I found the answer: for those who are looking for the same, this is how I did it. You should not consider uploading images to a database, instead you can save the name of the downloaded file in your database, and then get the file name and use it where you want the image to be displayed.
HTML code
<input type="file" name="imageUpload" id="imageUpload">
PHP code
if(isset($_POST['submit'])) { //Process the image that is uploaded by the user $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."; } $image=basename( $_FILES["imageUpload"]["name"],".jpg"); // used to store the filename in a variable //storind the data in your database $query= "INSERT INTO items VALUES ('$id','$title','$description','$price','$value','$contact','$image')"; mysql_query($query); require('heading.php'); echo "Your add has been submited, you will be redirected to your account page in 3 seconds...."; header( "Refresh:3; url=account.php", true, 303); }
CODE FOR IMAGE DISPLAY
while($row = mysql_fetch_row($result)) { echo "<tr>"; echo "<td><img src='uploads/$row[6].jpg' height='150px' width='300px'></td>"; echo "</tr>\n"; }
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