在将json数组解码为普通的php数组后，使用键获取图像url

Question

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 } ?>

Answer 1

change your php code to iterate over the fetched array, and remove var_dump() from your code as

<?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'];
        $image = json_decode($json, true);
        foreach($image as $key =>$val){
    ?>
        <img src="<?php echo $val ?>" alt=" " class="img-responsive" />
   <?php }
   }
?>