基于使用php的mysql查询的自动回复电子邮件

Question

I am working on a simple form only 2 fields. Name and Email. What I have succeded in doing so far is.

1. I have the form send the data to a mysql database.
2. I can query the database in total and also query the database based on a form entry.
3. I have tested and I can send email using a basic mail script or from a form.

Where I am running to an issue is I need the data to also be sent to the submitter along with a third field from the database whish is the auto-encrement primary key as their member number. My question is how can I send email based on the database entry.

ie; Thank you "Submitter Name" for joining your Member ID is "Primary Key". This is my current code.

Index.html

<form method="post" action="doit.php">
    Name<input name="Name" type="text" /><br>
    Email<input name="Email" type="text" /><br>
    <input name="Submit" type="submit" value="submit" />
</form>

doit.php

<?php
$con = mysqli_connect('localhost','DB_Username','DB_Password');

if(!$con)
{
    echo 'Not Connected To Server';
}
if(!mysqli_select_db($con,'DB_Name'))
{
    echo 'Database Not Selected';
}
$Name = $_POST['Name'];
$Email = $_POST['Email'];

$sql = "INSERT INTO Table_name (Name,Email) VALUES ('$Name','$Email')";

if(!mysqli_query($con,$sql))
{
    echo 'Not Inserted';
}
else
{
    echo 'Inserted';
}
header ("refresh:2; url=auto-response.php");


?>

auto-response.php

<?php
if ($_SERVER["REQUEST_METHOD"}=="POST") {
    $connnection=mysqli_connect("localhost","DB_username","DB_password","DB_Name");
    // Check connection
    if ($connection) {
        echo "connection successfull! <br>;"
    } else {
        die("connection failed. Reason ". mysqli_connect_err())
}
{

$Email=$_POST["Email"];
$Name=$_POST["Name"];
$ID=$_POST["ID"];

$sql="SELECT ID,Email,Name FROM Table_Name WHERE Email='".$Email."'";
$result = mysqli_query($connection,$sql);

$to      = 'Email';
$subject = 'Thank you For Joining';
$message = 'Thank you for joining "Email" Your Member ID is "ID"';
$headers = 'From: [e-mail]' . "
" .
    'Reply-To: [e-mail]' . "
" .
    'X-Mailer: PHP/' . phpversion();

mail($to, $subject, $message, $headers);
?> 

Answer 1

Mysql has its own function which will give you id. While you already have the name at the time of insert.

// Using the below line right after your insert you can get the primary id.
$id = $connection->insert_id; // Get latest inserted id.

Hope it helps.

Edit: Based on the edit you made i have arranged your code to how it should work. Try this now.

Index.html

<form method="post" action="doit.php">
    Name<input name="Name" type="text" /><br>
    Email<input name="Email" type="text" /><br>
    <input name="Submit" type="submit" value="submit" />
</form>

doit.php

$con = mysqli_connect('localhost','DB_Username','DB_Password');

if(!$con)
{
    echo 'Not Connected To Server';
}
if(!mysqli_select_db($con,'DB_Name'))
{
    echo 'Database Not Selected';
}

$Name = $_POST['Name'];
$Email = $_POST['Email'];

$sql = "INSERT INTO Table_name (Name,Email) VALUES ('$Name','$Email')";

if(!mysqli_query($con,$sql))
{
    echo 'Not Inserted';
}
else
{
    echo 'Inserted';
    $id = $con->insert_id; // Get latest inserted id.

    $to      = $Email;
    $subject = 'Thank you For Joining';
    $message = 'Thank you for joining "'.$Name.'" Your Member ID is "'.$id.'"';
    $headers = 'From: [e-mail]' . "
" .
        'Reply-To: [e-mail]' . "
" .
        'X-Mailer: PHP/' . phpversion();

    mail($to, $subject, $message, $headers);
}