Question

我正在尝试建立一个非常简单的表单以将数据提交到MySQL数据库（我以前从未真正设置过MySQL数据库或自己通过Web表单与之交互所需的连接）。

我的网页上有一个简单的HTML表单，允许用户输入姓名，电子邮件地址和他们想问的问题：

<!DOCTYPE html>
<html>
    <head>
        <meta content="text/html;charset=utf-8" http-equiv="Content-Type">
        <meta content="utf-8" http-equiv="encoding">
    Page title
    </head>
    <body>
        <h1>Customer Information</h1>

        <form id="customerInfoForm" accept-charset = "utf-8" action = "insert.php" method = "post" >
            <table id = "customerInfoTable">
                <tr><td>Name: </td><td><input type = "text" name = "customerName" ></input></td></tr>
                <tr><td>Email: </td><td><input type = "text" name = "customerEmail"></input></td></tr>
                <tr><td>Enquiry: </td><td><input type = "text" name = "enquiryText"></input></td></tr>
            </table>
            <br>
            <input type = "submit" value = "Sumit"></input>
        </form>
    </body>
</html>

<?php
$con = mysql_connect('localhost','[retracted]','[retracted]');

if(!$con) {
    echo 'Not connected to server!';
}

$Name = $_POST['customerName'];
$Email = $_POST['customerEmail'];
$Query = $_POST['enquiryText'];

$sql = "INSERT INTO Enquiries(Name, Email, Query) VALUES ('$Name', '$Email', '$Query')";
if(!mysql_query($con, $sql)) {
    echo 'Data not inserted';
} else {
    echo 'Data inserted';
}

?>

在单击表单上的“提交”按钮时调用的insert.php文件中，我具有以下内容：

<?php
    //Check if data has been entered
    if( isset( $_POST['data'] ) && !empty( $_POST['data'] ) )
    {
        $data = $_POST['data'];
    } else {
        header( 'location: form.html' );
        exit();
    }

    //set up mysql
    $sql_server = 'localhost';
    $sql_user = 'user';
    $sql_pwd = 'password';
    $sql_db = 'database';

    //Connect to sql database
    $mysqli = new mysqli( $sql_server, $sql_user, $sql_pwd, $sql_db) or die( $mysqli->error );

    //Insert details into table
    $insert = $mysqli->query( "INSERT INTO Enquiries ('data') VALUE ( '$data' )" );

    //Close mysqli connection
    $mysqli->close;
?>

但是，当前，当我单击表单上的“提交”按钮时，浏览器中显示错误：

error）; //将详细信息插入表$ insert = $ mysqli-> query（“ INSERT INTO查询（'data'）VALUE（'$ data'）”）; //关闭mysqli连接$ mysqli-> close; ？>

即上面的错误显示为网页的内容，因此看来我在数据库连接方面做错了，但我不确定是什么...有人可以向我指出正确的方向吗？

设置向MySQL数据库的简单表单提交

0 个答案: