Php PDO,准备使用 if($\u POST[“type”]=“checkEmail”) { $sql=“从`user`中选择*,其中user_email=:email”; $statement=$mysql->prepare($sql); $email=$mysql->quote($_POST[“email”]); $statement->execute(数组(“:email”=>$email)); $re=$statement->fetchAll(); 如果(1) print_r(json_encode($re));//echo json_encode(“对不起,有些人已经听到了这么好的电子邮件:
不要使用->quote(),->prepare()已经在做了Php PDO,准备使用 if($\u POST[“type”]=“checkEmail”) { $sql=“从`user`中选择*,其中user_email=:email”; $statement=$mysql->prepare($sql); $email=$mysql->quote($_POST[“email”]); $statement->execute(数组(“:email”=>$email)); $re=$statement->fetchAll(); 如果(1) print_r(json_encode($re));//echo json_encode(“对不起,有些人已经听到了这么好的电子邮件:,php,mysql,pdo,Php,Mysql,Pdo,不要使用->quote(),->prepare()已经在做了 if($_POST["type"] == "checkEmail") { $sql = "SELECT * FROM `user` WHERE user_email=:email"; $statement = $mysql->prepare($sql); $email = $mysql->quote($_POST["e
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($re));
}
为什么要使用嵌套的
if else
块?当您说prepare不起作用时,它是否返回false
?检查:DON't->quote()
您的值。DB/prepared语句已经为您解决了这个问题。实际上,您是在重复引用。
if($_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($re));
}
if($statement->execute(Array(":email" => $email))) {
// your code
} else {
print_r(json_encode($statement->errorInfo());
}