PHP：如何从目录中获取单个文件而不扫描整个目录？ [英] PHP: How can I grab a single file from a directory without scanning entire directory?

问题描述

我有一个130万个文件的目录，我需要移动到一个数据库。我只需要从目录中获取单个文件名，而不扫描整个目录。我抓住哪个文件无关紧要，当我完成它后，我将删除它，然后继续下一步。这可能吗？我可以找到的所有示例似乎将整个目录列表扫描到一个数组中。我只需要一次抓住一个来处理...每次不是130万。

解决方案

这应该做到这一点：

 <？php 
 $ h = opendir（'./'）; //打开当前目录
 while（false！==（$ entry = readdir（$ h）））{
 if（$ entry！='。&& $ entry！=' ..'）{//跳过。 
 echo $ entry; //做任何你需要做的文件
 break; //退出循环，所以没有更多的文件被读取
} 
} 
？> 
  

readdir

返回目录中下一个条目的名称。条目按文件系统存储的顺序返回。

I have a directory with 1.3 Million files that I need to move into a database. I just need to grab a single filename from the directory WITHOUT scanning the whole directory. It does not matter which file I grab as I will delete it when I am done with it and then move on to the next. Is this possible? All the examples I can find seem to scan the whole directory listing into an array. I only need to grab one at a time for processing... not 1.3 Million every time.

解决方案

This should do it:

<?php
$h = opendir('./'); //Open the current directory
while (false !== ($entry = readdir($h))) {
    if($entry != '.' && $entry != '..') { //Skips over . and ..
        echo $entry; //Do whatever you need to do with the file
        break; //Exit the loop so no more files are read
    }
}
?>

readdir

Returns the name of the next entry in the directory. The entries are returned in the order in which they are stored by the filesystem.

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