添加表单时,PHP重新加载页面(也许需要ajax)

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很抱歉,该主题的标题可能不正确,但这是我想出的最好的结果.

Sorry for maybe incorrect title for the topic, but this is the best that I came up with.

因此,我正在为网站构建管理面板.

So, I'm building admin panel for a website.

我有一个页面,并且在页面的某些部分中,我想刷新它并加载其他表单.

I have a page, and in some part of the page, i'd like to refresh it and load another form.

让我们添加一个时间表,然后在页面的某个位置上,我希望在单击链接后立即显示此表单.

let's say add a schedule, and somewhere down on the page I'd like to have this form displayed as soon as the link is clicked.

当用户保存它时,我希望该表格消失,而不是让它具有显示所有时间表的列表.

when a user saves it, I'd like that form to disappear and and in stead of that to have a list displaying all of the schedules.

我不想使用框架-我不是框架的支持者.该面板是使用PHP构建的.

I don't want to use frames - I'm not a supporter of frames. The panel is built using PHP.

也许使用Ajax可以实现吗?如果是->怎么样?任何指向好的示例或教程的链接.

Maybe this might be achived with Ajax? If yes -> How? any link to good example or tutorial.

推荐答案

是的,这将使用ajax解决.

yes this will be solved with ajax.

这是应该刷新页面的代码示例

Here is a code example when the page is supposed to refresh

$('#button').click(function() { $.ajax({ url: 'path/to/script.php', type: 'post', dataType: 'html', // depends on what you want to return, json, xml, html? // we'll say html for this example data: formData, // if you are passing data to your php script, needed with a post request success: function(data, textStatus, jqXHR) { console.log(data); // the console will tell use if we're returning data $('#update-menu').html(data); // update the element with the returned data }, error: function(textStatus, errorThrown, jqXHR) { console.log(errorThrown); // the console will tell us if there are any problems } }); //end ajax return false; // prevent default button behavior }); // end click

jQuery Ajax

jQuery Ajax

api.jquery/jQuery.ajax/

脚本说明了.

1-用户单击按钮.

2-单击功能向服务器发起XHR调用.

2 - Click function initiates an XHR call to the server.

3-url是php脚本,它将根据发布的值处理我们正在发送的数据.

3 - The url is the php script that will process the data we are sending based on the values posted.

4-类型为POST请求,需要数据才能返回数据.

4 - The type is a POST request, which needs data to return data.

5-在这种情况下,数据类型将为html.

5 - The dataType in this case will be html.

6-我们发送给脚本的数据可能是分配给变量formData的form元素的序列化.

6 - The data that we are sending to the script will probably be a serialization of the form element that is assigned to the variable formData.

7-如果XHR返回200,则在控制台中登录返回的数据,以便我们知道我们正在使用什么.然后将数据以html格式放置在所选元素中(#update-menu).

7 - If the XHR returns 200, then log in the console the returned data so we know what we are working with. Then place that data as html inside the selected element (#update-menu).

8-如果有错误,请让控制台为我们记录该错误.

8 - If there is an error have the console log the error for us.

9-返回false以防止出现默认行为.

9 - Return false to prevent default behavior.

10-全部完成.

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添加表单时,PHP重新加载页面(也许需要ajax)

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