Member Avatar for sushilsth 
 <?php
$query=mysql_query("SELECT model_no,images1 FROM nokia");
echo"<div>";
    while ($img = mysql_fetch_array($query))
    {
        echo $img[0];
        echo "<img src = cms/pages/images/".$img[1].">";
        }
echo"</div>";

?>

what is actually wrong in my code.. its not displaying images($img[1]) but it displays the content 'model_no' ie. $img[0]. plz any sugesstion would be greatly helpful...

  • 6 Contributors
  • 5 Replies
  • 369 Views
  • 18 Hours Discussion Span
  • Latest Post Latest Post by kgavhane12

Recommended Answers

All 5 Replies

Member Avatar for darkagn

Here is a good approach for displaying an image from a MySQL database using PHP.

Member Avatar for kannan mangalar 
<?php
$query = $mysqli->query("SELECT model_no,images1 FROM nokia");
$fetch = $query->fetch_object();
$img = $fetch->imgname;
?>
<html>
    <body>
        <img width="100%" height="100%" src="images/<?php echo $img; ?>">
    </body>
</html>
Member Avatar for diafol

@sush

I can't see much wrong with your code. Have a look at the underlying html (view source) in your browser. Does the image src attribute show the value you expect it to?

//EDIT

Strike that - you've messed up the double quotes

echo "<img src = cms/pages/images/".$img[1].">";

should be

echo "<img src = 'cms/pages/images/{$img[1]}'>";
Edited by diafol
Member Avatar for gabrielcastillo

yes, like diafol pointed out, change $img[0] to $img[1]

Member Avatar for kgavhane12 
<?php 
$query=mysql_query("SELECT model_no,images1 FROM nokia");
  while ($img = mysql_fetch_array($query))

  @$mod_no=$img['model_no'];
  @$image=$img['images1'];
  echo $mod_no."\n";
  ?>
  <img src="cms/pages/images/<?php echo $image; ?>" width="218" height="172">
  <?php
  }

?>
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