image in db php mysql code example
Example: image upload in php code with databases
<?php
// Create database connection
$db = mysqli_connect("localhost", "root", "", "image_upload");
// Initialize message variable
$msg = "";
// If upload button is clicked ...
if (isset($_POST['upload'])) {
// Get image name
$image = $_FILES['image']['name'];
// Get text
$image_text = mysqli_real_escape_string($db, $_POST['image_text']);
// image file directory
$target = "images/".basename($image);
$sql = "INSERT INTO images (image, image_text) VALUES ('$image', '$image_text')";
// execute query
mysqli_query($db, $sql);
if (move_uploaded_file($_FILES['image']['tmp_name'], $target)) {
$msg = "Image uploaded successfully";
}else{
$msg = "Failed to upload image";
}
}
$result = mysqli_query($db, "SELECT * FROM images");
?>
<!DOCTYPE html>
<html>
<head>
<title>Image Upload</title>
<style type="text/css">
#content{
width: 50%;
margin: 20px auto;
border: 1px solid #cbcbcb;
}
form{
width: 50%;
margin: 20px auto;
}
form div{
margin-top: 5px;
}
#img_div{
width: 80%;
padding: 5px;
margin: 15px auto;
border: 1px solid #cbcbcb;
}
#img_div:after{
content: "";
display: block;
clear: both;
}
img{
float: left;
margin: 5px;
width: 300px;
height: 140px;
}
</style>
</head>
<body>
<div id="content">
<?php
while ($row = mysqli_fetch_array($result)) {
echo "<div id='img_div'>";
echo "<img src='images/".$row['image']."' >";
echo "<p>".$row['image_text']."</p>";
echo "</div>";
}
?>
<form method="POST" action="index.php" enctype="multipart/form-data">
<input type="hidden" name="size" value="1000000">
<div>
<input type="file" name="image">
</div>
<div>
<textarea
id="text"
cols="40"
rows="4"
name="image_text"
placeholder="Say something about this image..."></textarea>
</div>
<div>
<button type="submit" name="upload">POST</button>
</div>
</form>
</div>
</body>
</html>