Question

My PHP code looks like this:

$input = "City.name = 'New York'";
$literal_pattern = '/\'.[^\']*\'/';
preg_match($literal_pattern, $input, $token);
echo $token[0]; // prints 'New York'

My regex needs to grab literals with escaped single quotes like:

$input = "City.name = 'New \' York'";
$literal_pattern = ???????????;
preg_match($literal_pattern, $input, $token);
echo $token[0]; // should prints 'New \' York'

What wil be the reges for $literal_pattern ?

Answer 1

Without this condition, simple…

/('[^']*')/

…would suffice, of course: match all sequences of “single quote, followed by any number of non-single-quote symbols, followed by a single quote again”.

But as we need to be ready for two things here – both “normal” and “escaped” ones. So we should add some spice to our pattern:

/('[^'\\]*(?:\\.[^'\\]*)*')/

It might look odd (and it is), but it’s actually pretty simple too: match sequences of…

• single quote symbol…
• …followed by zero or more “normal” characters (not ' or \),
• …followed by a subexpression of (“escaped” symbol, then zero or more “normal” ones), repeated 0 or more times…
• followed by a single quote symbol.

Example:

$input   = "City.name = 'New \\' York (And Some Backslash Fun)\\\\'\\'"; 
# ...as \' in any string literal will be parsed as a _single_ quote

$pattern = "/('[^'\\\\]*(?:\\\\.[^'\\\\]*)*')/";
# ... a choice: escape either slashes or single quotes; I choose the former

preg_match($pattern, $input, $token);
echo $token[0]; // 'New \' York (And Some Backslash Fun)\\'