正则表达式提取任何单词not（single | double）引用分隔

Question

I have a string describing a [ variable operator value ] structure like this:

type == 'prova' && padposition == "stefano" or 10>var_name

I need to build a regular expression to extract the variable name list:

type
padposition
var_name

to apply a post processing on them (basically converting them into key of a PHP array):

$arr_name['type']
$arr_name['padposition']
$arr_name['var_name']

I've found the way to match string delimited by single or double quotes:

('|")(\w*\w)('|")

but I'm not able (I'm too ignorant!) to negate it or simply to extract any word non single or double quote delimited.

Answer 1

A way to do it (highly readable and easy to maintain):

$str = 'type == \'prova\' && padposition == "stefano" or 10>var_name';

$pattern = <<<'EOD'
~
# you define first the basic elements (as for a lexer) with named groups
(?(DEFINE)
    (?<var> [a-z_]\w* ) # variable name

    (?<dqstr> (?<=") [^\\"]*+ (?s:\\.[^\\"]*)*+ (?=") ) # double quoted string
    (?<sqstr> (?<=') [^\\']*+ (?s:\\.[^\\']*)*+ (?=') ) # single quoted string
    (?<string> \g<dqstr> | \g<sqstr> ) # any string

    (?<num> [0-9]+(?:\.[0-9]+)? ) # a number

    (?<value> \g<string> | \g<num> ) # any value

    (?<comp> [!><=]= | =?[><] ) # comparison operator
)

# Then you write the pattern using these named groups

(?J) # allow duplicate named groups

# variable op value
(?<key> \g<var> ) \h* \g<comp> \h* ["']? (?<val> \g<value> ) ['"]? 
| # OR
# value op variable
["']? (?<val> \g<value> ) ['"]? \h* \g<comp> \h* (?<key> \g<var> ) 
~xi
EOD;

if (preg_match_all($pattern, $str, $matches, PREG_SET_ORDER)) {
    $arr_name = [];
    foreach($matches as $m) {
        $arr_name[$m['key']] = $m['val'];
    }
    print_r($arr_name);
}

Pattern demo