不调用 Ajax 请求

我正在向 onmouseover 函数发送一个 ajax 请求。这是我的代码

<div id="notifications" class="design" onmouseover="show_drop_downn()" onmouseout="hide_drop_downn()">
    <button id="notify"></button>
    <?php
        if($count_notify>0)
        { ?>
        <span id="alert"  > <?php echo $count_notify; ?> </span>
        <div id="drop_downn" onmouseover="show_drop_downn()" onmouseout="hide_drop_downn()">
            <?php foreach($values as $row): ?>
                <div id="request_display">
                    <span id="full_name">
                <?php
                echo $row->fname;
                echo " ";
                echo $row->lname;
                if($row->status=='1')
                {
                    echo " has accepted your friend request";
                }
                elseif($row->status=='2')
                {
                    echo " has declined your friend request";
                }
                ?>
                <br>
                <div id="image_short_fake">
                <img   id ="img_s" src="<?php echo $this->config->item('base_url'); ?><?php echo '/application/css/'. $row->filename?>"/>
                </div>
                  </span>
                 <!-- <button id="accept" onclick='setSelected_accept(this)' type="submit" value="<?php echo $row->userid; ?>">Accept</button>
                  <?php ?>
                  <button id="decline" onclick='setSelected_decline(this)' type="submit" value="<?php echo $row->userid; ?>">Decline</button>
                 -->
                 </div>

            <?php endforeach; ?>
        </div>

这是我的show_drop_downn（）函数

function show_drop_downn()
{
    document.getElementById("drop_downn").style.visibility = "visible"; 
    $.ajax
    ({
            type: "POST",
            url: "http://localhost/ok/index.php/search/ajax_delete_ntify",
        success: function(){ alert('ok'); },
        error:function(x,e)
        {
            if(x.status==0){
            alert('You are offline!!'n Please Check Your Network.');
            }else if(x.status==404){
            alert('Requested URL not found.');
            }else if(x.status==500){
            alert('Internel Server Error.');
            }else if(e=='parsererror'){
            alert('Error.'nParsing JSON Request failed.');
            }else if(e=='timeout'){
            alert('Request Time out.');
            }else {
            alert('Unknow Error.'n'+x.responseText);
            }
        }
    }); 
 }

我已经通过警报检查了我的show_drop_downn()函数被调用。但是 ajax 方法没有调用http://localhost/ok/index.php/search/ajax_delete_ntify我已经用回声证实了这一点。但是，当我在浏览器中编写函数地址时，它可以调用该函数。我不知道我的代码出了什么问题。请帮助我。如果有任何不清楚的地方，请告诉我您需要什么更多信息。再次：我的问题是 AJAX 请求不起作用。