Upload File Image to server Failure - php

Question

what i am trying to do is when an admin tries to create a new product to upload to the server an image of that and as it's name to be the product id.jpg. I have searched google with no result. The code seems to be right. Can someone help me please? I am using lamp as local test server.
<?php
// Parse the form data and add inventory item to the system
if (isset($_POST['product_name'])) {
$product_name = mysql_real_escape_string($_POST['product_name']);
$price = mysql_real_escape_string($_POST['price']);
$details = mysql_real_escape_string($_POST['details']);
$category = mysql_real_escape_string($_POST['category_choice']);
$condition= mysql_real_escape_string($_POST['condition']);
$supplier_choice= mysql_real_escape_string($_POST['supplier_choice']);
$handling_time= mysql_real_escape_string($_POST['handling_time']);
$weight= mysql_real_escape_string($_POST['weight']);
$information_box= $_POST['information'];
$pid = mysql_insert_id();
// Place image in the folder
$newname = "$pid.jpg";
move_uploaded_file( $_FILES['my_photo']['tmp_name'], "../inventory_images/$newname");
header("location: products.php");
exit();
}
?>
<form action="add_product.php" enctype="multipart/form-data" name="my_Form" id="my_Form" method="post">
<tr>
<td>Φωτογραφία</td>
<td>
<label>
<input type="file" name="my_photo" id="my_photo"/>
</label>
</td>
</tr>
<tr>
<td> </td>
<td>
<label>
<input type="submit" name="button" id="button" value="Αποθήκευση" />
</label>
</td>
</tr>
</table>
</form>

Answer 1

As you said, you are using lamp, check for the permission for the directory in which you are uploading file. In addition, also check for the permission for all the folder that you have mentioned in the destination path.
The directory and all the directory in the path, in which you are uploading file must have write permission.
If move_uploaded_file does not work then try out copy function.
Also check for the path where you are uploading is correct.

Answer 2

I think you'll have better luck getting answers if you thin down your code to contain only portions that are relevant to the problem. You'll probably see the answer to your own question after you do this.
For example, is the file being successfully uploaded? If not, then the mysql_xxx code and most of your form is irrelevant for your question.