Php mySQLi查询产生空值，尽管查询在函数外部工作

这件事快把我逼疯了，我就是做不出来。在我尝试回显最终的mysqli查询zen_id之前，该脚本工作正常。我在if和while语句之外测试了该查询，它工作正常。之前的查询在检索$model_数组['model']时也没有问题，并完美地对其进行了响应。我没有收到任何错误，只是没有输出任何内容。$id_array['zen_system_id']的vardump给了我null，正如$id_array，$result_2也以这种格式生成大量null值：objectmysqli_result5 5{[current_field]=>int0[field_count]=>int1[length]=>null[num_rows]=>int0[type]=>int0}我想我可能在这里漏掉了一些很明显的东西，我现在迷路了，可能与while函数或其他什么有关

//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'];
    }

}

---

尝试封装传递给sencond查询的数据，同时，为了检查查询是否会返回结果，您只需对获取本身设置一个if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\u数组['model']；我只是想去那里是为了调试

更新

由于您只需要获取oc_product_option_值中不存在的产品id，因此可以将查询更改为

$result = mysqli_query($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>';
}

---

避免此类故障的简单示例：

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

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

---

嗯，我没有找到产品也许我只是睡眠不足，但这对我有什么帮助？这将有助于了解问题所在，$result查询工作不正常，更改为尝试打印更改为$sql=SELECT product_id` FROM oc_product_option_value where product_id='。$product_id'；echo$sql$结果=mysqli_查询$mysqli，$sql；但是在这里回显$result=mysqli\u query$mysqli没有问题，从oc\u product\u选项\u值中选择product\u id，其中product\u id=$product\u id；echo$rows=mysqli\u fetch\u array$result；echo$rows['product_id']；我试图更改查询，但仍然得到一个未找到的产品错误：S，这对我来说毫无意义。这是一个非常糟糕的建议。它让你免于失败，就像鸵鸟让自己免于危险一样。