cannot display multiple image - php

Question

Image cannot be displayed in html page.
for example code :
<?php
$id = $_GET['id'];
$data =mysql_query("select * from img_homestay WHERE id='$id'");
while ($row =mysql_fetch_array($data))
{
$location = $row['location'];
echo '<img src="'.$location.'" width=30% height=10%>';
echo '<td><div align="center">Delete</div></td>';
echo "<br>";
}
?>

Answer 1

One obvious mistake in your code is that the closing tag '>' is missing for img. So it needs to be
echo "<img src='$location' width='30%' height='10%'>";
Apart from this be sure $location var has the correct absolute or relative path picked from DB to show the image. Then, you are using $id in the query. Be sure this is not the unique ID in the img_homestay table, as that will only return you one row. I believe you want to fetch all the images for a particular post id, so ensure that you are using that ID only for the correct field in the query.
Another suggestion that done switch between double quote and single quote string notation in PHP. This will make your code hard to read and comprehend. In one echo statement you are using
echo " ";
and in next statement you are using:
echo ' ';

Answer 2

Try with this
<?php
$id = $_GET['id'];
$data =mysql_query("select * from img_homestay WHERE id='".$id."'");
while ($row =mysql_fetch_array($data))
{
$location = $row['location'];
?>
<img src="<?php echo $location;?>" width="30%" height="10%">
<td><div align="center">Delete</div></td>
<br>
<?php }
?>

Answer 3

<?php
$id = $_GET['id'];
$data =mysql_query("select * from img_homestay WHERE id='".$id."'");
while ($row =mysql_fetch_array($data))
{
$location = $row['location'];
echo '<img src="'.$location.'" width=30% height=10%>';
echo '<td><div align="center">Delete</div></td>';
echo "<br>";
}
?>

Answer 4

Your processing is wrong. Your query is select * from img_homestay WHERE id='$id' which means it's going to fetch a single row (I assume, because you're making a query based on an ID which, I again assume, is a unique key) so you actually don't need to use the while loop (if you intend to use single image).
Still, if that's not the case, you may need to use <tr> for each row, so maybe, try this:
$id = $_GET['id'];
$data =mysql_query("select * from img_homestay WHERE id=$id");
while($row=mysql_fetch_array($data)){
$location = $row['location'];
echo "<tr><td><img src='$location' width=30% height=10%>";
echo "<td><div align='center'><a href='#' imgid='$row[imgid]' class='delbutton' title='Click To Delete'>Delete</a></div></td>";
echo "</tr>";
}
?>
Also, I see, you're using td here which means, you're using table but I see no table tags. My guess is, there is some issue in table structure. A better check to see if you're image is being parsed or not would be to check the source code of the rendered HTML page. I'm sure you're getting the img tag in resulting HTML but it is not being rendered due to error in HTML structure so you better check and correct your HTML with respect to the table you're using.