Php 如何将图像上载添加到现有表单？

我正在构建一个表单，将表单内容上传到数据库（mysql）中，还有一个单独的页面来显示数据库的内容。到目前为止，一切都很顺利。我需要用户能够上传一个与表单一起的图像文件，我需要图像本身与数据库内容一起显示在页面上

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT id, yourname, email, phone, sponsorname, sponsorshiplevel,    logofile FROM sponsors2015";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
    echo "ID: " . $row["id"]. "<br>". " Name: " . $row["yourname"]. "<br>". "  Email: " . $row["email"]. "<br>". " Phone: " . $row["phone"]. "<br>". "Sponsor Name: " . $row["sponsorname"]. "<br>". "Sponsorship Level: " . $row["sponsorshiplevel"]. "<br>". "<hr>";
}
} else {
echo "0 results";
}
$conn->close();
?>

如何通过修改现有代码来实现这一点？我已经包括了表单代码、发布到数据库的代码和显示以下内容的页面代码

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT id, yourname, email, phone, sponsorname, sponsorshiplevel,    logofile FROM sponsors2015";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
    echo "ID: " . $row["id"]. "<br>". " Name: " . $row["yourname"]. "<br>". "  Email: " . $row["email"]. "<br>". " Phone: " . $row["phone"]. "<br>". "Sponsor Name: " . $row["sponsorname"]. "<br>". "Sponsorship Level: " . $row["sponsorshiplevel"]. "<br>". "<hr>";
}
} else {
echo "0 results";
}
$conn->close();
?>

谢谢

表格：

你的名字：


电邮：


电话


赞助商名称：


赞助水平：

发布到数据库：

<?php //start php tag
//include connect.php page for database connection
include('connect.php');
//if submit is not blanked i.e. it is clicked.
{
$sql="insert into      sponsors2015(yourname,email,phone,sponsorname,sponsorshiplevel,logofile) values('".$_REQUEST['yourname']."', '".$_REQUEST['email']."', '".$_REQUEST['phone']."', '".$_REQUEST['sponsorname']."', '".$_REQUEST['sponsorshiplevel']."')";


$res=mysql_query($sql);
if($res)



{
Echo header('Location: sponsor-registration-success.php');
}
Else
{
Echo header('Location: sponsor-registration-problem.php');
}

}

?>

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT id, yourname, email, phone, sponsorname, sponsorshiplevel,    logofile FROM sponsors2015";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
    echo "ID: " . $row["id"]. "<br>". " Name: " . $row["yourname"]. "<br>". "  Email: " . $row["email"]. "<br>". " Phone: " . $row["phone"]. "<br>". "Sponsor Name: " . $row["sponsorname"]. "<br>". "Sponsorship Level: " . $row["sponsorshiplevel"]. "<br>". "<hr>";
}
} else {
echo "0 results";
}
$conn->close();
?>

---

要添加图像，需要文件输入标记

在此之后，您需要按如下方式处理文件

这似乎是你可以在谷歌上搜索到的东西，网上有很多关于如何做到这一点的教程。非常感谢！！我已经开始玩弄你的例子，并将适应我的需要。永远感谢大家的快速响应和知识！非常感谢。说到PHP，我仍然认为自己是新手，我是一个初学者。我一直在努力学习，而我似乎通过弄脏我的手学习得最好。。。也就是说，使用代码并学习它对更改的反应（以及我可以找到的好教程）。
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
} 

$sql = "SELECT id, yourname, email, phone, sponsorname, sponsorshiplevel,    logofile FROM sponsors2015";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
    echo "ID: " . $row["id"]. "<br>". " Name: " . $row["yourname"]. "<br>". "  Email: " . $row["email"]. "<br>". " Phone: " . $row["phone"]. "<br>". "Sponsor Name: " . $row["sponsorname"]. "<br>". "Sponsorship Level: " . $row["sponsorshiplevel"]. "<br>". "<hr>";
}
} else {
echo "0 results";
}
$conn->close();
?>