0

这个让我在拐弯处开车，只是无法解决。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。它之前的查询也没有任何问题可以检索$model_array['model']并完美地回显它。我没有收到任何错误，只是没有输出任何内容。一个 vardump$id_array['zen_system_id']给我 null ，同样$id_array，$result_2也会产生大量这种格式的 null 值：object(mysqli_result)#5 (5) { ["current_field"]=> int(0) ["field_count"]=> int(1) ["lengths"]=> NULL ["num_rows"]=> int(0) ["type"]=> int(0) }我想我可能在这里遗漏了一些非常明显的东西我现在只是迷路了，可能与 while 函数有关或者其他的东西？？

//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'];
    }

}
4

2 回答 2

2

尝试封装您传递给第二个查询的数据，此外，为了检查查询是否会返回结果，您可以if对 fetch 本身设置一个条件

$result = mysqli_query($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';
}

我已经改变了这条线$model = $model_array['model'] . "</br>";，我只是<br>为了调试目的。

UPDATE

由于您只需要获取其中不存在的产品 ID，因此oc_product_option_value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>';
}
于 2013-04-19T05:38:25.717 回答
0

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();
}
于 2016-11-29T13:03:17.647 回答