php - PHP警告:mysqli_num_rows()和数据库创建

Question

我正在尝试使用以下代码从WooCommerce Wordpress订单获取信息，然后将详细信息传递给数据库。应该执行此操作的代码如下：

//Pass udid to database
add_action('woocommerce_payment_complete', 'send_to_db');

function send_to_db( $order_id ) {
$order = new WC_Order( $order_id ); 
$udid = $order->customer_note;
$email = $order->billing_email;
$db_hostname = 'MYIP';
$db_database = 'MYDATABASE';
$db_username = 'MYUSER';
$db_password = 'MYPASSWORD';

// Connect to server.
$link = mysqli_connect($db_hostname, $db_username, $db_password);
if (!$link) {
die('Not connected : ' . mysqli_error());
}

// Select the database. 
$db_selected = mysqli_select_db($link, $db_database);
if (!$db_selected) {
die ('Can\'t use database : ' . mysqli_error());
}
//find package type
$items = $order->get_items();
foreach($items as $item) {
$type = '1';
if(preg_match("/Basic/i", implode($item))) {
        $type = '0';
    }

}
//put it into the db
$query= "INSERT INTO udid_orders (udid, email, type) VALUES ('$udid','$email','$type')";
$sql = mysqli_query($link, $query);
}
mysqli_close($link);
//end pass to db

执行此代码后，它将填充数据库，然后另一个帮助PHP将从数据库中获取信息。但是我有两个问题。第一个问题是，我认为我没有正确创建数据库。我可以成功连接，但是下订单后，信息不会存储在数据库中。为了创建数据库，我只是做了一个简单的
     CREATE DATABASE MYDATABASE;并创建数据库。我假设执行此代码后，表和行将自行创建。如果我错了，请纠正我。我的第二个问题是当我运行以下帮助程序代码时：

<?php
$x = '1';
//include 'db.php';
$db_hostname = 'MYIP';
$db_database = 'MYDATABASE';
$db_username = 'MYUSERNAME';
$db_password = 'MYPASSWORD';

// Connect to server.
$link = mysqli_connect($db_hostname, $db_username, $db_password);
if (!$link) {
    die('Not connected : ' . mysqli_error());
}

// Select the database. 
$db_selected = mysqli_select_db($link, $db_database);
if (!$db_selected) {
    die ('Can\'t use database : ' . mysqli_error());
}

//get udid from database
while($x = '1'){
$query = "SELECT id, udid, email, premium FROM udid_orders ORDER BY timestamp DESC LIMIT 1";
$sql = mysqli_query($link, $query);
if(mysqli_num_rows($sql) >= 1){
    echo 'registering';
    $row = mysqli_fetch_array($sql);
    $number = $row['id'];
    $udidsql = $row['udid'];
    $email = $row['email'];
    $is_premium = $row['premium'];
    $service_port = '622';
    $address = 'MYIPHERE';
    $socket = socket_create(AF_INET, SOCK_STREAM, SOL_TCP);
    if ($socket === false) {
        echo 'Failed to create socket';
        return;
    }
    $result = socket_connect($socket, $address, $service_port);
    if ($result === false) {
        echo 'Failed to connect socket';
        return;
    }

    $in = "weTi3xJEy7kbldDYWdXe";
    $out = '';

    socket_write($socket, $in, strlen($in));
        while ($out = socket_read($socket, 1024)) {
        break;
    }

    //get us a random number for the name 
    $request = 'OrderInfo ';
    //$random = rand(1, 2000);
    //name that nigga
    $name = $number;
    $request .= $name;
    $request .= " ";        
    //$request .= $order->billing_email;
    $request .= $email;
    $request .= " ";
    //$is_premium = '1';
    //  if(preg_match("/Bronze/i", implode($item))) {
    //      $is_premium = '0';
    //  }
    //  $request .= $is_premium;
    //}
    $request .= $is_premium;
    $out = '';
    socket_write($socket, $request, strlen($request)); 
    while($out = socket_read($socket, 1024)) {
        break;
    }
    $udids = explode(',', $udidsql);
    foreach($udids as $udid){
        $udid_request = "UdidRegister ";

        $udid = $udidsql;

        $udid_request .= $udid;

        $udid_status = "";
        socket_write($socket, $udid_request, strlen($udid_request));
    }
    socket_close($socket);

    $update = "UPDATE udid_orders SET status='1' WHERE udid='$udidsql'";
    $sql_update = mysqli_query($link, $update);

  }


else{
    echo 'sleeping';
    sleep(5);
    }   

}   
// close mysql
mysqli_close($link);
?>

我收到以下错误


  PHP警告：mysqli_num_rows（）期望参数1为mysqli_result，在/link.php的第25行中给出布尔值


我什至不知道这意味着什么，更不用说如何解决它了。

我对PHP和MySQL都还很陌生，因此正确方向的任何指针都将非常有帮助。

Answer 1

您的mysqli_query()返回FALSE，因为查询失败。您应在调用mysqli_num_rows()之前检查查询是否成功执行。