使用php获取BLOB图像并在html中显示

Question

I am facing an issue getting an image from mysql database and displaying on the web page. I have a user_image table with userid, image (blob datatype), caption, imagename(name of the image, example, 0101.jpg), time(timestamp of the image upload). Users can upload image to database and view their images acc to the timestamp. Any inputs on how to display the image on web page ? Here is the code:

<?php
session_start();
$con = mysqli_connect("localhost", "root", "","login") or die("Unable to connect to MySQL");
$userid = $_SESSION['userid'];
$sql = "SELECT image,image_name,caption FROM user_image WHERE userid=$userid";
if($query = mysqli_query($con, $sql)) {
    while($row=mysqli_fetch_array($query)) {
        header('Content-type: image/jpeg');
        echo $row["image"];
    }
}else{
    echo "Error";
}
$con->close();
?>

Here is the code for uploading image into database:

<?php
session_start(); 
if (isset($_POST['submit'])){
echo $_SESSION["userid"];
$userid = $_SESSION["userid"];
$con=mysqli_connect("localhost","root","","login")or die("Unable to connect to MySQL");
 $caption = $_POST['Caption'];


 $imageName = mysqli_real_escape_string($con,$_FILES["fileToUpload"]["name"]);
$imageData = mysqli_real_escape_string($con,file_get_contents($_FILES["fileToUpload"]["tmp_name"]));
$imageType  = mysqli_real_escape_string($con,$_FILES["fileToUpload"]["type"]);
echo $imageName;
echo $imageType;
if(substr($imageType,0,5) == "image"){
$stmt = mysqli_prepare($con, "INSERT INTO user_image (userid,timestamp, image, image_name,caption) VALUES (?,now(), ?, ?,?)");
if ($stmt === false) {
trigger_error('Statement failed! ' . htmlspecialchars(mysqli_error($con)), E_USER_ERROR);
}
$bind = mysqli_stmt_bind_param($stmt, "isss", $userid,$imageData, $imageName, $caption);
if ($bind === false) {
trigger_error('Bind param failed!', E_USER_ERROR);}
$exec = mysqli_stmt_execute($stmt);
if ($exec === false) {
trigger_error('Statement execute failed! ' . htmlspecialchars(mysqli_stmt_error($stmt)), E_USER_ERROR); }
 }
else
{
echo " only images are allowed";}
}
$con->close();
?>

Answer 1

i use this for my own site echo '<img src="data:image/jpeg;base64,'.base64_encode($customer->GetCustomer($user)["Customer_image"]).'">'

where Customer_image is blob the function GetCustomer($user) is for all information of customer

for my upload i use this

if(isset($_POST["send"])){
    $tmpName = $_FILES['image']['tmp_name'];
    $name = $_FILES['image']['name'];
    $fp = fopen($tmpName, 'r');
    $data = fread($fp, filesize($tmpName));
    fclose($fp);
    $product->UploadImageProduct($id, $name, $data);
}