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我是html/javascript的新手,我正在開發並試圖弄清楚如何使用this link中的代碼。根據以前的下拉選擇顯示第二個下拉列表
我能看到的解決方案呈現在下面的鏈接,但不知道如何在HTML頁面的上下文應用: http://jsfiddle.net/3UWk2/3/
我抄一個HTML網頁中的HTML和JavaScript函數的內容,但它基本上顯示所有的行和選擇框...任何想法這是如何工作的?我是否應該在不同的文件中使用HTML代碼和JavaScript函數分別設置html頁面,並以某種方式將它們一起調用?
<select size="1" id="Rank" title="" name="Rank">
<option value="">-Select Your Rank-</option>
<option value="airman">Airman</option>
<option value="senior-airman">Senior Airman</option>
</select>
<div class="container">
<div class="airman">
<select class="second-level-select">
<option value="">-Select Your Rank-</option>
<option value="basic-ore-1">Basic Ore Miner - Level 1</option>
<option value="basic-ore-2">Basic Ore Miner - Level 2</option>
</select>
</div>
<div class="senior-airman">
<select class="second-level-select">
<option value="">-Select Your Rank-</option>
<option value="omber-miner-1">Omber Miner - Level 1</option>
<option value="omber-miner-2">Omber Miner - Level 2</option>
</select>
</div>
</div>
<div class="second-level-container">
<div class="basic-ore-1">
Line of text for basic ore miner 1
</div>
<div class="basic-ore-2">
Line of text for basic ore miner 2
</div>
<div class="omber-miner-1">
Line of text for omber miner 1
</div>
<div class="omber-miner-2">
Line of text for omber miner 2
</div>
</div>
$(document).ready(function() {
$('#Rank').bind('change', function() {
var elements = $('div.container').children().hide(); // hide all the elements
var value = $(this).val();
if (value.length) { // if somethings' selected
elements.filter('.' + value).show(); // show the ones we want
}
}).trigger('change');
$('.second-level-select').bind('change', function() {
var elements = $('div.second-level-container').children().hide(); // hide all the elements
var value = $(this).val();
if (value.length) { // if somethings' selected
elements.filter('.' + value).show(); // show the ones we want
}
}).trigger('change');
});
嗨,你好,我已經試過以上並複製你的代碼放到一個HTML頁面,但似乎並沒有工作。 ..我仍然看到所有的行和過濾器不工作:( – user3652574
@ user3652574更改//ajax.googleapis.com到http://ajax.googleapis.com – deadulya
HI ...看起來像下面的鏈接似乎工作: 「http://ajax.googleapis.com/ajax/libs/jquery/1.7.1/jquery.min.js」 – user3652574