Question

我被困在这里它给了我这个错误=（任何帮助？

$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API和样式。

mysql_connect和mysql_select_db与MySQLi的库不同。

尝试：

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

而不是：

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

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

然后将准备好的陈述改为：

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

从：

开始

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

警告：mysqli :: prepare（）只需要1个参数，给出2

1 个答案: