Question

我正在写这个剧本

<form method="post" id="form" action="join_config.php">
    <label>First Name</label>
    <input name="first_name" type="text" id="firstname">

    <label>Last Name</label>
    <input name="last_name" type="text" id="lastname">

    <label>Email</label>
    <input name="email" type="email">

    <label>Twitter Handle</label>
    <input name="twitter" type="text">

    <label>Country</label>
    <input name="country" type="text">

    <label>City</label>
    <input name="city" type="text">

    <label>Phone</label>
    <input name="phone" type="checkbox">iPhone
    <input name="phone" type="checkbox">Samsung
    <input name="phone" type="checkbox">HTC
    <input name="phone" type="checkbox">Nokia

    <p style="margin-bottom:0;">Others:</p><input name="phone" type="text">
        <label>Availability</label>
        <input name="availability" type="text">
        <input name="submit" type="submit" value="Submit" id="#submit">
</form>

我的PHP mysqli代码

<?php 

    $firstname = $_POST['first_name'];
    $lastname = $_POST['last_name'];
    $email = $_POST['email'];
    $twitter = $_POST['twitter'];
    $country = $_POST['country'];
    $city = $_POST['city'];
    $phone = $_POST['phone'];
    $availability = $_POST['availability'];
    $date = date("d.m.y");

    $SQL = "INSERT INTO join(first_name, last_name, email, twitter, country, city, phone, availability, date)VALUES('$firstname','$lastname','$email','$twitter','$country','$city','$phone','$availability','$date')";
    echo $SQL . "<br>";

    include_once('inc/config.php');

    $stmt = $db->prepare("INSERT INTO join VALUES (?,?)");
    $stmt->bind_param('ss', $firstname, $lastname);
    $stmt->execute();
    $stmt->close();

?>

当我正在运行时，我收到此错误：

Fatal error: Call to a member function bind_param() on a non-object in /join_config.php on line 17

我在Mac OSX Lion上运行，稍后会包含安全线程，但首先需要插入我的数据库。

任何帮助或建议？

致命错误：在第17行的非对象上调用成员函数bind_param（）

0 个答案: