PHP Default Function Parameter values, how to 'pass default value' for 'not last' parameters?

Question

Most of us know the following syntax:

function funcName($param='value'){     echo $param; } funcName();  Result: "value" 

We were wondering how to pass default values for the 'not last' paramater? I know this terminology is way off, but a simple example would be:

function funcName($param1='value1',$param2='value2'){     echo $param1."\n";     echo $param2."\n"; } 

How do we accomplsh the following:

funcName(---default value of param1---,'non default'); 

Result:

value1 not default 

Hope this makes sense, we want to basically assume default values for the paramaters which are not last.

Thanks.

Answer 1

PHP doesn't support what you're trying to do. The usual solution to this problem is to pass an array of arguments:

function funcName($params = array()) {     $defaults = array( // the defaults will be overidden if set in $params         'value1' => '1',         'value2' => '2',     );      $params = array_merge($defaults, $params);      echo $params['value1'] . ', ' . $params['value2']; } 

Example Usage:

funcName(array('value1' => 'one'));                    // outputs: one, 2 funcName(array('value2' => 'two'));                    // outputs: 1, two funcName(array('value1' => '1st', 'value2' => '2nd')); // outputs: 1st, 2nd funcName();                                            // outputs: 1, 2 

Using this, all arguments are optional. By passing an array of arguments, anything that is in the array will override the defaults. This is possible through the use of array_merge() which merges two arrays, overriding the first array with any duplicate elements in the second array.