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PHP开发文件系统实例讲解_PHP

Jun 01, 2016 pm 12:32 PM
Example develop document system explain

PHP中有许多与文件系统有关的函数,这些函数不仅可以打开文件,还可以显示目录中的内容、移动文件和其他一些功能,许多人甚至用PHP开发基于互联网的文件资源管理器。



下面的脚本样例可以显示一个目录清单,注释已经包含在代码中:
<? /把要读取的目录的全路径名存入一个名字为$dir_name的变量中。/
$dir_name = "/home/me/";
/ 创建一个句柄,其值是打开一个给定目录的结果/
$dir = opendir($dir_name);
/ 建立一个文字块,用以放置列表元素(文件名字)/
$file_list = "<ul>";
/ 使用一个while语句,读取已经打开的目录中的所有元素,如果文件的名字不是“.”或“..”,则显示列表中的名字/
while ($file_name = readdir($dir)) {
if (($file_name != ".") &&&& ($file_name != "..")) {
$file_list .= "<li>$file_name";
}
}
$file_list .= "</ul>";
/关闭打开的目录,结束PHP模块/
closedir($dir);
?>
<!-- Start your HTML -->
<HTML>
<HEAD>
<TITLE>Directory Listing</TITLE>
</HEAD>
<BODY>
<!-- Use PHP to print the name of the directory you read -->
<P>Files in: <? echo "$dir_name"; ?></p>
<!-- Use PHP to print the directory listing -->
<? echo "$file_list"; ?>
</BODY>
</HTML>
好了,我们已经得到了一个目录清单。需要注意的是,要读取一个文件(稍后我们将进行讲解)或目录的内容,PHP运行的系统上的用户必须至少有读取文件的权限。



下面是一个如何拷贝文件的例子:
<? /把你要拷贝的源文件的全路径赋给一个名字为$original的变量中,把拷贝的文件的全路径赋予一个名字为$copied的变量中/
$original = "/home/me/mydatabasedump";
$copied = "/archive/mydatabasedumo_1010";
/ 使用copy()函数拷贝源文件,如果拷贝没有完成则会显示一个错误信息/
@copy($original, $copied) or die("Couldn"t copy file.");
?>
这个例子是一个文件备份系统的原型。在这段脚本运行时,它把文件拷贝到一个不同的位置进行保存。稍微修改一下守护程序,就可以在一天中你指定的时刻执行它,而无需用户的干预。


摘自:中国BS网
作者:邱宗英

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