我试图找出如何解决这个问题:
$('#username').blur(function(){
$.post('register/isUsernameAvailable',
{"username":$('#username').val()},
function(data){
if(data.username == "found"){
alert('username already in use');
}
}, 'json');
});
接近这个:
rules: {
username: {
minlength: 6,
maxlength: 12,
remote: {
url: 'register/isUsernameAvailable',
type: 'post',
data: {
'username': $('#username').val()
}
}
}
但是我很难完成它.我想要的是代替警报,让它显示错误消息,但我可以在实际的jquery验证消息中设置消息.
http://docs.jquery.com/Plugins/Validation/Methods/remote#options
更新:
出于某种原因,它没有把它作为一个POST做它作为GET请求并且不确定原因.这是更新的代码:
rules: {
username: {
minlength: 6,
maxlength: 12,
remote: {
url: 'register/isUsernameAvailable',
dataType: 'post',
data: {
'username': $('#username').val()
},
success: function(data) {
if (data.username == 'found')
{
message: {
username: 'The username is already in use!'
}
} …出于某种原因,我的三元运算符赋值不适用于我的数组的第二部分.有谁看到我做错了什么？它应该只看是否有永久链接字段的值,如果没有,则插入link_url数组.
function getSiteMap()
{
$this->db->select('site_menu_structures_links.id, site_menu_structures_links.link_name');
$this->db->from('site_menu_structures_links');
$this->db->where('site_menu_structures_links.status_id', 1);
$this->db->where('site_menu_structures_links.is_category', 'Yes');
$this->db->order_by('site_menu_structures_links.sort_order');
$catQuery = $this->db->get();
if ($catQuery->num_rows())
{
foreach ($catQuery->result() as $cats)
{
// Set the Main Category into the array
$testArray[$cats->id] = array(
'id' => $cats->id,
'name' =>$cats->link_name
);
$this->db->select('site_content_pages.permalink, site_menu_structures_links_children.id, site_menu_structures_links_children.link_url, site_menu_structures_links_children.link_name');
$this->db->from('site_menu_structures_links_children');
$this->db->join('site_content_pages', 'site_content_pages.id = site_menu_structures_links_children.site_content_pages_id');
$this->db->where('site_menu_structures_links_id', $cats->id);
$this->db->where('site_menu_structures_links_children.status_id', 1);
$this->db->order_by('site_menu_structures_links_children.sort_order');
$childrenQuery = $this->db->get();
if ($childrenQuery->num_rows())
{
foreach ($childrenQuery->result() as $child)
{
$testArray[$cats->id]['children'][$child->id] = array(
'linke_url' => (empty($child->permalink)) ? $child->link_url : $child->permalink,
'link_name' => $child->link_name, …我正在datatables.net网站上查看一些说明或文档,而不是如果您在一个页面上有多个表并希望以不同方式处理每个表,该怎么办.
我试过了明显的事.将每个id分配给不同的id然后在我的js中为每个id执行代码,但由于某种原因它不允许它.我没有收到错误,但数据表本身中断并且不执行任何操作.
$(document).ready(function() {
var oTable = $('#inbox').dataTable( {
"bAutoWidth": false,
"aoColumnDefs": [
{ "bSortable": false, "aTargets": [ 0, -1 ] },
{ "sWidth": "20px", "aTargets": [ 0, -1 ] },
{ "sWidth": "100px", "aTargets": [ 1 ] },
{ "sWidth": "150px", "aTargets": [ 3 ] }
]
} );
var oTable = $('#sent').dataTable( {
"bAutoWidth": false,
"aoColumnDefs": [
{ "bSortable": false, "aTargets": [ 0, -1 ] },
{ "sWidth": "20px", "aTargets": [ 0, -1 ] },
{ "sWidth": "100px", …这似乎是一个愚蠢的问题,并且是基本的HTML,但无论如何我都会问它.我已经看到一些帖子提出了这个类似的问题,并没有帮助解决我的问题.我正在使用本地服务器进行测试.我有这个用于我的语法.
