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'];
}
}
最佳答案
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>';
}
关于php - 尽管查询在函数之外工作,但 mySQLi 查询生成空值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16097599/