Filtering Rows of a 2D Array Based on Row Intersections

In PHP, the array_diff_assoc() function is designed to find the difference between two arrays while prioritizing key-value pairs. However, while using this function to filter rows from a 2D array based on intersection with another 2D array, it may not always yield the expected results.

Understanding the Issue

The problem arises due to the strict comparison performed by array_diff_assoc(). It compares string representations of key-value pairs during comparison. This means that even if two key-value pairs contain the same values, they are not considered equal unless their string representations are identical.

Sample Data

Consider the following sample data:

$array1 = [
    [12 => 'new q sets'],
    [11 => 'common set']
];

$array2 = [
    [11 => 'common set']
];

Incorrect Output

When we attempt to use array_diff_assoc() to filter $array1 based on the rows in $array2, we get an incorrect output:

$output = array_diff_assoc($array1, $array2);

print_r($output);
// Output: [
//     [11 => 'common set']
// ]

This output shows that the common row is present in the result, while the intended output should contain the exclusive row from $array1.

Cause of the Issue

As mentioned earlier, the problem lies in the strict comparison performed by array_diff_assoc(). When comparing the following two arrays:

Array ( [0] => "Array" [1] => "Array" )
Array ( [0] => "Array" )

the function returns the different key-value pair as the result because the key-value pairs are not string-identical.

Correct Approach

To address this issue, we can use a slightly different approach that checks for the existence of specific key-values in the arrays:

$filteredRows = array_filter($array1, function($row) use ($array2) {
    return !in_array($row, $array2);
});

print_r($filteredRows);
// Output: [
//     [12 => 'new q sets']
// ]

This approach uses in_array() to check if each row from $array1 is present in $array2. If a row is not present in $array2, it is included in the filtered results.