man,aging the outputs , need urgent help

webtuto · 03-12-2008, 04:24 PM

hey i have a problem with organizing the outputs from database

for exemple i want to be in a row 5 names then come to the other ligne and put other 5 rows

here is my code

PHP Code:


			
while($row = mysql_fetch_array($reesul))

 {

    echo $row['name'];

    echo "<img src='$row[photo]'>";

 }

thans in advance , i really need the answer thanks

freenity · 03-12-2008, 05:19 PM

maybe this??

PHP Code:


			
while($row = mysql_fetch_array($reesul))
 {
    for($i=0;$i<5;$i++)
    {
        $row = mysql_fetch_array($reesul);
        echo $row['name'];
        echo "<img src='$row[photo]'>";
    }
 }

webtuto · 03-12-2008, 05:23 PM

no , for exemple
the name of the singer then his photo under the name
5 photos and name on one ligne
then it come to the other ligne and post 5 other , .....

freenity · 03-12-2008, 05:25 PM

ah, then I think you have to while the whole results put names in one array, and photos in another.
After that just print 5 names from the first array, then 5 photos on the other lines from the second array, and so on...

ReSpawN · 03-12-2008, 06:46 PM

Then I could suggest to make it a 3D array.

array contains an array for the first value of the first array, and then in that array the information from the value you're trying to access. I will 'map' it out later, got to go!

DeMo · 03-12-2008, 08:53 PM

From your first post it seems the photos also come from the database.
How is the data returned by your SQL query organized?

If it returns 5 rows for each artist (name repeats but the photo changes) you could do it like this:

PHP Code:


			
// Some helper variables

$counter = 0;

$last_name = '';



while ($row = mysql_fetch_array($reesul) {

  ++$counter;



  // Check if we have a new name

  if ($row['name'] != $last_name) {

    // Yes, we have a new name so we print it

    echo $row['name'], '<br />';

    // And we also update $last_name to the current name.

    $last_name = $row['name'];

  }

  

  // Here we print the image tag

  echo '<img src="', $row['photo'], '" />';



  // If we looped through 5 records it's time to print a line break (<br />)

  if ($counter % 5 == 0) {

    echo '<br />';

  }

}

xenon · 03-12-2008, 09:16 PM

I'd go with CSS. Something like:

PHP Code:


			
echo '<ul id="listing">';
while( $row = mysql_fetch_array($result) )
{
    // this would be a loop through all the entries
    echo '<li>
<div>'.$row['title'].'</div>
<div><img src="'.$row['photo'].'" /></div>
</li>';
}
echo '<li class="clear">&nbsp;</li></ul>';

And the CSS:

Code:

#listing { width:600px;}
#listing li { width:100px; float:left;}
#listing li.clear { float:none; clear:both; height:1px; overflow:hidden; line-height:1px;}

webtuto · 03-13-2008, 07:40 PM

thanks but the name and photo changes