我有一个奇怪的问题,我只是没有找到一个解决方案。问题是,准备好的sql语句没有绑定值、参数,甚至没有通过execute函数传递它们。相反,它会插入':blah'占位符。正如我所说,我已经尝试过bindParam, bindValue和这个方法都没有结果。但是,我现在要再试一次。
我输出了在执行调用之前发送的参数。
Array ( [:username] => schenn [:salt] => NW5552wekj5155cNr52O54q56 [:hashpass] => 5e54240aec6294873d11d6ac3e5b135136a1b671 [:email] => monkey@monkey.com [:state] => OR [:country] => USA [:last_login] => 12/08/2011 )
代码如下:
$query = "INSERT INTO player_acct (username, salt, hashpass, email, state, country, last_login)
VALUES (':username', ':salt', ':hashpass', ':email', ':state', ':country', ':last_login')";
$stmt = $pdoI->prepare($query);
$params = array(":username" => $this->username, ":salt" => $this->salt, ":hashpass" => $this->hashpass,
":email" => $this->email, ":state" => $this->state, ":country" => $this->country, ":last_login" => $this->last_login );
$stmt->execute($params);
您不应该在SQL中引用占位符。尝试以下语句作为SQL字符串:
$query = "INSERT INTO player_acct (username, salt, hashpass, email, state, country,
last_login) VALUES (:username, :salt, :hashpass, :email, :state, :country, :last_login)";
绑定变量时,不能在SQL语句中引用绑定值
$query = "INSERT INTO player_acct (username, salt, hashpass, email, state, country, last_login) VALUES (:username, :salt, :hashpass, :email, :state, :country, :last_login)";
还要确保$this->email
等…