php - 尽管查询在函数之外工作，但 mySQLi 查询生成空值

Question

这个问题让我不知所措，但就是无法解决。该脚本工作正常，直到我尝试回显最终的 mysqli 查询(zen_id)。我已经在 if 和 while 语句之外测试了查询，它工作正常。之前的查询也可以轻松检索 $model_array['model'] 并完美地回显它。我没有收到任何错误，只是没有输出任何内容。 $id_array['zen_system_id'] 的 vardump 给我 null，$id_array 也是如此，$result_2 也会产生大量 null 值格式如下: object(mysqli_result)#5 (5) { ["current_field"]=> int(0) ["field_count"]=> int(1) ["lengths"]=> NULL ["num_rows""]=> int(0) ["type"]=> int(0) } 我想我可能在这里遗漏了一些非常明显的东西，我现在只是迷失了方向，可能与while 函数什么的？？

//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'];
    }

}

Answer 1

尝试封装您传递给第二个查询的数据，另外，要检查查询是否会返回结果，您可以创建 if获取本身的条件

$result = mysqli_query($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';
}

我更改了这一行 $model = $model_array['model'] . "</br>";我只是想 <br>是为了调试目的。

UPDATE

因为您只需要获取oc_product_option_value中不存在的产品id您可以将查询更改为此

$result = mysqli_query($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>';
}