Echo mysql results in a loop?

Posted by Roy D. Porter on Stack Overflow See other posts from Stack Overflow or by Roy D. Porter
Published on 2013-10-31T04:43:32Z Indexed on 2013/10/31 9:54 UTC
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Filed under:

php

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html

|

algorithm

|

mysqli

I am using turn.js to make a book. Every div within the 'deathnote' div becomes a new page.

<div id="deathnote"> //starts book
<div style="background-image:url(images/coverpage.jpg);"></div> //creates new page
<div style="background-image:url(images/paper.jpg);"></div> //creates new page
<div style="background-image:url(images/paper.jpg);"></div> //creates new page
</div> //ends book

What I am doing is trying to get 3 'content' (content being a name and cause of death) divs onto 1 page, and then generate a new page.

So here is what i want:

<div id="deathnote"> //starts book
<div style="background-image:url(images/coverpage.jpg);"></div> //creates new page
<div style="background-image:url(images/paper.jpg);"></div> //creates new page
<div style="background-image:url(images/paper.jpg);"> //creates new page but leaves it open
<div> CONTENT </div>
<div> CONTENT </div>
<div> CONTENT </div>
</div> //ends the page
    </div> //ends book

Seems simple enough, however the content is data from a MySQL DB, so i have to echo it in using PHP. Here is what i have so far

<div id="deathnote">
    <div style="background-image:url(images/coverpage.jpg);"></div>
    <div style="background-image:url(images/paper.jpg);"></div>
    <div style="background-image:url(images/paper.jpg);"></div>
    <div style="background-image:url(images/paper.jpg);"></div>
    <div style="background-image:url(images/paper.jpg);"></div>
    <div style="background-image:url(images/paper.jpg);"></div>
        <?php
$pagecount = 0;
$db = new mysqli('localhost', 'username', 'passw', 'DB');

if($db->connect_errno > 0){
    die('Unable to connect to database [' . $db->connect_error . ']');
}

$sql = <<<SQL
    SELECT *
    FROM `TABLE`
SQL;

if(!$result = $db->query($sql)){
    die('There was an error running the query [' . $db->error . ']');
}
//IGNORE ALL OF THE GARBAGE ABOVE. IT IS SIMPLE CONNECTING SCRIPT THAT I KNOW WORKS
//THE METHOD I AM HAVING TROUBLE WITH IS BELOW
 $pagecount = 0;
while($row = $result->fetch_assoc()){ //GETS THE VALUE (and makes sure it isn't nothing
echo '<div style="background-image:url(images/paper.jpg);">'; //THIS OPENS A NEW PAGE
while ($pagecount !== 3) { //KEEPS COUNT OF HOW MUCH CONTENT DIVS IS ON THE PAGE
                        while($row = $result->fetch_assoc()){ 


//START A CONTENT DIV
echo '<div class="content"><div class="name">'  . $row['victim'] . '</div><div class="cod">'  . $row['cod'] . '</div></div>';
//END A CONTENT DIV
                            $pagecount++; //UP THE PAGE COUNT
                        }
                                                }
                        $pagecount=0; //PUT IT BACK TO 0
echo '</div>'; //END PAGE
}

$db->close();
?>
    <div style="background-image:url(images/backpage.jpg);"></div> //BACK PAGE
</div>

At the moment i seem to be causing and infinite loop so the page won't load. The problem resides within the while loops.

Any help is greatly appreciated. Thanks in advance guys. :)

Developer IT