Php 使用stmt绑定参数插入问题

我的SQL还有一个问题。下面您可以看到我正在使用的代码：

<?php
$con=mysqli_connect("localhost","user","pass","my_db");
// Check connection
if (mysqli_connect_errno()) {
  echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

// escape variables for security
$name = mysqli_real_escape_string($con, $_POST['name']);
$contactpersonname = mysqli_real_escape_string($con, $_POST['contactpersonname']);
$departmentname = mysqli_real_escape_string($con, $_POST['departmentname']);
$title = mysqli_real_escape_string($con, $_POST['title']);
$email= mysqli_real_escape_string($con, $_POST['email']);

$stmt = $con -> prepare("INSERT INTO Contacts (Name, ContactPerson, Department, Title, EmailAddress) VALUES ('?', '?', '?', '?', '?')");
$stmt -> bind_param("sssss", $name, $contactpersonname, $departmentname, $title, $email);
$stmt -> execute();


echo "1 record added";

mysqli_close($con);
?>

单引号（
”
）表示SQL中的varchar文本。通过在
？
周围加引号，可以防止它们被绑定。只需删除引号，您就可以开始了：

$stmt = $con -> prepare("INSERT INTO Contacts (Name, ContactPerson, Department, Title, EmailAddress) VALUES (?, ?, ?, ?, ?)");

删除占位符“
”？”
->
？
周围的引号。去掉空格：
$con->prepare
到
$con->prepare
，然后使用bind命令作为：
绑定参数（$stmt，'ssss'，vars…
），并在您的查询中从
？
中删除引号。谢谢大家！！！终于成功了！！！：D