在下面的代码中,我试图使用Jquery ajax和php制作一个帖子报告系统。当您点击帖子的.reportit
按钮时,您必须获取帖子ID然后显示弹出窗体,然后在按下此表单的.report
按钮时,您必须在获取后执行ajax帖子的ID以及将发送给report_send_ajax.php
的其他信息。
如何将以下代码组合起来呢?
$('.reportit').click(function(){
var i = $('.sikayet-et-alani').attr('id', $('.reportit').attr('id'));
$.ajax({
type: "POST",
url: 'report_send_ajax.php',
data: i,
beforeSend: function(){$("#posting").html('<img src="icons/ajaxloader.gif"/>'); },
success: function(html) {
//Do Something
}
});
});
使用此代码:
$(document).ready(function(){
$('.reportit').click(function(){
$('.vduzalani').animate({'opacity':'.50'}, 300, 'linear');
$('.sikayet-et-alani').animate({'opacity':'1.00'}, 300, 'linear');
$('.vduzalani, .sikayet-et-alani').css('display', 'block');
});
$('.close').click(function(){
close_box();
});
});
function close_box()
{
$('.vduzalani, .sikayet-et-alani').animate({'opacity':'0'}, 300, 'linear', function(){
$('.vduzalani, .sikayet-et-alani').css('display', 'none');
});
}
来自 codepen.io
的JavaScript DEMO答案 0 :(得分:0)
试试这个
$(document).ready(function(){
var i;
$('.reportit').click(function(){
$('.vduzalani').animate({'opacity':'.50'}, 300, 'linear');
$('.sikayet-et-alani').animate({'opacity':'1.00'}, 300, 'linear');
$('.vduzalani, .sikayet-et-alani').css('display', 'block');
//get id
i = $(this).attr('id');
});
//click .report button
$('.bildir').click(function(e){
e.preventDefault();
$.ajax({
type: "POST",
url: 'report_send_ajax.php',
data: {id:i, sikayet_bildirimi:$('input[name="sikayet_bildirimi"]:checked').val()},
beforeSend: function(){$("#posting").html('<img src="icons/ajaxloader.gif"/>'); },
success: function(html) {
//Do Something
}
});
return false;
});
$('.close').click(function(){
close_box();
});
});
function close_box()
{
$('.vduzalani, .sikayet-et-alani').animate({'opacity':'0'}, 300, 'linear', function(){
$('.vduzalani, .sikayet-et-alani').css('display', 'none');
});
$('input[name="sikayet_bildirimi"]').attr('checked',false);
}