mySQLi查询产生空值，尽管查询在函数之外工作

This one's driving me round the bend, and just can't work it out. The script works fine up until when I try to echo the final mysqli query (zen_id). I have tested the query outside of the if and while statements and it works fine. The query before it also has no troubles retrieving $model_array['model'] and echoes it flawlessly. I don't get any errors, simply nothing is outputted. A vardump of $id_array['zen_system_id'] gives me null, as does $id_array, $result_2 also produces a large number of null values in this format: object(mysqli_result)#5 (5) { ["current_field"]=> int(0) ["field_count"]=> int(1) ["lengths"]=> NULL ["num_rows"]=> int(0) ["type"]=> int(0) } I think i'm probably misssing something quite obvious here I'm just lost at the moment, possibly something to do with the while function or something??

//Connect to Database
$mysqli = mysqli_connect("localhost", "login", "user", "database");
//Check Connection
if ($mysqli->connect_errno) {
    echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
}

$result_1 = mysqli_query($mysqli,"SELECT `product_id` FROM oc_product");


while ($rows_1 = mysqli_fetch_array($result_1))
{ 

    $product_id = $rows_1['product_id'];

    $result = mysqli_query($mysqli,"SELECT `product_id` FROM `oc_product_option_value` WHERE `product_id`=$product_id");

    $rows = mysqli_fetch_array($result);

    //echo $product_id . " " . $rows['product_id'] . "</br>";

    if($rows['product_id'] == null) 
    {

        $result = mysqli_query($mysqli,"SELECT `model` FROM `oc_product` WHERE `product_id`=$product_id");
        $model_array = mysqli_fetch_array($result);
        $model = $model_array['model'] . "</br>";

        //echo $model;

        $result_2 = mysqli_query($mysqli,"SELECT `zen_system_id` FROM `oc_wholesale_link` WHERE `model_id`='" . $model . "'");
        $id_array = mysqli_fetch_array($result_2);
        echo $zen_id = $id_array['zen_system_id'];
    }

}

try to encapsulate data which you are passing to sencond query, also, to check if query would return result you can just make an if condition to the fetch itself

$result = mysqli_query($mysqli,"SELECT `product_id` FROM `oc_product_option_value` WHERE `product_id`= '".$product_id."'");
if($rows = mysqli_fetch_array($result))
{
    $result = mysqli_query($mysqli,"SELECT `model` FROM `oc_product` WHERE `product_id`=  '".$product_id."'");
    $model_array = mysqli_fetch_array($result);
    $model = $model_array['model'];

    //echo $model;

    $result_2 = mysqli_query($mysqli,"SELECT `zen_system_id` FROM `oc_wholesale_link` WHERE `model_id`='" . $model . "'");
    $id_array = mysqli_fetch_array($result_2);
    echo $zen_id = $id_array['zen_system_id'];
}
else
{
    echo 'Product Not Found';
}

I've changed this line $model = $model_array['model'] . "</br>"; i just supposed <br> was there for debuggin purpose.

UPDATE

Since you only need to get products id wich doesn't exist in oc_product_option_value you can change the query to this

$result = mysqli_query($mysqli,"SELECT `product_id` FROM `oc_product_option_value` WHERE `product_id`= '".$product_id."'");
if($rows = mysqli_fetch_array($result))
{
    $result = mysqli_query($mysqli,"SELECT `model` FROM `oc_product` WHERE `product_id`=  '".$product_id."'");
    $model_array = mysqli_fetch_array($result);
    $model = $model_array['model'];

    //echo $model;

    $result_2 = mysqli_query($mysqli,"SELECT `zen_system_id` FROM `oc_wholesale_link` WHERE `model_id`='" . $model . "'");
    $id_array = mysqli_fetch_array($result_2);
    echo $zen_id = $id_array['zen_system_id'];
}
else
{
    echo $product_id . '<br>';
}

Simple example to avoid such failure:

/* Select queries return a resultset */
$result = mysqli_query($link, "SELECT Name FROM City LIMIT 10")
if (!is_null($result)) {
    printf("Select returned %d rows.
", mysqli_num_rows($result));

    /* free result set */
    mysqli_free_result($result);
} else {
    echo "Mysqli query failed: " . mysqli_error();
}