MySQLi等效的MySQL代码

Question

你可以给我这个相当于MySQLi的代码吗？不能正确的。

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

编辑：护理向我解释什么是错的？

$mysqli = new mysqli($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

那么，在OO意义上说，它会去从：

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

为（假定数字用户ID：

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

为（假设字符串用户ID）：

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

为（假定准备好的发言）：

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

现在，这是假设你正在使用库MySQLi的OOP版本（你应该恕我直言，因为它使生活在许多方面更容易）。

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 
} 

---

办法，我会做到这一点：

$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

我会做这种方式：

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

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