正则表达式从函数参数中获取字符串？

Question

Can anyone help me with regex problem. Im making a script to go through all my .php files and get all strings passed to certain function. I need to match this cases:

/* Double quotes */
function("some string"); // Match: some string
function("some \"string\""); // Match: some "string"
function("some 'string'"); // Match: some 'string'

/* Single quotes */
function('some string'); // Match: some string
function('some \'string\''); // Match: some 'string'
function('some "string"'); // Match: some "string"

Function can also accept parameters after string, so it also needs to match this cases:

/* Additional parameters */
function("some string", "param"); // Match: some string
function("some string", $param); // Match: some string

So essentially, param can be either a string (quoted or double quoted) or unquoted variable. I need to get string only from first parameter of function, regardless if second parameter exists or is quoted in any way...

Thanks in advance...

Answer 1

Your particular example was successfully processed with...

preg_match_all('#\\(\\s*("((\\\\.|[^"])+)"|\'((\\\\.|[^\'])+)\'),?#s', $test, $matches);

Here's ideone demo.

Explanation: we try to match opening parenthesis (thankfully, it's PHP; it'd be far more difficult in Ruby), followed by any number of whitespace characters, followed by...

• either "(\\.|[^"])+"
• or '(\\.|[^'])+'

... followed by optional comma.

Each of this sequences covers both 'special characters' (escaped with slash) and 'normal ones' (that are not the same as delimiters).