php array_merge without erasing values?

Question

Background: Trevor is working with a PHP implementation of a standard algorithm: take a main set of default name-value pairs, and update those name-value pa

Answer 1

array_replace_recursive($array, $array2);

This is the solution.

Answer 2

Well, if you want a "clever" way to do it, here it is, but it may not be as readable as simply doing a loop.

$merged = array_merge(array_filter($foo, 'strval'), array_filter($bar, 'strval'));

edit: or using +...

Answer 3

Try this:

$merged = array_map(
    create_function('$foo,$bar','return ($bar?$bar:$foo);'),
    $foobar,$feebar
);

Not the most readable solution, but it should replace only non-empty values, regardless of which order the arrays are passed..

Answer 4

This will put duplicates into a new array, I don't know if this is what you want though.

<?php
  $foobar =   Array('firstname' => 'peter','age' => '33',);
  $feebar =   Array('firstname' => '','lastname' => 'griffin',);
  $merged=$foobar;
  foreach($feebar as $k=>$v){
    if(isset($foobar[$k]))$merged[$k]=array($v,$foobar[$k]);
    else $merged[$k]=$v;
  }
  print_r($merged);
?>

This will simply assure that feebar will never blank out a value in foobar:

<?php
  $foobar =   Array('firstname' => 'peter','age' => '33',);
  $feebar =   Array('firstname' => '','lastname' => 'griffin',);
  $merged=$foobar;
  foreach($feebar as $k=>$v) if($v)$merged[$k]=$v;
  print_r($merged);
?>

or ofcourse,

<?
  function cool_merge($array1,$array2){
    $result=$array1;
    foreach($array2 as $k=>$v) if($v)$result[$k]=$v;
    return $result;
  }

  $foobar =   Array('firstname' => 'peter','age' => '33',);
  $feebar =   Array('firstname' => '','lastname' => 'griffin',);
  print_r(cool_merge($foobar,$feebar));
?>

Answer 5

If you also want to keep the values that are blank in both arrays:

array_filter($foo) + array_filter($bar) + $foo + $bar

Answer 6

Adjust to your needs:

# Replace keys in $foo
foreach ($foo as $key => $value) {
    if ($value != '' || !isset($bar[$key])) continue;
    $foo[$key] = $bar[$key];
}

# Add other keys in $bar
# Will not overwrite existing keys in $foo
$foo += $bar;