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Editorial Team
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Asked: 2026-06-17T20:15:36+00:00

In the top-voted answer to this fantastic question , the following regular expression is

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In the top-voted answer to this fantastic question, the following regular expression is used in a preg_replace call (from the answer’s auto_version function):

'{\\.([^./]+)$}'

The end goal of this regular expression is to extract the file’s extension from the given filename. However, I’m confused about why the very beginning of this regular expression works. Namely:

Why does \\. match the same way as \. in a regex?

Shouldn’t the former match (a) one literal backslash, followed by (b) any character, while the second matches one literal period? The rules for single quoted strings state that \\ yields a literal backslash.

Consider this simple example:

$regex1 = '{\.([^./]+)$}';  // Variant 1 (one backslash)
$regex2 = '{\\.([^./]+)$}'; // Variant 2 (two backslashes)

$subject1 = '/css/foobar.css';   // Regular path
$subject2 = '/css/foobar\\.css'; // Literal backslash before period

echo "<pre>\n";
echo "Subject 1: $subject1\n";
echo "Subject 2: $subject2\n\n";

echo "Regex 1: $regex1\n";
echo "Regex 2: $regex2\n\n";

// Test Variant 1
echo preg_replace($regex1, "-test.\$1", $subject1) . "\n";
echo preg_replace($regex1, "-test.\$1", $subject2) . "\n\n";

// Test Variant 2
echo preg_replace($regex2, "-test.\$1", $subject1) . "\n";
echo preg_replace($regex2, "-test.\$1", $subject2) . "\n\n";
echo "</pre>\n";

The output is:

Subject 1: /css/foobar.css
Subject 2: /css/foobar\.css

Regex 1: {\.([^./]+)$}  <-- Output matches regex 2
Regex 2: {\.([^./]+)$}  <-- Output matches regex 1

/css/foobar-test.css
/css/foobar\-test.css

/css/foobar-test.css
/css/foobar\-test.css

Long story short: why should \\. yield the same matched results in a preg_replace call as \.?

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1 Answer

  1. Editorial Team

    Consider that there is double escaping going on: PHP sees \\. and says “OK, this is really \.“. Then the regex engine sees \. and says “OK, this means a literal dot”.

    If you remove the first backslash, PHP sees \. and says “this is a backslash followed by a random character — not a single quote or a backslash as per the spec — so it remains \.“. The regex engine again sees \. and gives the same result as above.

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