start PHP echo over every fourth mysql table row - php

Question

I have a set of image url's in a database which I am echoing into an <img>tag to display on a website.
I am using bootstrap and have my basic model set up like this:
<div class="row-fluid gallery">
<div class="span3">
<img src="some fancy url here for the image">
</div>
</div>
Now if you have ever used bootstrap you know once that span3 reaches 12 (basically when 4 images are displayed in the row). You must start the above all over to keep the images all in line.
I have a PHP script below that echoes out the image source and the above layout. The problem is, I have more than 4 images to echo out. I removed credentials for security purposes.
$con=mysqli_connect("localhost","username","password","db_name");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM gallery");
while($row = mysqli_fetch_array($result))
{
echo "<div class='row-fluid gallery'>";
echo "<div class='span3'>";
echo"<img src='". row['image_url']."'>";
echo "</div>";
echo "</div>";
}
mysqli_close($con);
My question is how do you do something like:
for every 4th image {
<div class="row-fluid gallery">
<div class="span3">
<img src="some fancy url here for the image">
</div>
</div>
}
Basically, I can say it in English and know what I need. But I can't say it in PHP and have it know what I need. Any help would be appreciated.

Answer 1

Create a counter variable outside the loop and check every fourth
$i = 1;
while($row = mysqli_fetch_array($result))
{
// do every time
if($i % 4 == 0)
{
// do only each 4th time
}
$i++;
}

Answer 2

$counter=0;
while($row = mysqli_fetch_array($result))
{
echo "<div class='row-fluid gallery'>";
echo "<div class='span3'>";
echo"<img src='". row['image_url']."'>";
echo "</div>";
echo "</div>";
$counter++;
if (!($counter%4))
// do your fancy staff, this is the forth image
}