Displaying image name under image using PHP - php

Question

I have developed a page where i display image from the database along with its image name.
I managed to display the names and imagenames from the database successfully, but the images and imagenames are aligned like:
IMG1 IMG2 IMG3 IMGname1 IMGname2 IMGname3
The code i tried to display the images and its name from the database:
<div class='badgeicon'>
<?php
$sql = "SELECT * FROM badges WHERE badge_id='$id_badge'";
$result = mysqli_query($db,$sql) or die(mysqli_error($db));
while($row = mysqli_fetch_array($result)){
$image_src2 = $row['badge_image'];
$imagedes = $row['description'];
?>
<img src= "<?php echo $image_src2; ?>" width="130px" height="140px">
<?php echo $imagedes; ?>
I expect the output to be :
IMG1 IMG2 IMG3
IMGname1 IMGname2 IMGname3
i.e., with the imagenames directly under the respective images.

Answer 1

I think your best bet is to format with css.
<div class="image">
<img src= "<?php echo $image_src2; ?>" width="130px" height="140px">
<span class="description"><?php echo $imagedes; ?></span>
</div>
And here for the style
.image{
display: inline-block;
}
.description{
bottom: 0;
}
You can do a lot more depending on exactly what you are trying to achieve.

Answer 2

I think this should be what you are looking for, keeping the img and name in a loop:
<div class='badgeicon'>
<?php
$sql = "SELECT * FROM badges WHERE badge_id='$id_badge'";
$result = mysqli_query($db,$sql) or die(mysqli_error($db));
$fetchresult = mysqli_fetch_array($result);
foreach($fetchresult as $row){
echo '<img src="'.$row['badge_image'].'" width="130px" height="140px">';
echo $row['description'];
}
?>