CODEWITHSUNDEEP
phpmysql
James Osguthorpe
James Osguthorpe

Reputation: 171

MySQL - PHP: Display Results In Table Rows (5 results per row)

I have 10 images in my database. I would like to display 5 images per table row.

image1 - image 2 - image 3 - image 4 - image 5

(new row)

image6 - image7 - image8 - image9 - image10

How am I able to do this using MySQL and PHP?

My Code:

$query = mysql_query("SELECT * FROM gallery WHERE gallery_number='$gallery_number'");
while($fetch = mysql_fetch_assoc($query)){

  $image_name = $fetch['image_name'];
  $image_location = $fetch['image_location'];

  echo '<table width="960">';
  echo '<tr>';
  echo '<td>'.'<img src="'.$image_location.'" alt="'.$image_name.'"/>'.'</td>'; 
  echo '</tr>';
  echo '</table>';  
}

As you can see, the results only display in one huge row. The thing I must add is the rows will need to expand with the result.

Example; there maybe more than 10 images per result. There could be, say, 20 images, this would mean I would need 4 rows...

Upvotes: 0

Views: 7865

Answers (1)

przeqpiciel
przeqpiciel

Reputation: 328

    $query = mysql_query("SELECT * FROM gallery WHERE gallery_number='$gallery_number'");

    echo '<table width="960">';
    $i = 0; //first, i set a counter 
    while($fetch = mysql_fetch_assoc($query)){
    //counter is zero then we are start new row  
        if ($i==0){
            echo '<tr>';
        }
        //here we creating normal cells <td></td>
        $image_name = $fetch['image_name'];
        $image_location = $fetch['image_location'];
        echo '<td>'.'<img src="'.$image_location.'" alt="'.$image_name.'"/>'.'</td>';   
        //there is a magic - if our counter is greater then 5 we set counter to zero and close tr tag  
        if ($i>5){
            $i=0;
            echo '</tr>';
        };  
        $i++; //$i = $i + 1 - counter + 1
    }
    echo '</table>';

Upvotes: 1

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