Question

我已经尝试过查看类似的问题，但是，我正在试图了解自己的错误。

我的设置：

XAMPP GymPHP（编写代码） localhost上的数据库叫做健身房数据库中的表称为包含9列的记录

我创建了一个表单，现在正尝试在表单和数据库之间建立连接，以便将数据从表单移动到数据库中。

我的代码：

form.php的：

<html>

<head>

    <title>Gym Form</title>
    <meta charset="UTF-8">
    <meta name="viewport" content="width=device-width, initial-scale=1.0">


</head>

<body>

    <form action="Login.php" method="post">

    <span>Gym Membership Registration</span><br><br>

    <Span>Title: </Span><input type ="text" Value =" " name ="Title" /><br>

    <Span>First Name: </Span><input type ="text" Value =" " name ="Fname" /><br>

    <Span>Last Name: </Span><input type ="text" Value =" " name ="Lname" /><br><br>

    <Span>Gender: </Span><select name ="Gender">

        <option value ="Junior">Male</option>
        <option value ="Adult">Female</option>
        <option value ="Senior">Private</option>

    </select><br>

    <Span>DOB: </Span><input type ="date" name ="DOB" /><br><br>

    <Span>MembershipExpiry: </Span> <input type ="date" name ="MemX" /><br>

    <Span>MembershipType: </Span><select name = "MemType">

        <option value ="Junior">Junior</option>
        <option value ="Adult">Adult</option>
        <option value ="Senior">Senior</option>

    </select><br><br>

    <Span>Email Address: </Span><input type ="email" name ="Email" /><br><br>

    <input type="Submit" name="submit" value ="Submit Form">




    </form>

</body>

Connect.php：

    <?php

$conn = mysql_connect("localhost", "root", "");

mysql_select_db("gym");

if(!$conn)
    echo"Error Connecting to Database!";
else
    echo"Connected to Database!";

的login.php：

include "Connect.php";

$title = $_POST['Title'];
$fname = $_POST['Fname'];
$lname = $_POST['Lname'];
$gender = $_POST['Gender'];
$dob = $_POST['DOB'];
$memx = $_POST['MemX'];
$memtype = $_POST['MemType'];
$email = $_POST['Email'];

$sql = mysql_query("INSERT INTO records (Title, Fname, Lname, Gender, DOB,                   MemX, MemType, Email) values ('$title', '$fname', '$lname', '$gender', '$dob', '$memx', '$memtype', '$email')", $conn);

我填写表单并收到一条消息，说明已建立连接，但是，当我检查数据库时...没有添加记录？请帮助，谢谢。

Answer 1

以下代码将帮助您找到错误并解决弃用错误。

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

// Perform queries 
mysqli_query($con, "INSERT INTO records (Title, Fname, Lname, Gender, DOB,MemX, MemType, Email) values ('" . $title . "', '" . $fname . "', '" . $lname . "', '" . $gender . "', '" . $dob . "', '" . $memx . "', '" . $memtype . "', '" . $email . "');" or die(mysqli_error($con));

mysqli_close($con);
?>

XAMPP - 表单到数据库

1 个答案: