如何在PHP中解决var_dump（$ testArray）

Question

What is the problem with the code below? What will it output? How can it be fixed?

$referenceTable = array();
$referenceTable['val1'] = array(1, 2);
 $referenceTable['val2'] = 3;
$referenceTable['val3'] = array(4, 5);

 $testArray = array();

 $testArray = array_merge($testArray, $referenceTable['val1']);
 var_dump($testArray);
 $testArray = array_merge($testArray, $referenceTable['val2']);
var_dump($testArray);
  $testArray = array_merge($testArray, $referenceTable['val3']);
   var_dump($testArray);

Thanks

Answer 1

The output will be as follows:

array(2) { [0]=> int(1) [1]=> int(2) } NULL NULL

You may also see two warnings generated, similar to the following:

Warning: array_merge(): Argument #2 is not an array

Warning: array_merge(): Argument #1 is not an array.

The issue here is that, if either the first or second argument to array_merge() is not an array, the return value will be NULL. For example, although one might reasonably expect that a call such as array_merge($someValidArray, NULL)would simply return $someValidArray, it instead returns NULL! (And to make matters worse, this is not documented well at all in the PHP documentation.)

As a result, the call to $testArray = array_merge($testArray, $referenceTable['val2']) evaluates to $testArray = array_merge($testArray, 3)and, since 3 is not of type array, this call to array_merge() returns NULL, which in turn ends up setting $testArray equal to NULL. Then, when we get to the next call to array_merge(), $testArray is now NULL so array_merge() again returns NULL. (This also explains why the first warning complains about argument #2 and the second warning complains about argument #1.)

The fix for this is straightforward. If we simply typecast the second argument to an array, we will get the desired results.