从路径中提取文件名 [英] extract filename from path

问题描述

我需要从文件路径字符串获取文件名.例如，从该字符串\abc\def\filename.txt我需要获取filename.txt

i need to get filename from file path string. For example, from this string \abc\def\filename.txt i need to get filename.txt

尝试通过正则表达式执行此操作:

trying to do this with regexp:

$filepath="abc\filename.txt";
$filename = preg_replace("/.+\\/","",$filepath);

但是它给了我一个错误.我应该使用什么正则表达式来解决这个问题?

but it gives me an error. What regex i should use to solve this?

推荐答案

您应该改用函数基名:

$filepath = 'abc\filename.txt';
$filename = basename($filepath);

重要说明，当字符串中包含反斜杠时，您需要使用单引号，否则请正确地对它们进行转义.

edit: important note, you need to use single quotes when you have backslashes in your strings, else escape them properly.

注意:此将不起作用:

$filepath = "abc\filename.txt";
$filename = basename($filepath);

因为变量$ filepath实际上成立:

because you're variable $filepath infact holds:

abc[special char here equalling \f]ilename.txt

另一个 此正则表达式也适用..

another edit: this regex works too..

$filepath = '\def\abc\filename.txt';
$basename = preg_replace('/^.+\\\\/', '', $filepath);

原始文件的所有错误之处在于，您使用双引号而不是单引号，并且反斜杠需要转义两次(\\而不是\).

all that was wrong with your original was that you had double-quotes rather than single, and backslash needs double escaped (\\ rather than \).

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