如何在AJAX调用中隐藏元素？

I've got a page that has a dropdown menu, and upon selection, creates a new element (a table) using an AJAX call. Firebug shows this on action:

我有一个带有下拉菜单的页面,在选择后,使用AJAX调用创建一个新元素(表)。 Firebug在行动中显示:

GET http://www.site.com/page.php?q=category

The actual code for this is:

实际代码是:

select name="category" id="category" onchange="load(this.value)

However, in addition to it creating a new table, I'd like it to destroy/ not display another table that's already on the page. The table has class="table2" id="PR". What additional code do I put into that select tag above? Thanks

但是,除了创建一个新表之外,我还希望它能够销毁/不显示已经在页面上的另一个表。该表具有class =“table2”id =“PR”。我在上面的select标签中添加了哪些附加代码?谢谢

3 个解决方案

#1

Try with the following:

尝试使用以下内容:

$('#PR').remove();

But maybe you'll need to make a function to call onchange:

但也许你需要创建一个函数来调用onchange:

function mychange(event)
{
  load( event.currentTarget.value );
  $('#PR').remove();
}

select name="category" id="category" onchange="mychange(event)"

#2

I would change it so you have this code instead:

我会改变它,所以你有这个代码:

$(function() {
    $('#category').change(function() {
         load($(this).val());
         $('#PR').hide(); // or .remove() if you want to completely remove it;
    });
});

You can then remove the onchange from the select tag as it is wired up via the jquery event.

然后,您可以在通过jquery事件连接时从select标记中删除onchange。

#3

Without jQuery try:

没有jQuery尝试:

onchange="load(this.value); document.getElementById('table1').style.display='none';"

#1