问题描述
我正在尝试使用jquery POST函数,但它正在以AJAX样式处理请求。我的意思是它实际上没有进入我要告诉它的页面。
i am trying to use the jquery POST function but it is handling the request in AJAX style. i mean its not actually going to the page I am telling it to go.
$("#see_comments").click(function() {
$.post(
"comments.php",
{aid: imgnum},
function(data){
});
});
这个函数应该带有辅助值到comments.php页面。它的发布很好但没有重定向到comments.php。
this function should go to comments.php page with the aid value in hand. its posting fine but not redirecting to comments.php.
@Doug Neiner
澄清:
1我有15个链接(图片)。我点击一个链接,它加载我的JavaScript。 JS知道我打开了wat imgnum。我想在comments.php中找到这个imgnum。我必须使用这个JavaScript,没有其他方法可以做到这一点。 JS是强制性的。
2.您的方法成功发布援助值。但在comments.php中,当我尝试回显该值时,它什么也没显示。
3.我正在使用Firebug。在控制台中,它显示了我在步骤(2)中成功完成的回显REQUEST。
@Doug Neinerclarification:
1. I have 15 links (images). i click on a link and it loads my javascript. the JS knows wat imgnum i opened. this imgnum i want in the comments.php. i have to use this javascript and no other means can do the trick. the JS is mandatory.2. your method successfully POSTs the aid value. but in the comments.php when i try to echo that value, it displays nothing.3. I am using Firebug. in the Console, it shows the echo REQUEST i made in Step (2) successfully.
推荐答案
我知道你要做什么,但它不是你想要的。
I know what you are trying to do, but its not what you want.
首先,除非您在服务器上更改数据,否则请勿使用 POST
请求。只需 #see_comments
是正常的< a href ='/ comments.php?aid = 1'> ...
First, unless you are changing data on the server, don't use a POST
request. Just have #see_comments
be a normal <a href='/comments.php?aid=1'>...
如果 使用 POST
,那么这样做就可以了跟随您的通话的页面:
If you have to use POST
, then do this to get the page to follow your call:
$("#see_comments").click(function() {
$('<form action="comments.php" method="POST">' +
'<input type="hidden" name="aid" value="' + imgnum + '">' +
'</form>').submit();
});
这实际上是如何运作的。
第一个 $。post
仅一个AJAX方法,不能用来做传统的表单
提交就像你描述的那样。因此,为了能够发布值和导航到新页面,我们需要模拟表单
帖子。
First $.post
is only an AJAX method and cannot be used to do a traditional form
submit like you are describing. So, to be able to post a value and navigate to the new page, we need to simulate a form
post.
所以流程如下:
- 你点击图片,你的JS代码得到了
imgnum
- 接下来,有人点击
#see_comments
- 我们创建一个临时
表单
,其中imgnum
值作为隐藏字段 - 我们提交该表单,其中发布值和加载
comments.php
页 - 您的
comments.php
页面将有权访问已发布的变量(即在PHP中它将是$ _ POST ['aid']
)
- You click on the image, and your JS code gets the
imgnum
- Next, someone clicks on
#see_comments
- We create a temporary
form
with theimgnum
value in it as a hidden field - We submit that form, which posts the value and loads the
comments.php
page - Your
comments.php
page will have access to the posted variable (i.e. in PHP it would be$_POST['aid']
)
这篇关于非AJAX jquery POST请求的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!