mysql 中的图像未显示在 php 中

Question

i am trying to display database value in php, my php code is below

<?php
 $servername = "localhost";
 $username = "root";
 $password = "";
 $dbname = "teia";

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

$sql = "SELECT * from registers ORDER BY ID DESC limit 1 ";
$result = $conn->query($sql);

if (!empty($result) && $result->num_rows > 0) {
    // output data of each row
    while($row = $result->fetch_assoc()) {
        $name     = $row['firstname'];
        $lastname = " $row[lastname] ";
        $mobile   = " $row[mobilenumber] ";
        $exp      = " $row[experience] ";
        $photo    = " $row[Photo] ";
    }
} else {
    echo "0 results";
}

this is what i tried to get image

<?php echo '<img src="data:image/jpeg;base64,'.base64_encode( $row['Photo'] ).'"/>';?>

every data is being displayed except the image, may i know what is the reason for it?

enter image description here enter image description here

Answer 1

A general practice is to store images in directories on the file system and store references to the images in the database. eg path to the image,the image name, etc.. Or alternatively, you may even store images on a content delivery network (CDN) or numerous hosts across some great expanse of physical territory, and store references to access those resources in the database.

As you mentioned that, you have store just title of the image. As well as, you have to store the image to particular path.

    $target_dir = "uploads/";
    $target_file = $target_dir . basename($_FILES["imageUpload"]["name"]);
    $uploadOk = 1;
    $imageFileType = pathinfo($target_file,PATHINFO_EXTENSION);

    if (move_uploaded_file($_FILES["imageUpload"]["tmp_name"], $target_file)) {
        echo "The file ". basename( $_FILES["imageUpload"]["name"]). " has been uploaded.";
    } else {
        echo "Sorry, there was an error uploading your file.";
    }

You can store the name of the image like:

$image=basename( $_FILES["imageUpload"]["name"],".jpg"); 

When you want to render then just do like this:

<img src='uploads/$image_name' height='150px' width='300px'></td>"

Where $image_name will be retrieve from the database

Answer 2

For Inserting image into database as blob format:

$db = mysqli_connect("localhost","root","","DbName"); //keep your db name
$image = addslashes(file_get_contents($_FILES['images']['tmp_name']));
//you keep your column name setting for insertion. I keep image type Blob.
$query = "INSERT INTO products (id,image) VALUES('','$image')";  
$qry = mysqli_query($db, $query);

Try to retrieve the image like this:

$db = mysqli_connect("localhost","root","","DbName"); //keep your db name
$sql = "SELECT * FROM products WHERE id = $id";
$sth = $db->query($sql);
$result=mysqli_fetch_array($sth);
echo '<img src="data:image/jpeg;base64,'.base64_encode( $result['image'] ).'"/>';