Question

我正在使用mysqli_insert_id测试获取最后一次auto_increment_ID的功能。当我发现如果我使用2种不同的方法时，我感到很困惑，结果是不同的。

方法1

<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";

// Create connection
$conn = mysqli_connect($servername, $username, $password, $dbname);
// Check connection
if (!$conn) {
die("Connection failed: " . mysqli_connect_error());
}

$sql = "INSERT into item(uid,item_id,item_name,item_price,item_quantity) 
                    VALUES('1','0','hhh','23','23');";

if (mysqli_query($conn, $sql)) {
$last_id = mysqli_insert_id($conn);
echo "New record created successfully. Last inserted ID is: " . $last_id;
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}

mysqli_close($conn);
?>

此程序方式正常，因为我可以获取最后一个ID。

方法2

db.php中

class db{

protected $db_host;
protected $db_name;
protected $db_user_name;
protected $db_pass;

public function __construct($host,$db_n,$user_n,$pass) {
    $this->db_host=$host;
    $this->db_name=$db_n;
    $this->db_user_name=$user_n;
    $this->db_pass=$pass;
}

public function conn(){
    return mysqli_connect($this->db_host, $this->db_user_name, $this->db_pass, $this->db_name);
}
}

test.php的

require "db.php";

$db=new db('localhost','bs','root','');

$sql = "INSERT into item(uid,item_id,item_name,item_price,item_quantity) 
                    VALUES('1','0','hhh','23','23');";

if (mysqli_query($db->conn(), $sql)) {
$last_id = mysqli_insert_id($db->conn());
echo "New record created successfully. Last inserted ID is: " . $last_id;
} else {
echo "Error: " . $sql . "<br>" . mysqli_error($db->conn());
}

mysqli_close($db->conn());

这种方法根本不起作用......它会得到结果0。谁知道我哪里出错了？

Answer 1

只是因为你两次调用$ db-＆gt; conn（），你需要将$ db-＆gt; conn（）分配给变量。如下：

require "db.php";

$db=new db('localhost','bs','root','');

$sql = "INSERT into item(uid,item_id,item_name,item_price,item_quantity) 
                VALUES('1','0','hhh','23','23');";

$conn = $db->conn();
if (mysqli_query($conn, $sql)) {
    $last_id = mysqli_insert_id($conn);
    echo "New record created successfully. Last inserted ID is: " . $last_id;
} else {
   echo "Error: " . $sql . "<br>" . mysqli_error($conn);
}

mysqli_close($conn);

Answer 2

只初始化->conn()方法一次，然后重复使用它。每次调用都会创建一个新的：

$db = new db('localhost','bs','root','');
$connection = $db->conn(); // initialize once
$sql = "INSERT into item(uid,item_id,item_name,item_price,item_quantity) 
                    VALUES('1','0','hhh','23','23');";

if (mysqli_query($connection, $sql)) {
    $last_id = mysqli_insert_id($connection);
    echo "New record created successfully. Last inserted ID is: " . $last_id;
} else {
    echo "Error: " . $sql . "<br>" . $connection->error;
}

$connection->close();

或使用面向对象的接口（->insert_id属性）：

$db = new db('localhost','bs','root','');
$connection = $db->conn(); // initialize once
$sql = "INSERT into item(uid,item_id,item_name,item_price,item_quantity) 
                    VALUES('1','0','hhh','23','23');";

if($connection->query($sql)) {
    $last_id = $connection->insert_id;
    echo "New record created successfully. Last inserted ID is: " . $last_id;
} else {    
    echo "Error: " . $sql . "<br>" . $connection->error;
}

$connection->close();

mysqli_insert_id不能处理php OOP方法

2 个答案: