Use preg_replace() to add two backslashes before each match-CodePudding

I have code below, what I need to change to get result mercedes\\-benz instead of mercedes\-benz

$value = 'mercedes-benz';
$pattern = '/(\ |-|\/|&&|\|\||!|\(|\)|\{|}|\[|]|\^|"|~|\*|\?|:|\\\)/';
$replace = '\\\\${1}';
echo preg_replace($pattern, $replace, $value);

CodePudding user response：

I can't be sure that I've 100% translated your original attempt, but this works for your lone sample input.

The pattern uses a character class and curly braced quantifiers to improve readability and brevity. Using \K eliminates the need for the reference in the replacement string.

Code: (Demo)

$value = 'mercedes-benz';
$pattern = '`&{2}|\|{2}|[- /!(){}[\]^"~*?:\\\]\K`';
$replace = '\\\\\\';
echo preg_replace($pattern, $replace, $value);

Ultimately, the trick was to keep adding backslashes to the replacement to get them to show up.

CodePudding user response：

Welcome to the joys of "leaning toothpick syndrome" - backslash is such a commonly used escape character that it frequently requires escaping multiple times. Let's have a look at your case:

Required output (presumably because of some other escaping context): \\
Escape each \ with an additional \ for use in the PCRE regex engine: \\\\
Escape each \ there for use in a PHP string: \\\\\\\\

$value = 'mercedes-benz';
$pattern = '/(\ |-|\/|&&|\|\||!|\(|\)|\{|}|\[|]|\^|"|~|\*|\?|:|\\\)/';
$replace = '\\\\\\\\${1}';
echo preg_replace($pattern, $replace, $value);

As mickmackusa points out, you can get away with six rather than eight backslashes in some cases, such as a replacement of '\\\\\\'; this works because the regex engine sees \\\, which is an escaped backslash (\\) followed by a single backslash (\) that can't be escaping anything because it's the end of the string. Simply doubling for each "layer" of escaping is probably safer than learning when this short-cut is and isn't valid, though.