Question

这个让我绕过弯道，但是无法解决这个问题。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，它运行正常。之前的查询也没有检索$model_array['model']的麻烦并且完美地回应它。我没有任何错误，根本没有输出。 $id_array['zen_system_id']的vardump给我null，$id_array也是如此，$result_2也会以这种格式产生大量的空值：object(mysqli_result)#5 (5) { ["current_field"]=> int(0) ["field_count"]=> int(1) ["lengths"]=> NULL ["num_rows"]=> int(0) ["type"]=> int(0) }我想我可能会误会某些东西很明显，我现在只是迷失了，可能与while函数有什么关系？

//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，同时检查查询是否会返回结果，您只能为获取本身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';
}

我已更改此行$model = $model_array['model'] . "</br>";我只是假设<br>用于调试目的。

UPDATE

由于您只需要获取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>';
}

Answer 2

避免此类失败的简单示例：

/* 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();
}

mySQLi查询产生空值，尽管查询在函数之外工作

2 个答案: