我认为我的jQuery导入排序如下:
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.6.4/jquery.min.js"></script>
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jqueryui/1.8.16/jquery-ui.min.js" />
然后我有这样的整个jQuery代码:
<script type="text/javascript" >
$(function()
{
$("input[type=submit]").click(function()
{
var name = $("#problem_name").val();
var problem_blurb = $("#problem_blurb").val();
var dataString = 'problem_name='+ name + '&problem_blurb=' + problem_blurb;
if(name=='' || problem_blurb == '')
{
$('.success').fadeOut(200).hide();
$('.error').fadeOut(200).show();
}
else
{
$.ajax({
type: "POST",
url: "/problems/add_problem.php",
dataType: "json",
data: dataString,
success: function(json)
{
$('.success').fadeIn(200).show();
$('.error').fadeOut(200).hide();
// Here can update the right side of the screen with the newly entered information
//alert (json);
new_string = "<h2>Most Recently Added Problems</h2>";
// Have to figure out how to make this work with the DOM.
for (var i = 0, l = json.length; i < l; ++i)
{
title = json[i]['problem_title'];
member_id = json[i]['creator_member_id'];
description = json[i]['problem_description'];
problem_date = json[i]['problem_date'];
upvotes = json[i]['upvotes'];
downvotes = json[i]['downvotes'];
problem_id = json[i]['problem_id'];
new_string = new_string + "<p>Problem name: <a href='http://www.problemio.com/problems/problem.php?problem_id=" + problem_id + "'>" + title + "</a></p>";
new_string = new_string + "<p>Problem description: " + description + "</p>";
new_string = new_string + "<p>Entered date " + problem_date + "</p>";
new_string = new_string + "<a href='/problems/edit_problem.php?problem_id=" + problem_id + "'>Edit</a>";
new_string = new_string + "<hr />";
}
$("#recent_problems").replaceWith( new_string ) ;
}
});
}
return false;
});
});
</script>
<script type="text/javascript">
$(document).ready(function()
{
var $dialog = $('<div></div>')
.html('This dialog will show every time!')
.dialog({
autoOpen: false,
title: 'Basic Dialog'
});
$('.vote_up').click(function()
{
problem_id = $(this).attr("data-problem_id");
var dataString = 'problem_id='+ problem_id + '&vote=+';
$.ajax({
type: "POST",
url: "/problems/vote.php",
dataType: "json",
data: dataString,
success: function(data)
{
// ? :)
alert (data);
},
error : function(data)
{
//alert("ajax error, json: " + data.responseText);
errorMessage = data.responseText;
if ( errorMessage == "not_logged_in" )
{
alert ("errr");
// Try to create the popup that asks user to log in.
//$(".dialog").dialog();
$dialog.dialog('open');
// prevent the default action, e.g., following a link
return false;
}
else
{
alert ("not");
}
//alert(JSON.stringify(data));
}
});
//Return false to prevent page navigation
return false;
});
$('.vote_down').click(function()
{
alert("down");
problem_id = $(this).attr("data-problem_id");
var dataString = 'problem_id='+ problem_id + '&vote=-';
//Return false to prevent page navigation
return false;
});
});
</script>
当在此页面上时:http://www.problemio.com我按下“upvote”链接,我没有弹出jQuery对话框。而且没有错误。但是有$ dialog.dialog('open')的行;应该打开我的对话框吧?
或者我有两个地方检查文件是否准备好是一个问题?我粘贴了我的整个jQuery代码,以防我出现一些新手错误。
谢谢!
答案 0 :(得分:3)
您没有包含jQuery UI CSS,因为我在链接中看到对话框出现但未格式化。
在head部分中包含该行(最好在包含JS文件之前使用):
<link href="http://ajax.googleapis.com/ajax/libs/jqueryui/1.8.16/themes/black-tie/jquery-ui.css" rel="stylesheet" type="text/css" />
另请正确关闭script
代码
替换:
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jqueryui/1.8.16/jquery-ui.min.js" />
使用:
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jqueryui/1.8.16/jquery-ui.min.js" ></script>
答案 1 :(得分:2)
您在DOM中创建了一个DIV,但您从未将其添加到现有结构中。试试这个:
var $dialog = $('<div>');
$('body').append($dialog);