I am very new to PHP and need to figure this out for a project - I've made an image submit form in HTML that changes the img within a div to the image that selected using the form. Here is how I accomplish this:
echo '<div class="img-container">';
echo '<img class="userimg" src="../images/backgroundplanet.png" />';
echo '<img class="testimg" src="" />'; //stores image from php file
echo '</div>';
echo '<div class="upload-button">Edit Profile</div>';
echo '<input class="file-upload" enctype="multipart/form-data" type="file" name="submit" accept="image/*"/>';
echo '</div>';
and
$(document).ready(function() {
var readURL = function(input) {
if (input.files && input.files[0]) {
var reader = new FileReader();
reader.onload = function (e) {
$('.userimg').attr('src', e.target.result);
}
reader.readAsDataURL(input.files[0]);
}
}
$(".file-upload").on('change', function(){
readURL(this);
});
$(".upload-button").on('click', function() {
$(".file-upload").click();
});
});
I need to upload this image to a column I have created for medium BLOBs in a MySql data base for each individual user. I've tried a number of different methods (tried $sql = new mysql, etc) but have no idea what I'm doing.
I'm now following this tutorial - http://www.sevenkb.com/php/how-to-insert-upload-image-into-mysql-database-using-php-and-how-to-display-an-image-in-php-from-mysql-database/ and have written the following that is supposed to upload an image to the database:
if(!isset($_FILES['upload_image']))
{
echo '<p>Please Select Image to Upload</p>';
}
else
{
try {
upload();
echo '<p>Image Uploaded into MySQL Database as LONGBLOB Using PHP </p>';
}
catch(Exception $e)
{
echo '<h4>'.$e->getMessage().'</h4>';
}
}
function upload(){
/*** check if a file was uploaded ***/
echo '<p>you uploaded</p>';
if(is_uploaded_file($_FILES['upload_image']['tmp_name']) && getimagesize($_FILES['upload_image']['tmp_name']) != false)
{
/*** get the image info. ***/
$size = getimagesize($_FILES['upload_image']['tmp_name']);
/*** assign our variables ***/
$type = $size['mime'];
$imgfp = fopen($_FILES['upload_image']['tmp_name'], 'rb');
$size = $size[3];
$name = $_FILES['upload_image']['name'];
$maxsize = 99999999;
/*** check the file is less than the maximum file size ***/
if($_FILES['upload_image']['size'] < $maxsize )
{
/*** connect to db ***/
$dbh = new PDO("mysql:host=localhost;dbname=sqlserver", 'username', 'password');
/*** set the error mode ***/
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
/*** our sql query ***/
$stmt = $dbh->prepare("INSERT INTO img (image_type ,image, image_size, image_name) VALUES (? ,?, ?, ?)");
/*** bind the params ***/
$stmt->bindParam(1, $type);
$stmt->bindParam(2, $imgfp, PDO::PARAM_LOB);
$stmt->bindParam(3, $size);
$stmt->bindParam(4, $name);
/*** execute the query ***/
$stmt->execute();
}
else
{
/*** throw an exception is image is not of type ***/
throw new Exception("File Size Error");
}
}
else
{
// if the file is not less than the maximum allowed, print an error
throw new Exception("Unsupported Image Format!");
}
}
But the image isn't uploading and does not appear in the new column I made in the database. The only thing displayed on the page is "Please Select Image to Upload"
What's wrong here? Ultimately I need to echo this in a div.. What should I do with this?
When trying to run example:
Errors when entering in my table name:
When going to phpadmin: 