<div id="top5">
<img src="images/top5.png" alt="Top 5 Rankings" />
<ol>
<li class="top5rankings"><a href="roster/kidwonder.php">Kid Wonder</a></li>
<li class="top5rankings"><a href="roster/kidwonder.php">Kid Wonder</a></li>
<li class="top5rankings"><a href="roster/kidwonder.php">Kid Wonder</a></li>
<li class="top5rankings"><a href="roster/kidwonder.php">Kid Wonder</a></li>
<li class="top5rankings"><a href="roster/kidwonder.php">Kid Wonder</a></li>
</ol>
</div>
但是它没有为有序列表输入数字,只是看看是否有人能看到我的问题.这是我在我的css中应用于渲染页面的内容:
li.top5rankings {
color: red;
display: block;
text-align: center;
}
styles.css(第103行)
* {
margin: 0;
padding: 0;
}
styles.css(第4行)继承自ol
ol {
list-style-position: inside;
list-style-type: decimal;
}
出于某种原因,这个小代码阻止用户检查实际的复选框并让它在其中放置复选标记,并且检查它的唯一方法是单击该行.
$('table tr').click(function() {
checkBox = $(this).children('td').children('input[type=checkbox]');
if(checkBox.attr('checked'))
checkBox.removeAttr('checked');
else
checkBox.attr('checked', 'checked');
});
我想知道我在这里做错了什么.我想date_published在我的查询中格式化该字段,并且我t_string syntax error在我的IDE中获得了一个.
$this->db->select('site_news_articles.article_title, site_news_articles.is_sticky,' date_format('site_news_articles.date_published, 'f jS, Y')');
更新:
function getNewsTitles($category_id) {
$this->db->select('site_news_articles.article_title, site_news_articles.is_sticky');
$this->db->select("DATE_FORMAT(site_news_articles.date_published, '%M %e, %Y') as formatted_date", TRUE);
$this->db->from('site_news_articles');
$this->db->where('site_news_articles.news_category_id', $category_id);
$this->db->where('site_news_articles.is_approved', 'Yes');
$this->db->where('site_news_articles.status_id', 1);
$this->db->order_by('site_news_articles.date_published', 'desc');
$this->db->limit(10);
$query = $this->db->get();
return $query->result_array();
}
错误号码:1064
您的SQL语法有错误; 检查与MySQL服务器版本对应的手册,以便在'FROM(site_news_articles)WHERE 附近使用正确的语法site_news_articles.news_category_id= 2 A'在第2行
选择site_news_articles.article_title,site_news_articles.is_sticky,DATE_FORMAT(site_news_articles.date_published,'%M%E,%Y')如FORMATTED_DATE FROM(site_news_articles)WHERE site_news_articles.news_category_id= 2, site_news_articles.is_approved= '是' …
我在代码点火器内部工作,并认为如果语句将通过但它给我一个意外=错误.
if($title !=== FALSE){
} else{
}
我绝对没有看到为什么这不起作用的问题.对可能性的任何想法？
$this->db->select(CONCAT_WS(' ', 'users.first_name', 'users.last_name') 'AS name');
编辑:
我更新了建议的行但由于某种原因我仍然收到错误.
function getAllMessages($user_id)
{
$this->db->select('pm.id');
$this->db->select('pm.subject');
$this->db->select("CONCAT_WS(' ', users.first_name, users.last_name) AS name");
$this->db->select("DATE_FORMAT('pm.date_sent', '%M %D, %Y'");
$this->db->select('pm.message_read');
$this->db->from('users_personal_messages AS pm');
$this->db->join('users', 'users.user_id = pm.sender_id');
$this->db->where('recipient_id', $user_id);
$query = $this->db->get();
if ($query->num_rows() > 0)
{
return $query->result();;
}
else
{
return 0;
}
}
更新:
function getAllMessages($user_id)
{
$this->db->select('pm.id');
$this->db->select('pm.subject');
$this->db->select("CONCAT_WS(' ', users.first_name, users.last_name) AS name");
$this->db->select("DATE_FORMAT('pm.date_sent', '%M %D, %Y')");
$this->db->select('pm.message_read');
$this->db->from('users_personal_messages AS pm');
$this->db->join('users', 'users.user_id = pm.sender_id');
$this->db->where('recipient_id', $user_id);
$query = $this->db->get(); …