PHP Generate Array from Function Arguments

Question

I have a function that I want to generate an array with. The format of the array is specific and only values need to be changed based on the arguments of my function. Here's the function:

 function generatearray($name1, $field1, $name2, $field2) {
  $language = ($user->language) ? $user->language : 'und';
   $edit = array(
    $name1 => array(
     $language => array(
      0 => array(
       'value' => $field1,
     ),
    ),
   ), // ..... other elements $name2
  );
  return $edit;
}

$name2 and $field2 are optional arguments.

It may not answer your question, but you're not defining $user anywhere - you need to either pass it in as a parameter, or declare it as a global. — andrewsi
– andrewsi, Commented Sep 20, 2012 at 19:18
Thanks for the help! I ended up defining $user as a 5th argument. — Ken J
– Ken J, Commented Sep 20, 2012 at 20:42

ews2001 · Accepted Answer · 2012-09-20 19:36:18Z

1

Not sure what the question is...but this code will do exactly what you're wanting to know about...a way to build an array based on up to 2 name/value pairs provided to a function.

printarray(generatearray('name_one','field_one','name_two','field_two'));

function generatearray($name1, $field1, $name2 = null, $field2 = null) {
$new_array = array();
$language = isset($user->language) ? $user->language : 'und';
$new_array[$name1][$language] = $field1;
if (!is_null($name2) && !is_null($field2)) {
    $new_array[$name2][$language] = $field2;
}
return $new_array;
}

function printarray($arr) {
foreach ($arr as $user_key => $user_value) {
    echo "User: ".$user_key."<br />";
    foreach ($user_value as $language_key=>$language_value) {
        echo "Language: ".$language_value."<br />";
    }
}
}

answered Sep 20, 2012 at 19:36

ews2001

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