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HomeBackend DevelopmentPHP TutorialPHP 如何判断映射目录是否存在?

PHP 如何判断映射目录是否存在?
如果存在,如何获得此目录的大小和修改时间
 
目录为(\\apguds\note\x-section\05C62\) 开始运行中可直接打开
不知道PHP中代码对于这个路径是要怎么写的,试了半天没成功 


回复讨论(解决方案)

你这个映射目录是否需要密码访问?

<?php// Define the parameters for the shell command$location = "\\servername\sharename";$user = "USERNAME";$pass = "PASSWORD";$letter = "Z";// Map the drivesystem("net use ".$letter.": \"".$location."\" ".$pass." /user:".$user." /persistent:no>nul 2>&1");$dir = $letter.":/";// Check the directoryif(is_dir($dir) && is_readable($dir)) {    // 获取目录大小}

你这个映射目录是否需要密码访问?

<?php// Define the parameters for the shell command$location = "\\servername\sharename";$user = "USERNAME";$pass = "PASSWORD";$letter = "Z";// Map the drivesystem("net use ".$letter.": \"".$location."\" ".$pass." /user:".$user." /persistent:no>nul 2>&1");$dir = $letter.":/";// Check the directoryif(is_dir($dir) && is_readable($dir)) {    // 获取目录大小}




我这个是不用密码的,可直接打开,现在就是想得到如何判断是否存在,因为是对分目录进行操作
然后存在就获得目录大小和修改时间
谢谢了

那你就这样用好了

<?php // Define the parameters for the shell command$location = "\\servername\sharename";$letter = "Z"; // Map the drivesystem("net use ".$letter.": \"".$location."\" /persistent:no>nul 2>&1"); $dir = $letter.":/"; // Check the directoryif(is_dir($dir) && is_readable($dir)) {    // 获取目录大小}

还是没有用,你这个是要创建一个映射盘? 我电脑好像没这个(运行后也没有创建个个),我是直接在运行输入后就可以打开的

<?php// Define the parameters for the shell command$location = "\\apguds02\mi_note\x-section\04c692";$letter = "Z"; // Map the drivesystem("net use ".$letter.": \"".$location."\" /persistent:no>nul 2>&1"); $dir = $letter.":/"; // Check the directoryif(is_dir($dir) && is_readable($dir)) {    echo '目录存在';}else{  echo '目录不存在';}?>


这个目录是存在的,可是输出是不存在, 另外存在的时候如何获得大小和日期啊 

看下效果图

不知道为什么,我用这个还是不行,已经用另一种方法解决了,
感谢你的一直关注的技术支持,希望社区多一些像你这样热心的人

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