2015-01-16 82 views
-2

我的PHP滑塊有點工作,我設法讓它鏈接到我的數據庫。但是,我需要它循環滑塊內的所有圖像,我的代碼不起作用,但我相信我錯過了循環查詢?!如何使我的PHP滑塊工作

這是我的代碼在我的滑塊:

<div class="theme-dark 16 columns"> 
     <div id="slider" class="nivoSlider"> 

      <img src="<?php print $row['image']?>"/> 
      <!-- <img src="images/nivo/arts.png" alt="the grand theatre and nothern ballet"> 
      <img src="images/nivo/slider3.png" alt="leeds night light slider image"> 
      <img src="images/nivo/slider2.png" alt="Leeds Trinity slider image"> 
      <img src="images/nivo/slider4.png" alt="leeds art hotels"> 
      <img src="images/nivo/slider1.png" alt="leeds art slider image"> --> 
     </div> 
</div> 

這是我的代碼運行我的滑塊的時刻:

<?php 

$myQuery = "SELECT * FROM SliderImg"; 

$result = $con->query($myQuery); 
if (!$result) die('Query error: ' . mysqli_error($con)); 


$row = mysqli_fetch_array($result); 
?> 

回答

0

你需要在你的PHP循環:

<?php 
$myQuery = "SELECT * FROM SliderImg"; 

$result = $con->query($myQuery); 
if (!$result) die('Query error: ' . mysqli_error($con)); 

$rows = array(); 

while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) { 
    $rows[] = array(
     'image' => $row['image']; 
    ) 
} 
?> 

並在您的HTML上:

<div class="theme-dark 16 columns"> 
    <div id="slider" class="nivoSlider"> 
     <?php 
     foreach($rows as $row) { ?> 
      <img src="<?php print $row['image']?>"/> 
     <?php 
     } ?> 
    </div> 
</div> 
0
<?php  
    $myQuery = "SELECT * FROM SliderImg"; 
    $result = $con->query($myQuery); 
    if (!$result) die('Query error: ' . mysqli_error($con)); 
?> 
<div class="theme-dark 16 columns"> 
    <div id="slider" class="nivoSlider"> 
     // use a while here 
     <?php while ($row = mysqli_fetch_array($result, MYSQLI_ASSOC)){ ?> 
      <img src="<?php print $row['image']?>"/> 
     <?php } ?> 
    </div> 
</div> 
+0

負荷的mysql是不是一個好的做法。 – goldlife

+0

是的,但答案是根據請求和問題中編寫的代碼。 – Nepal12

0

你可以通過你的數據使用的foreach循環,這樣

<div class="theme-dark 16 columns"> 
    <div id="slider" class="nivoSlider"> 

     <?php foreach ($row as $key => $item): ?> 
      <img src="<?php echo $item['image']?>"/> 
     <?php endforeach; ?> 

    </div> 
</div> 

希望幫助..在意見