[RESOLVED] Gallery layout help

unitedintruth

I have used the following code to select the data for photos from mysql database:

<?php
$con = mysql_connect("localhost","uname","password");
if (!$con)
{
  die('Error on connecting: ' . mysql_error());
}

mysql_select_db("my_db", $con);
$result = mysql_query("SELECT * FROM _gallery");

while($row = mysql_fetch_array($result))
{
  echo $row['name'] . " <br /> " . $row['caption'] ."<br /> " . $row['mage'] ."<br />";
}

mysql_close($con);

?>

now I need to layout the images on the page to lay 3 images wide. I think if I use this:

$colsPerRow = 3;

if that is in fact correct, I still don't know where to put it. Can someone please help me lay out this data correctly?

slick101

are you trying to create a table?

echo "<table><tr>"; // start table

$i = 0; // for iteration purpose
while($row = mysql_fetch_array($result))
{
  if($i % 3 == 0)
     echo "</tr><tr>"; // have 3 columns 

  echo '<td><a href="' . $row['image'] . '" title="' . $row['name'] . '>' . $row['caption'] . '</a></td>';
  $i++;
}
echo "</tr></table"; // end table

unitedintruth

Yes, creating the table is exactly what I am trying to do.

I had to change my code to get it to work properly prior to adding your suggestions., would you please tell me how I would do it considering the changes I needed to make?

My code is:

<?php
        $con = mysql_connect("localhost","username","password");
        if (!$con)
        {
        die('Could not connect: ' . mysql_error());
        }
        mysql_select_db("mydb", $con);
        $result = mysql_query("SELECT * FROM puppies
        WHERE approved='yes'");

    echo "<table border='1'>";
    while($row = mysql_fetch_array($result))
    {
    echo "<tr>";
    echo "<td width=\"150\" align=\"center\"><img src=\"http://www.americanmi-kiregistryassociation.com/images/puppy_pics/" . $row['photo'] . "\"><br />";
    echo "Date of Birth: " . $row['dob'] . "<br />";
    echo "Date Available: " . $row['doa'] . "<br />";
    echo "Puppy Number: " . $row['nbr'] . "</td>";
    echo "</tr>";
    }
    echo "</table>";

    mysql_close($con);
  ?>

again, thank you. You guys save my life pretty often.

Thanks,
Donnie

unitedintruth

anyone got any ideas for me?

Thanks,
Donnie

ShawnK

What you want to do is add an $i variable. For example Row 1 Image 1 would have an i value of 1, then increase for R1I2, and then to R1I3. Once it gets to three have it insert a new <tr>