I really have no idea how to do this. Can't figure it out. Any help would be great, please and thank you.
JSON code (it is stored in 'images' column in tblproducts table in the database)
{
"200x200":"http://img.fkcdn.com/image/mobile/p/s/u/lenovo-k6-power-k33a42-200x200-imaezt6hypjzhdug.jpeg",
"400x400":"http://img.fkcdn.com/image/mobile/p/s/u/lenovo-k6-power-k33a42-400x400-imaezt6hypjzhdug.jpeg",
"800x800":"http://img.fkcdn.com/image/mobile/p/s/u/lenovo-k6-power-k33a42-800x800-imaezt6hypjzhdug.jpeg",
"unknown":"http://img.fkcdn.com/image/mobile/p/s/u/lenovo-k6-power-k33a42-original-imaezt6hypjzhdug.jpeg"
}
I need to decode that json array into normal php array using json_decode
function and get the image url using key, fetch each of the image and display it in different tag.
<?php
$category_id = $_GET['category_id'];
$result = mysql_query("select * from tblproducts where category_id = '$category_id");
while($row = mysql_fetch_array($result, MYSQL_ASSOC)){
$json = $row['images'];
$image = var_dump(json_decode($json, true));
?>
<img src="<?php echo $image[0] ?>" alt=" " class="img-responsive" />
<img src="<?php echo $image[1] ?>" alt=" " class="img-responsive" />
<img src="<?php echo $image[2] ?>" alt=" " class="img-responsive" />
<?php } ?>