if($_POST["type"] == "checkEmail")
{
$sql = "SELECT * FROM `user` WHERE user_email=:email";
$statement = $mysql->prepare($sql);
$email = $mysql->quote($_POST["email"]);
$statement->execute(Array(":email"=>$email));
$re = $statement->fetchAll();
if(1)
print_r(json_encode($re));//echo json_encode("Sorry,Some has tood that good e-mail:<");
else
print_r(json_encode($re));
}
我尝试使用查询功能,它成功了!但准备功能不起作用!
最佳答案
不要使用 ->quote(),->prepare() 已经这样做了。
if($_POST["type"] == "checkEmail")
{
$sql = "SELECT * FROM `user` WHERE user_email=:email";
$statement = $mysql->prepare($sql);
$email = $_POST["email"];
$statement->execute(Array(":email"=>$email));
$re = $statement->fetchAll();
if(1)
print_r(json_encode($re));//echo json_encode("Sorry,Some has tood that good e-mail:<");
else
print_r(json_encode($re));
}
如果这不是问题,那么您可能应该发布您收到的错误:
if($statement->execute(Array(":email" => $email))) {
// your code
} else {
print_r(json_encode($statement->errorInfo());
}
关于php - PDO,准备使用,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/26592174/