jQuery的工具 - 一個鏈接的作品，即使旋轉：假

Question

當我點擊.backward錨是第一選項卡上，它顯示最後一個。即使rotate設置爲false。

$(document).ready(function(){ 

    $('.agenda-tabs').tabs(".agenda-terms", { 
     effect: 'fade', 
     rotate: false 
    }).slideshow({ 
     clickable: false 
    }); 
}); 

如何阻止它從第一個「旋轉」到最後一個選項卡？

的HTML代碼是：

<div class="wrapper"> 
     <div class="agenda-navigation"> 
      <select class="agenda-track-select"> 
       <option value="track-1">Track 1</option> 
       <option value="track-2">Track 2</option> 
       <option value="track-3">Track 3</option> 
       <option value="track-4">Track 4</option> 
       <option value="track-5">Track 5</option> 
       <option value="track-6">Track 6</option> 
       <option value="track-7">Track 7</option> 
       <option value="track-8">Track 8</option> 
      </select> 
      <div class="buttons"> 
       <button class="backward">&laquo; Previous</button> 
       <button class="forward">Next &raquo;</button> 
      </div> 
     </div> 
     <div class="agenda"> 
      <div class="agenda-term-list"> 
       <div class="agenda-term-time">11:00 - 12:00</div> 
       <div class="agenda-term-time">12:00 - 13:00</div> 
       <div class="agenda-term-time">11:00 - 12:00</div> 
       <div class="agenda-term-time">12:00 - 13:00</div> 
      </div> 

      <div class="items"> 

       <div class="agenda-terms"> 
        <div class="agenda-term-row"> 
         <div class="agenda-term-point tracks-all"> 
          Lorem ipsum register now! 
         </div> 
        </div> 
        <div class="agenda-term-row"> 
         <div class="agenda-term-point track-1 current"> 
          Lorem ipsum t1 
         </div> 
         <div class="agenda-term-point track-2"> 
          Lorem ipsum dolor sit amet... t2 
         </div> 
         <div class="agenda-term-point track-3"> 
          Lorem ipsum t3 
         </div> 
         <div class="agenda-term-point track-4"> 
          Lorem ipsum dolor sit amet... t4 
         </div> 
        </div> 
        <div class="agenda-term-row"> 
         <div class="agenda-term-point tracks-all"> 
          Lorem ipsum register now! 
         </div> 
        </div> 
        <div class="agenda-term-row"> 
         <div class="agenda-term-point track-1 current"> 
          Lorem ipsum t1 
         </div> 
         <div class="agenda-term-point track-2"> 
          Lorem ipsum dolor sit amet... t2 
         </div> 
         <div class="agenda-term-point track-3"> 
          Lorem ipsum t3 
         </div> 
         <div class="agenda-term-point track-4"> 
          Lorem ipsum dolor sit amet... t4 
         </div> 
        </div> 
       </div> 
       <div class="agenda-terms"> 
        <div class="agenda-term-row"> 
         <div class="agenda-term-point tracks-all"> 
          Lorem ipsum register now! 
         </div> 
        </div> 
        <div class="agenda-term-row"> 
         <div class="agenda-term-point track-5 current"> 
          Lorem ipsum t5 
         </div> 
         <div class="agenda-term-point track-6"> 
          Lorem ipsum dolor sit amet... t6 
         </div> 
         <div class="agenda-term-point track-7"> 
          Lorem ipsum t7 
         </div> 
         <div class="agenda-term-point track-8"> 
          Lorem ipsum dolor sit amet... t8 
         </div> 
        </div> 
        <div class="agenda-term-row"> 
         <div class="agenda-term-point tracks-all"> 
          Lorem ipsum register now! 
         </div> 
        </div> 
        <div class="agenda-term-row"> 
         <div class="agenda-term-point track-5 current"> 
          Lorem ipsum t5 
         </div> 
         <div class="agenda-term-point track-6"> 
          Lorem ipsum dolor sit amet... t6 
         </div> 
         <div class="agenda-term-point track-7"> 
          Lorem ipsum t7 
         </div> 
         <div class="agenda-term-point track-8"> 
          Lorem ipsum dolor sit amet... t8 
         </div> 
        </div> 
       </div> 
      </div> 
     </div> 
    </div> 
    <div class="agenda-tabs"> 
     <a href="#"></a> 
     <a href="#"></a> 
    </div> 

Answer 1

，我發現這是jQuery的工具的錯誤。請使用this version我改變自己，而他們沒有在他們的來源修復它。

我改變了這一點：

if (b.rotate) { 
    var k = g.length - 1; 
    if (0 > a) return e.click(k, h); 
    if (a > k) return e.click(0, h) 
} 

要這樣：

var k = g.length - 1; 
if (0 > a) if (b.rotate) return e.click(k, h); else a = 0; 
if (a > k) if (b.rotate) return e.click(0, h); else a = k; 

也取得了撥弄你的代碼，以顯示它的工作：http://jsfiddle.net/bortao/k2f57/

不得不刪除一個<a>因爲你有兩個標籤，而不是三個。