警告：mysqli::prepare() 只需要 1 个参数，2 个在

Question

我被困在这里它给了我这个错误=（有什么帮助吗？

$link = mysql_connect("localhost","odyssexxxxxx","xxxxxxxx")
or die ("Could not connect :" . mysql_error());

// db
mysql_select_db("odysseus_matchcode",$link);


// ejecucion del query

if ($insert_stmt = $mysqli->prepare("UPDATE members SET (nombre, apellido, username, email, password, salt, telefono) VALUES (?, ?, ?, ?, ?, ?, ?) WHERE `cedula` = '$cedula'",$link)) {    
   $insert_stmt->bind_param('sssssss', $nombre, $apellido, $username, $email, $password, $random_salt, $telefono); 
    // Execute the prepared query.
   $insert_stmt->execute();
}

Answer 1

您正在混合您的 API 和样式。

mysql_connect 和 mysql_select_db 与 MySQLi 属于不同的库。

试试看：

$mysqli = new mysqli("host","user","password","database"); 

代替：

$link = mysql_connect("localhost","odyssexxxxxx","xxxxxxxx")
or die ("Could not connect :" . mysql_error());

// db
mysql_select_db("odysseus_matchcode",$link);

然后将你准备好的语句更改为：

if ($insert_stmt = $mysqli->prepare("UPDATE members SET (nombre, apellido, username, email, password, salt, telefono) VALUES (?, ?, ?, ?, ?, ?, ?) WHERE `cedula` = ?")) { 

   $insert_stmt->bind_param('ssssssss', $nombre, $apellido, $username, $email, $password, $random_salt, $telefono,$cedula); 

我还建议阅读手册：http://uk1.php.net/manual/en/book.mysqli.php

开始于：

Mysqli::Construct 了解如何正确初始化 MySQLi 类，然后继续准备语句 stmt::FunctionName