簡體   English   中英

Spry選項卡式面板未顯示動態數據

[英]Spry tabbed panel not showing dynamic data

嗨,我正在嘗試使用動態數據設置spry選項卡式面板,我想使用單個查詢來完成這項工作,我只在“ Tab 1”,“ Tab 2”,“ Tab 3”中顯示結果,但不顯示任何數據,我認為我在同一個查詢中重復使用三次循環區域,這就是為什么這種情況。我嘗試了foreach循環,但沒有幫助,到目前為止,這是我的測試代碼,請提出正確的方法這個。

謝謝。

我敏捷的選項卡式面板代碼是這樣的。

<div id="TabbedPanels1" class="TabbedPanels">
<ul class="TabbedPanelsTabGroup">
<li class="TabbedPanelsTab" tabindex="0">Tab 1</li>
<li class="TabbedPanelsTab" tabindex="0">Tab 2</li>
<li class="TabbedPanelsTab" tabindex="0">Tab 3</li>
</ul>
<div class="TabbedPanelsContentGroup">

<div class="TabbedPanelsContent">
<?php while ($list = mysql_fetch_assoc($queryresult)): ?>
<div class="category-main-container">
<div class="category-main-desc"> <?php echo $list['title'];?></a>
</div><!--end of category-desc-->
</div><!--end of category-container-->
<?php endwhile;?>
</div>

<div class="TabbedPanelsContent">
<?php while ($list = mysql_fetch_assoc($queryresult)): ?>
<div class="category-main-container">
<div class="category-main-desc"> <?php echo $list['id'];?></a>
</div><!--end of category-desc-->
</div><!--end of category-container-->
<?php endwhile;?>
</div>

<div class="TabbedPanelsContent">
<?php while ($list = mysql_fetch_assoc($queryresult)): ?>
<div class="category-main-container">
<div class="category-main-desc"> <?php echo $list['rating'];?></a>
</div><!--end of category-desc-->
</div><!--end of category-container-->
<?php endwhile;?>
</div>

</div>
</div>
<script type="text/javascript">
var TabbedPanels1 = new Spry.Widget.TabbedPanels("TabbedPanels1");
</script>

和PHP查詢是這樣的。

$offset = ($currentpage - 1) * $rowsperpage;
$sqlquery = "SELECT * FROM $tablename ORDER BY $orderby $sortby LIMIT $rowsperpage OFFSET $offset";
$queryresult = mysql_query($sqlquery, $connection) or trigger_error("SQL", E_USER_ERROR);

更簡單的方法是對每個標簽使用單獨的MySQL查詢,例如

$sqlquery1 = "SELECT id FROM table_name ORDER BY id ASC";
$queryresult1 = mysql_query($sqlquery1, $connection) or trigger_error("SQL", E_USER_ERROR);
$sqlquery2 = "SELECT title FROM table_name ORDER BY id ASC";
$queryresult2 = mysql_query($sqlquery2, $connection) or trigger_error("SQL", E_USER_ERROR);
$sqlquery3 = "SELECT rating FROM table_name ORDER BY id ASC";
$queryresult3 = mysql_query($sqlquery3, $connection) or trigger_error("SQL", E_USER_ERROR);

並在選項卡式菜單中這樣做。

<div class="TabbedPanelsContent">
<?php while ($list = mysql_fetch_assoc($queryresult1)): ?>
<div class="category-main-container">
<div class="category-main-desc"> <?php echo $list['id'];?></a>
</div><!--end of category-desc-->
</div><!--end of category-container-->
<?php endwhile;?>
</div>

<div class="TabbedPanelsContent">
<?php while ($list = mysql_fetch_assoc($queryresult2)): ?>
<div class="category-main-container">
<div class="category-main-desc"> <?php echo $list['title'];?></a>
</div><!--end of category-desc-->
</div><!--end of category-container-->
<?php endwhile;?>
</div>  

<div class="TabbedPanelsContent">
<?php while ($list = mysql_fetch_assoc($queryresult3)): ?>
<div class="category-main-container">
<div class="category-main-desc"> <?php echo $list['rating'];?></a>
</div><!--end of category-desc-->
</div><!--end of category-container-->
<?php endwhile;?>
</div>

這應該可以解決您的問題。

暫無
暫無

聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.

 
粵ICP備18138465號  © 2020-2024 STACKOOM.COM