', $dirr);79 if(!$flag)80 {81 if($dira[0] == '')array_shift($dira);82 foreach($dirr as $kk=>&$v)83 ($kk+1) != count($dirr)?84 ($v ?85 $v = '..' :86 $v = '.'87 ) :88 $v = implode('/', array_slice($dira, $k)) ;89 $result = $dtag.implode('/', $dirr);90 }91 return $result;92}