我有以下ajax:
$.ajax({
type: 'POST',
url: myBaseUrl + 'Products/ajax_get_subcategories',
dataType: 'json',
data: {
id: id
},
success: function (data) {
var length = data.length;
var div_subcategory = $('#subcategory');
div_subcategory.html('');
div_subcategory.append(
"<select id='subcategory' name='data[Product][subcategory_id]'>"
);
for (var i = 0; i < length; i++) {
var id = data[i]['Subcategory']['id'];
var name = data[i]['Subcategory']['name'];
$('#subcategory').append(
"<option value=''+id>" + name + "</option>"
);
}
div_subcategory.append("</select>");
}
});
现在,您可以看到它将
select
附加到div
块中。但!有一个问题是调用了ajax之后的HTML输出:
div id="subcategory" class="subcategory">
<select id="subcategory" name="data[Product][subcategory_id]"></select>
<option +id="" value="">Telte</option>
<option +id="" value="">Toilet</option>
<option +id="" value="">Service</option>
<option +id="" value="">Borde</option>
<option +id="" value="">Stole</option>
<option +id="" value="">Lyd og lys</option>
</div>
如您所见,它会在添加选项之前关闭select标记。
谁能告诉我为什么会这样吗?
最佳答案
当你写:
div_subcategory.append("<select id='subcategory' name='data[Product][subcategory_id]'>");
jQuery将插入
<select id='subcategory' name='data[Product][subcategory_id]'></select>
并且因为
div_subcategory
将具有与您要匹配div的选择相同的ID。相反,我将通过在字符串中创建html并一次注入所有内容来编写此代码。
var html += "<select id='subcategorysel' name='data[Product][subcategory_id]'>";
for (var i = 0; i < length; i++) {
var id = data[i]['Subcategory']['id'];
var name = data[i]['Subcategory']['name'];
html += "<option value=''+id>" + name + "</option>";
}
html += "</select>";
div_subcategory.append(html);
此代码段更新您的代码以使用不同的ID,并一次性添加所有html内容,这应该会更快。