Question

我正在尝试创建一个小表单，供人们访问我的网站，只需输入以下信息：

名称
姓
电子邮件地址

它应该将此信息从我的数据库发回到我的表中。

现在这里的事情变得“复杂”（......好吧，对我来说，好吗？）

如果我们的表中确实存在电子邮件地址，那么它必须回显“已添加电子邮件地址。谢谢！”。否则，如果它不存在，那么它必须回显“电子邮件地址已经存在！”

出于某种原因，它没有这样做。它正在吐出这个错误：

Fatal error: Call to a member function fetch_assoc() on a non-object in C:\wamp\www\addemail.php on line 46

我附上了我的php代码（如下）：

<!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN">

<html lang="en">

<head>

<script LANGUAGE="JavaScript">
<!--
function redirect () { setTimeout("go_now()",3000); }
function go_now ()   { window.location.href = "newsletter_frame.html"; }
//-->
</script>

<!-- This is the link to the external CSS file-->
<link rel="stylesheet" type="text/css" href="css/mystyle.css">

<!-- This is the link to the external Javascript file-->
<script type="text/javascript" src="js/myjs.js"></script>

<title>Page</title>

<meta http-equiv="Content-Type" content="text/html; charset=ISO-8859-1">
<meta name="verify-v1" content="gDSxHR1Tk8vWtty9GoRHABGFEH+Bh2VHYCHv0Cx1/Ek=">
<meta name="copyright" content="Burger man">
<meta name="keywords" content="afro, afrodeep, soulful, funky, deep, vocal house, broken beats, disco">
<meta name="description" content="Burgers and chips">
<meta name="page-type" content="Information">

</head>
    <body  onLoad="redirect()" id="iframe_newsl_spec">

        <?php
            $dbc = mysqli_connect('localhost', 'root', '', 'si_store') or die('Error connecting to MySQL server.');

                if ( isset($_POST['email_address']) && isset($_POST['firstname']) && isset($_POST['surname']) )
                    {
                        $email_address = $_POST['email_address'];
                        $firstname = $_POST['firstname'];
                        $surname = $_POST['surname'];

                        $query = mysqli_query("SELECT * FROM email_list WHERE email_address = '$email_address'");

                        if ( mysql_num_rows($query) > 0)
                        {
                            echo 'Email address already exists!';
                        }
                        else
                        {
                            mysqli_query("INSERT INTO email_list(email_address, firstname, surname) VALUES ('$email_address','$firstname','$surname')");

                            echo 'Email Address Added. Thank you!';
                        }   
                    }

            //mysqli_query($dbc, $query) or die ('Error connecting to database.');

            mysqli_close($dbc);
        ?>



    </body>
</html>

Answer 1

修复mysql_*个函数并将其更改为mysqli_*，和之后，您已更改了代码，以便使用预处理语句< / em>，您应该添加以下代码：

$stmt = mysqli_prepare('SELECT * FROM email_list WHERE email_address = ?'); //BTW, if all you want to check if a record exists, don't use SELECT * //Just select one field, not the entire row mysqli_stmt_bind_param($stmt,'s',$_POST['email_address']); //bind param, to statement, as String, value ^^ mysqli_stmt_execute($stmt); //execute the query mysqli_stmt_bind_result($stmt,$col1,$col2,$col3); //bind variables that will be assigned the value of the fields while(mysqli_stmt_fetch($stmt)) {//get the values in question echo 'col1: '.$col1.'<br/>col2: '.$col2.'<br/>col3: '.$col3.'<br/>'; }

要获得结果集的关联数组，您将不得不read the docs有一些函数可能需要使用。

致命错误：在非对象上调用成员函数Fetch_Assoc（）

1 个答案: