PHP警告：mysqli_num_rows（）和數據庫創建

Question

我正在嘗試使用以下代碼從WooCommerce Wordpress訂單獲取信息，然后將詳細信息傳遞給數據庫。 應該執行此操作的代碼如下：

//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

執行此代碼后，它應該填充數據庫，然后另一個幫助PHP將從數據庫中獲取信息。 但是我有兩個問題 。 第一個問題是，我認為我沒有正確創建數據庫。 我可以成功連接，但是下訂單后，信息不會存儲在數據庫中。 為了創建數據庫，我只做了一個簡單的CREATE DATABASE MYDATABASE; 並創建了數據庫。 我假設執行此代碼后，表和行將自行創建。 如果我錯了，請糾正我。 我的第二個問題是當我運行以下幫助程序代碼時：

<?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);
?>

我收到以下錯誤

PHP警告：mysqli_num_rows（）期望參數1為mysqli_result，在/link.php的第25行中給出布爾值

我什至不知道這意味着什么，更不用說如何解決它了。

我對PHP和MySQL都還很陌生，因此正確方向的任何指針都將非常有幫助。

Answer 1

您的mysqli_query()返回FALSE，因為查詢失敗。 您應在調用mysqli_num_rows()之前檢查查詢是否成功執行。

Answer 2

我猜在您的幫助腳本中，$ sql等於false（表示查詢失敗）。 此外，表不會自動創建，而必須明確創建-應該可以解決您的第一個問題。

Answer 3

為了解決此問題，我最終更改了幾行代碼以反映以下內容（根據文檔）

// Select the database. 
$db_selected = mysqli_select_db($link, $db_database);
if (!$db_selected) {
    die ('Can\'t use database : ' . mysqli_error());
}
    $query = $mysqli->prepare("SELECT * FROM udid_orders");
    $query->execute();
    $query->store_result();

    $rows = $query->num_rows;

    echo $rows;
//get udid from database
while($x = '1'){
$query = "SELECT id, udid, email, type FROM udid_orders";

$sql = mysqli_query($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['type'];
    $service_port = '622';
    $address = '107.170.105.37';
    $socket = socket_create(AF_INET, SOCK_STREAM, SOL_TCP);
    if ($socket === false) {
        echo 'Failed to create socket';
        return;
    }