editing file not taking default value=image_name - php

Question

View
<form class="m-form" action="<?php echo AURL;?>products/update_product/<?php echo $products['product_id'];?>" method="post" enctype="multipart/form-data">
<input type="file" name="product_image_name" class="form-control m-input dropify" placeholder="" data-default-file="<?php echo Website_Assets.'images/'.$products['product_image_name'];?>" value="<?php $products['product_image_name'];?>" data-max-file-size="2M" required>
</form>
In above code product_image_name not taking any value but showing the image picking the path and when I change image it post the image_name
Controller
public function update_product($product_id)
{
echo "<pre>";
print_r ($_FILES['product_image_name']);
echo "</pre>";
exit();
}
changing the image works ok but if i donot change image its not picking the default value of the image

Answer 1

Short answer: An input of type file can not have a default value.
Instead use an <img />-tag to show default images. For example:
<img src=" <?php echo Website_Assets.'images/'.$products['product_image_name'];?>" />
Normal answer: Assuming you are trying to upload an image, save it to the entity that is behind the form and later edit it; try to think around using just the one field. Instead perhaps a sequence like this:
Upload a file using the upload field
Save the uploaded file on the machine that runs PHP (server-side)
Store the path to the uploaded file in the entity
In the edit view of the entity show both:
the image that was uploaded (or a placeholder)
an upload field to upload a new image
Good luck!
Tip: You might also want to have a look at open source libraries like dropzonejs. Existing libraries often give examples and excellent documentation. This example visually combines the upload field with the display field.