MySQL的回聲導致循環？

Question

我正在使用turn.js做書。 “ deathnote” div每個div都會變成一個新頁面。

<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

我正在做的是嘗試將3個“內容”（內容是一個名稱和死因） divs一頁，然后生成一個新頁面。

所以這就是我想要的：

<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

看起來很簡單，但是內容是來自MySQL DB的數據，因此我必須在使用PHP時回顯它。 這是我到目前為止所擁有的

<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>

此刻，我似乎正在引起無限循環，因此該頁面無法加載。 問題出在while循環內。

任何幫助是極大的贊賞。 在此先感謝大家。 :)

Answer 1

請首先在本地實例中運行以下代碼，並確認以下代碼是您的預期概念。 之后，您可以更改結果集並告訴我，

  <?php 

       $Numbers = range(1, 25); // change your result set here 
       $Limit = 3;
       $i=0;
       foreach($Numbers as $number){

            if($i==$Limit){
                    echo "<hr>"; // new page
                    $i=0; // reset the limiter
            }

            echo $number; // Values to be displayed

            $i++;
        }             
      ?>

Answer 2

我為您重寫了循環，現在循環遍歷了整個數據庫，並在需要時放置了開始div和結束div。

<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>

<?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 . ']');
}

$pagecount=0;
while($row = $result->fetch_assoc()){
if($pagecount==0){ echo '<div style="background-image:url(images/paper.jpg);">';}
echo '<div class="content"><div class="name">'  . $row['victim'] . '</div><div class="cod">'  . $row['cod'] . '</div></div>';
if($pagecount==2){echo '</div>';}
$pagecount++;
if($pagecount==3){$pagecount=0;}
}
//this page count closes the last div (page) if you have less than three entries in it
if($pagecount==1||$pagecount==2){echo '</div>';}
$db->close();
?>
    <div style="background-image:url(images/backpage.jpg);"></div> //BACK PAGE
</div>