MySQLi等效的MySQL代碼

Question

你可以給我這個相當於MySQLi的代碼嗎？不能正確的。

<?php 
if(mysql_num_rows(mysql_query("SELECT userid FROM users WHERE userid = '$userid'"))){ 
//code to be exectued if user exists 
} 
?> 

編輯：護理向我解釋什麼是錯的？

$mysqli = new mysqli($host, $username, $pass, $db); 

if ($mysqli->connect_error) { 
    die('The Server Is Busy. Please Try Again Later.'); 
} 
$result = $mysqli->query("SELECT userid FROM users WHERE userid = '$userid'"); 
if ($result->num_rows) { 
    echo "<h1>AWESOME</h1>"; 
} 

Answer 1

那麼，在OO意義上說，它會去從：

if(mysql_num_rows(mysql_query("SELECT userid FROM users WHERE userid = '$userid'"))){ 
    //code to be exectued if user exists 
} 

爲（假定數字用戶ID：

$result = $mysqli->query("SELECT userid FROM users WHERE userid = ".(int) $userid); 
if ($result->num_rows) { 
    //code 
} 

爲（假設字符串用戶ID）：

$result = $mysqli->query("SELECT userid FROM users WHERE userid = '". $db->real_escape_string($userid) . "'); 
if ($result->num_rows) { 
    //code 
} 

爲（假定準備好的發言）：

$stmt = $mysqli->prepare("SELECT userid FROM users WHERE userid = ?"); 
$stmt->bind_param('s', $userid); 
$stmt->execute(); 
$stmt->store_result(); 
if ($stmt->num_rows) { 
    //code 
} 

現在，這是假設你正在使用庫MySQLi的OOP版本（你應該恕我直言，因爲它使生活在許多方面更容易）。

Answer 2

$connection = mysqli_connect(...); 

if(mysqli_num_rows(mysqli_query($connection,"SELECT userid FROM users WHERE userid = '$userid'"))){ 
//code to be exectued if user exists 
} 

---

辦法，我會做到這一點：

$connection = mysqli_connect(...); 

$result = mysqli_query(
    $connection, 
    "SELECT COUNT(*) AS cnt FROM users WHERE userid = '$userid'" 
) or die(mysqli_error()); //use proper error handling here: or die() used for brevity 

$row = mysqli_fetch_assoc($result); 

if($row['cnt'] > 0) { 
    //do your thing 
} 

Answer 3

我會做這種方式：

$result = $mysqli->query("SELECT userid FROM users WHERE userid = '$userid'"); 
$row = mysqli_fetch_assoc($result); 

if ($row['userid'] > 0) { 
    echo "<h1>AWESOME</h1>"; 
}