photo upload in php code example

Example 1: how to upload image in php and store in database and folder

// Get the name of images
  	$Get_image_name = $_FILES['image']['name'];
  	
  	// image Path
  	$image_Path = "images/".basename($Get_image_name);

  	$sql = "INSERT INTO student_table (imagename, contact) VALUES ('$Get_image_name', 'USA')";
  	
	// Run SQL query
  	mysqli_query($conn, $sql);

  	if (move_uploaded_file($_FILES['image']['tmp_name'], $image_Path)) {
  		echo "Your Image uploaded successfully";
  	}else{
  		echo  "Not Insert Image";
  	}
  }

Example 2: basic code for file upload in php

//This is the minimal code for an image upload for first time learners
//html portion
<!DOCTYPE html>
<html>
<head>
	<title>ImageUpload</title>
</head>
<body>
	<form action="upload.php" method="post" enctype="multipart/form-data">
		<label>Username</label>
		<input type="text" name="username">
		<br>
		<label>UploadImage</label>
		<input type="file" name='myfile'>
		<br/>
		<input type="submit" value="upload">
	</form>
</body>
</html>
  
 //php portion
  <?php
	$user=$_POST['username'];
	$image=$_FILES['myfile'];
	echo "Hello $user <br/>";
	echo "File Name<b>::</b> ".$image['name'];

	move_uploaded_file($image['tmp_name'],"photos/".$image['name']);
	//here the "photos" folder is in same folder as the upload.php, 
	//otherwise complete url has to be mentioned
	?>

Example 3: php image upload in database

<?php
error_reporting(0);
?>
<?php
$msg = "";

// If upload button is clicked ...
if (isset($_POST['upload'])) {

	$filename = $_FILES["uploadfile"]["name"];
	$tempname = $_FILES["uploadfile"]["tmp_name"];
		$folder = "image/".$filename;

	$db = mysqli_connect("localhost", "root", "", "image_upload");

		// Get all the submitted data from the form
		$sql = "INSERT INTO images (filename) VALUES ('$filename')";

		// Execute query
		mysqli_query($db, $sql);

		// Now let's move the uploaded image into the folder: image
		if (move_uploaded_file($tempname, $folder)) {
			$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>
<link rel="stylesheet" type= "text/css" href ="style.css"/>
<div id="content">

<form method="POST" action="" enctype="multipart/form-data">
	<input type="file" name="uploadfile" value=""/>

	<div>
		<button type="submit" name="upload">UPLOAD</button>
		</div>
</form>
</div>
</body>
</html>

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