由网友(浅梦)分享简介:我试图用code以下,但它不工作:最新工作:I'm trying to use the code below, but it's not working:UPDATED WORKING: $(document).ready(function() { $('.infor').click(function () {...
我试图用code以下,但它不工作: 最新工作:
I'm trying to use the code below, but it's not working: UPDATED WORKING:
$(document).ready(function() {
$('.infor').click(function () {
var datasend = $(this).html();
$.ajax({
type: 'POST',
url: 'http://domain.com/page.php',
data: 'im_id='+datasend',
success: function(data){
$('#test_holder').html(data);
}
});
});
});
正如你可以看到我用$ datasend作为无功派,但它不会返回它的价值,只是它的名字。
As you can see I used $datasend as the var to send but it doesn't return the value of it, only its name.
感谢。
推荐答案
我会改变
$ datasend = $(本)。html的;
至
VAR datasend = $(本)。html的();
I would change
$datasend = $(this).html;
to
var datasend = $(this).html();
接下来我会改变
数据:'im_id = $ datasend',
至
数据:'im_id ='+ datasend,
Next I would change
data: 'im_id=$datasend',
to
data: 'im_id='+datasend,
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