爆炸时打印阵列

Question

In the below code i am getting the output

Array ( [0] => 1 [1] => 2 )

but the expected output is

Array ( [0] => 1 [1] => 2 [2] => 3)  Array ( [0] => 1 [1] => 2 )   

why is it always executing the second if condition? although the first condition is also true. this is the code I have tried

    <?php 
$test_arr=array();
$temp_option_arr=array();
$option_arr=array();

$options_array_val = Array ( 0 => "animals:1", 1 => "animals:2", 2 => "animals:3", 3 => "birds:1", 4 => "birds:2" );

    foreach($options_array_val as $options_val)
    {
        $search_filter = explode(":", $options_val);
        print_r($search_filter);
        if(!in_array($search_filter[0],$option_arr))
        {
            array_push($temp_option_arr,$search_filter[1]);
            array_push($option_arr,$search_filter[0]);
            $temp_option_arr=array();

        }
        array_push($temp_option_arr,$search_filter[1]);
    }
    $test_arr[$search_filter[0]]=$temp_option_arr;
    $find_species = array();
    if(!empty($test_arr['animals']))
    {

        $find_species = $test_arr['animals'];
        print_r($find_species);

    }

    if(!empty($test_arr['birds']))
    {

        $find_species = $test_arr['birds'];
        print_r($find_species);
    }

?>     

Answer 1

The line

$test_arr[$search_filter[0]]=$temp_option_arr;

is out of the foreach scope, therefore it only sets the array for the birds and not for the animals, so you need to move it one line upper.

Also, you assign temp_option_arr to a value

array_push($temp_option_arr,$search_filter[1]);

then set it back to empty array without using it

$temp_option_arr=array();

You can remove the first one I guess