Update an image with one by one - php

Question

I want to update image in one by one.
Example : I want to update only IMG1.
img1 : img_id = 1, img_name = demo1.jpg
img2 : img_id = 2, img_name = demo2.jpg
When I press update button with only IMG1 it should send img_id = 1 to $id = $_POST['image_id']; But it send img_id = 2 Then will make IMG2 has been update not IMG1.
updateimage.php
<div class="col-md-4 col-sm-6 col-xs-12">
<div class="text-center">
<?php $result = mysql_query("SELECT img_id, img_name FROM fm_product_image WHERE p_id_img = '$ID'")or die(mysql_error()); ?>
<?php while ($imagerows = mysql_fetch_array($result)) { $imgid = $imagerows['img_id']; ?>
<img class="avatar img-circle img-thumbnail" style="border-radius: 2px; width: 70%;" src="upload/<?php echo $imagerows['img_name']; ?>">
<h6>Upload your product image.</h6>
<input type="file" name="image" class="text-center center-block well well-sm">
<input type="text" required value="<?php echo $imgid; ?>" name="image_id" class="text-center center-block well well-sm">
<button type="submit" name="image" class="btn btn-success" style="margin: auto;">Update</button>
<br>
<br>
<?php } ?>
</div>
</div>
<?php
$id = $_POST['image_id'];
if (isset($_POST['image'])) {
$image = $_FILES["image"]["name"];
$image_name= addslashes($_FILES['image']['name']);
$size = $_FILES["image"] ["size"];
move_uploaded_file($_FILES["image"]["tmp_name"],"upload/" . $_FILES["image"]["name"]);
$product_img = $_FILES["image"]["name"];
mysql_query(" UPDATE fm_product_image SET img_name = '$product_img' WHERE img_id = '$id'")or die(mysql_error());
header('location:product.php');
} ?>

Answer 1

The problem is in your loop. You are submitting the same form, and that form contains the text fields which each image's ID. So, even if you have ten images, the IDs for all ten will be submitted, and the server will most likely use the last ID, which would be 10, and update that one.
The solution is to create a form on each iteration of your loop, that way, when you hit the submit button, it submits only the form containing the ID of a given image. Here is a solution below:
<?php $result = mysql_query("SELECT img_id, img_name FROM fm_product_image WHERE p_id_img = '$ID'")or die(mysql_error()); ?>
<?php while ($imagerows = mysql_fetch_array($result)) : ?>
<?php $imgid = $imagerows['img_id']; ?>
<img class="avatar img-circle img-thumbnail" style="border-radius: 2px; width: 70%;" src="upload/<?php echo $imagerows['img_name']; ?>">
<h6>Upload your product image.</h6>
<form method="post" action="YOUR_ACTION_PAGE" enctype="multipart/form-data">
<input type="file" name="image" class="text-center center-block well well-sm">
<input type="text" required value="<?php echo $imgid; ?>" name="image_id" class="text-center center-block well well-sm">
<button type="submit" name="image" class="btn btn-success" style="margin: auto;">Update</button>
</form>
<br>
<br>
<?php endwhile; ?>
Notice I created a new form on each iteration through the loop.
I must also mention that you need to consider switching to the better and safer mysqli_* family of functions.

Answer 2

The problem is that you cannot have multiple inputs of type text / file with the same name attribute in the same form tag or in the DOM in there is no form tag around the inputs.
To solve this problem, right after you open the while, add a form with a different name (e.g. name of the image that you want updated) and right before closing the while loop, close the form tag.
EDIT:
This is what I was talking about.
<?php while ($imagerows = mysql_fetch_array($result)) { $imgid = $imagerows['img_id']; ?>
<form name="update_<?php echo $imagerows['img_name']; ?>_form" method="post">
<img class="avatar img-circle img-thumbnail" style="border-radius: 2px; width: 70%;" src="upload/<?php echo $imagerows['img_name']; ?>">
<h6>Upload your product image.</h6>
<input type="file" name="image" class="text-center center-block well well-sm">
<input type="text" required value="<?php echo $imgid; ?>" name="image_id" class="text-center center-block well well-sm">
<button type="submit" name="image" class="btn btn-success" style="margin: auto;">Update</button>
<br>
<br>
</form>
<?php } ?>