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>");
}
});
现在如您所见,它将select
附加到div
块中。
但是!这里有一个问题是调用 ajax 后 HTML 的输出:
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选择标签。
谁能告诉我为什么会这样?
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>
而且由于div_subcategory
将具有与选择相同的 id,因此您将匹配div。
相反,我会通过在字符串中创建 html 并一次注入所有内容来编写它。
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,并一次性附加所有 html,这应该更快。
尝试
更改成功代码
success: function (data) {
var length = data.length;
var div_subcategory = $('#subcategory');
div_subcategory.html('');
var select_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'];
select_append += "<option value=''" + id + ">" + name + "</option>";
}
select_append += "</select>"
div_subcategory.append(select_append);
}
创建一个变量select_append
,并在其中连接所有代码,并在最后附加该变量。
试试
$(div_subcategory).find('select').append(...)
当你编写这个语法时
div_subcategory.append(
"<select id='subcategory' name='data[Product][subcategory_id]'>"
);
DOM 会自动检测这样的 HTML 元素
<select id='subcategory' name='xxx'></select>
然后在此之后附加其他选项元素
所以解决方案是首先制作一个 HTML 字符串,然后附加像
var html = "<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'];
html += "<option value='' + id>" + name + "</option>";
}
div_subcategory.append(html);