Display MySQL Results as 3 column table

[NickG21]

Hey Everyone,

I'm creating a site that will show images uploaded for certain days working on a job site. Kind of a day-to-day photo journal for the customer. On the site, the user gets here, sees 3 large images, and a series of thumbnails if more than 3 images exist for that day (works fine).

However, underneath that I want to display a 3-4 column setup of "Archived Dates" that provide a link to the images of the other dates. I have this working correctly, but the results are displayed as follows:

 

Date 1: Date 2: Date 3:

 

etc....

 

I want them to display like this;

Day 1 Day 4

Day 2 Day 5

Day 3 Day 6

 

and so on..... in a 3 column format. Here is the code I have right now just looping through to display these link results, not the rest of the page. I am trying to do it tableless right now, but if that isn't the right way to go, please let me know.

Thanks to anyone in advance,

Nick

 

 $SQLRowe = "SELECT DISTINCT RoweImgDate from tblRowe WHERE RoweImgDate !='" . $_GET['date'] . "' Order by RoweImgDate DESC Limit 0, 30";
        //echo $SQLRowe;
        $rsSQLRowe = mysql_query($SQLRowe);

<span class="rowe">Archived Photos:</span><br/>
                    <div id="archive">
                    <?php
                    
                        while($row = mysql_fetch_array($rsSQLRowe)){

                            //echo "<a href='index.php?id="' . $row[RoweImgID] . '"' class='link'>$row[RoweImgDate]</a></br>";
                                    echo "<div id='archivedates'>";
                                    echo "<a href='index.php?date=" . $row[RoweImgDate] . "' class='link'>$row[RoweImgDate]</a>";
                                    echo "</div>";
                                    //echo "<img src='images/$row[RoweImage]'/><br/>";
                                    //echo "<span class='FeatDesc'><p>$row[RoweImgDesc]</p></span><br/>";
                                }
                    ?>
                    </div>

[BlueSkyIS]

i use this, i found on the net. you can adjust it for mysql results:

 

 

/// PUT DATA IN ROWS INSTEAD OF COLUMNS //////////$items = array('ALABAMA','ALASKA');//create an array with your data// Default # of Columns$numcols = 4;// Number of Items$numitems = count($items);// Number of Rows$numrows = ceil($numitems/$numcols);echo '<table>';for ($row=1; $row <= $numrows; $row++) {$cell = 0;echo ' <tr>'."\n";for ($col=1; $col <= $numcols; $col++) {	echo '  <td>'."\n";	if ($col===1) {		$cell += $row;		print $items[$cell - 1];	} else {		$cell += $numrows;		print $items[$cell - 1];	}	echo '  </td>'."\n";}echo ' </tr>'."\n";}echo '</table>';

 

[ignace]

This isn't something complex:

 

echo '<div style="float: left;">', "\n";

$i = 1; $total_per_column = 20;
while($row = mysql_fetch_assoc($result)) {
    if(($i % $total_per_column) == 0) {
        echo '</div>', "\n", '<div style="float: left;">', "\n";
    }
    // echo whatever you want it to look like
    ++$i;
}

echo '</div>';

 

It's adviced to calculate $total_per_column based on the result set (or limit the result set).

[NickG21]

hey guys thanks for the suggestions, unfortunately, i'm not able to get either of them working.  BlueSky, I can't seem to get the array to properly store all the values from the SQL result rows, only 1 value is being stored in the $items array.

 

ignace: it seems like you're code leaves at least one of the <div> tags open when it is executed, I got yours to begin displaying how I wanted but it would only display the first and 4th results, as i set the num of rows per column to only 3, displaying like this;

 

1                                                      4

 

instead of:

 

1                                                    4

2

3

thanks again for the help guys, ill keep plugging

[BlueSkyIS]

BlueSky, I can't seem to get the array to properly store all the values from the SQL result rows, only 1 value is being stored in the $items array.

 

loop over the results, loading them into an array, then build the table.