我正在尝试获取一个表单来动态添加更多带有id的输入字段,这些字段也是动态生成的。我有大部分工作,除了fieldset id取最后一个输入字段的名称而不是原始字段集(如果你检查字段集的id,它说“fenceType2”而不是“fenceDescription2”)。如果有人能解释我哪里出错了,我将非常感谢!非常感谢!这是代码的小提琴 - http://jsfiddle.net/gv0029/dyJjA/ - 这里是代码: HTML:
<div id="inputFence1" class="clonedInputFence">
<fieldset id="fenceDescripton">
<legend><strong>Fence Description</strong>
</legend>Fence Description:
<select name="fenceHeight" id="fenceHeight">
<option value="select">Select Fence Height</option>
<option value="6" id="fH6">6 Ft.</option>
</select>
<select name="fenceType" id="fenceType">
<option value="select">Select Fence Type</option>
<option value="wr1" id="wr1">WRC #1</option>
</select>
</fieldset>
</div>
<input type="button" id="btnAddFence" value="Add Another Fence" />
<input type="button" id="btnDelFence" value="Remove Fence" />
JS:
//Dynamic Fence Input Fields
$('#btnAddFence').click(function () {
var num = $('.clonedInputFence').length; // how many "duplicatable" input fields we currently have
var newNum = new Number(num + 1); // the numeric ID of the new input field being added
// create the new element via clone(), and manipulate it's ID using newNum value
var newElem = $('#inputFence' + num).clone().attr('id', 'inputFence' + newNum);
//Fence Description
newElem.children($("select[name=fenceHeight] option:selected")).attr('id', 'fenceHeight' + newNum).attr('name', 'fenceHeight' + newNum);
newElem.children($("select[name=fenceType] option:selected")).attr('id', 'fenceType' + newNum).attr('name', 'fenceType' + newNum);
$('#inputFence' + num).after(newElem);
// enable the "remove" button
//$('#btnDel').attr('disabled','');
$('#btnDelFence').removeAttr('disabled');
// business rule: you can only add 5 names
//if (newNum == 5)
//$('#btnAdd').attr('disabled','disabled');
});
$('#btnDelFence').click(function () {
var num = $('.clonedInputFence').length; // how many "duplicatable" input fields we currently have
$('#inputFence' + num).remove(); // remove the last element
// enable the "add" button
//$('#btnAdd').attr('disabled','');
$('#btnAddFence').removeAttr('disabled');
// if only one element remains, disable the "remove" button
if (num - 1 == 1) $('#btnDelFence').attr('disabled', 'disabled');
});
$('#btnDelFence').attr('disabled', 'disabled');
感谢任何和所有帮助!再次感谢!
答案 0 :(得分:1)
喜欢这个 DEMO
//Fence Description
newElem.children('fieldset').attr('id', 'fenceDescripton'+newNum);
newElem.children("select[name=fenceHeight] option:selected").attr('id', 'fenceHeight' + newNum).attr('name', 'fenceHeight' + newNum);
newElem.children("select[name=fenceType] option:selected").attr('id', 'fenceType' + newNum).attr('name', 'fenceType' + newNum);
$('#inputFence' + num).after(newElem);
答案 1 :(得分:1)
.children()方法与.find()的不同之处在于.children()只沿DOM树向下移动一个级别,而.find()可以遍历多个级别以选择后代元素(孙子等)同样。
由于您提供了$(...)
而不仅仅是选择器字符串,children
似乎选择了克隆div的直接子节点,即fieldset
。
所以,你应该进行两次修改
.find()
代替children
newElem.find("select[name=fenceHeight]").attr('id', 'fenceHeight' + newNum).attr('name', 'fenceHeight' + newNum);
newElem.find("select[name=fenceType]").attr('id', 'fenceType' + newNum).attr('name', 'fenceType' + newNum);
请参见修改后的JSFiddle
答案 2 :(得分:0)
嘿,我实际上想出了如何解决这个问题。 首先,我删除了代码的不同部分之间的单独ID,以使其更易于管理,然后我改变了我尝试创建newElements的方式:
newElem.children($("select[name=fenceHeight] option:selected")).attr('id', 'fenceHeight' + newNum).attr('name', 'fenceHeight' + newNum);
到:
newElem.find(':input[name="railQuantity"]').attr('id', 'railQuantity' + newNum).attr('name', 'railQuantity' + newNum);
主要问题显然是我的选择。谢谢你们的帮助!非常感谢!