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1

Given the following haystack and needle:

$haystack = array(
  'foo' => array(
    1 => 'one',
    2 => 'two',
    3 => 'three',
  ),
);

$needle = array(
  'foo' => array(
    1 => 'one',
    2 => 'two',
  ),
);

I want to check if all nested key-value pairs of the needle occur in the haystack (like it does in the example above), while ignoring any additional key-value pairs that may exist in the haystack (like $haystack['foo'][3] in the example).

There are many similar questions on SO but I haven't found a solution for this specific use case. Is there a (combination of) standard PHP functions to do this? What is the most elegant solution?

[update]

I didn't make clear yet that the arrays may not always have the same depth. Also, the keys of the elements in the arrays may be different every time.

share|improve this question
1  
You should show what you've tried. I have an answer posted but deleted as I prefer to answer questions that show effort. –  John Conde Mar 6 at 16:03
    
So, you want to check if $needle is inside $haystack ?? –  samaYo Mar 6 at 16:03
    
@JohnConde I know, I'll post what I came up with as an answer. –  marcvangend Mar 6 at 16:07
    
@sam_io yes, that is what I want to know, taking all nested levels of both arrays into account. –  marcvangend Mar 6 at 16:08
    
@marcvangend I tried :/ sadly it's too tricky, and me got no time. It is quite interesting challenge. good luck :) +1 for the question. –  samaYo Mar 6 at 16:59

2 Answers 2

up vote 2 down vote

array_intersect() will tell you what values match. Just make sure that matches your $needle.

echo $needle['foo'] === array_intersect($needle['foo'], $haystack['foo']);
share|improve this answer
    
If I'm not mistaken, this will only work with arrays that are one level deep, right? –  marcvangend Mar 6 at 16:09
    
If I change the values I still get a match. –  Jay Blanchard Mar 6 at 16:12
    
My updated answer solves that problem –  John Conde Mar 6 at 16:14
1  
If it is a nested array than this won't work and there is no "elegant" way to do this. The array key can obviously be dynamic in your code. –  John Conde Mar 6 at 16:27
1  
I think this is doable as the OP requested, but not without less than at-least 10 lines of code. However, this answer does not enable dynamic keys, or more depth. –  samaYo Mar 6 at 16:29
up vote 0 down vote accepted

This is what I came up with myself. It works, but it seems more complex than it should be...

/**
 * Helper function which recursively checks if the key-value pairs in one array
 * are all present in another array. If all key-value pairs in the needle are
 * present in the haystack, and the haystack also contains additional items,
 * the check wil still pass.
 *
 * @param array $needle
 *   The array with the key-value pairs to look for.
 * @param array $haystack
 *   The array in which to look for the key-value pairs.
 * @return bool
 *   TRUE if all key-value pairs of the needle occur in the haystack. FALSE if
 *   one or more keys or values are missing or different.
 */
function array_contains(array $needle, array $haystack) {
  // First, check if needle and haystack are identical. In that case it's easy.
  if ($needle === $haystack) {
    return TRUE;
  }
  foreach ($needle as $key => $value) {
    // If the key does not occur in the haystack, we're done.
    if (!isset($haystack[$key])) {
      return FALSE;
    }
    // If the value is an array...
    if (is_array($value)) {
      // ...see if the counterpart in $haystack is an array too...
      if (!is_array($haystack[$key])) {
        return FALSE;
      }
      // ...and if so, recurse.
      if (array_contains($value, $haystack[$key]) == FALSE) {
        return FALSE;
      }
    }
    // If the values are not arrays and not the same, the check fails.
    else if ($value != $haystack[$key]) {
      return FALSE;
    }
  }
  // If we still didn't fail, all tests have passed.
  return TRUE;
}
share|improve this answer
    
Modify your OP to add this. –  Jay Blanchard Mar 6 at 16:13
    
See my updated answer. It's a oneliner and works. –  John Conde Mar 6 at 16:14
    
@JayBlanchard I have been criticized by mods before for including a possible answer in my question, telling me that is should be an answer instead... –  marcvangend Mar 6 at 16:30

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