mysqli_stmt::bind_param()：变量数与

Question

我想运行多个mysql查询(不是并发的)。我是用事先准备好的声明来这样做的。这是我的代码要点：

    <?php



if(isset($_GET['username'])&&isset($_GET['activationid'])){

require_once("../database/db_connect.php");

$stmt= $mysqli->stmt_init();
$stmt->prepare("Select username FROM users where username= ? AND activationid= ?");
$username=$_GET['username'];
$activationid=$_GET['activationid'];
$stmt->bind_param("ss",$username,$activationid);
$stmt->execute();
$row=$stmt->get_result()->fetch_array(MYSQLI_ASSOC);

if(!strcmp($row['username'],$username)){
    echo 'you are registered successfully';



$stmt->prepare("UPDATE users SET active=yes where username = ?");
$stmt->bind_param("s",$username);
$stmt->execute();







}
}

?>

db_connect.php是：

<?php 





define('DBHOST','localhost');
define('DBUSER','root');
define('DBPASS','password');
define('DBNAME','Reminder');

    $mysqli= new mysqli(DBHOST,DBUSER,DBPASS,DBNAME) ;
    if($mysqli->connect_error) { 

    echo $mysqli->mysqli_connect_error();

    }
else {
echo "connected successfully";
}




?>

这给了我一个错误：

警告: mysqli_stmt::bind_param()：变量数与

有人能告诉我我做错了什么吗？

Answer 1

请按以下方式执行您的代码，并检查：-

//$stmt= $mysqli->stmt_init(); comment this line
$stmt = $mysqli->prepare("Select username FROM users where username= ? AND activationid= ?") or die( $mysqli->error);
$username=$_GET['username'];
$activationid=$_GET['activationid'];
$stmt->bind_param("ss",$userid,$activationid);
$stmt->execute();

至于第二个同样的问题:-

$stmt = $mysqli->prepare("UPDATE users SET active=yes where username = ?") or die($mysqli->error);
$stmt->bind_param("s",$username);
$stmt->execute();

注意：-请注意您的变量是正确定义和设置的。谢谢。

Answer 2

在准备好和绑定之后，(在执行之前)，执行

echo $mysqli->error;

没有什么比知道到底发生了什么