Php喜欢和不像使用JQUERY AJAX

Question

我有一个关于我的帖子喜欢和不同的问题。问题是当我点击.like_button <span id='you"+New_ID+"'><a href='#'>You</a> like this.</span>没有显示。

我仍然检查浏览器开发者控制台。但是当我点击像按钮像按钮将改变，但<span id='you"+New_ID+"'><a href='#'>You</a>, </span>没有显示。但如果我刷新页面<span id='you"+New_ID+"'><a href='#'>You</a>, </span>即将到来。

任何人都可以帮助我吗？

我使用这个代码像不像：

AJAX JQUERY：

$('.like_button').die('click').live("click", function() { 

    var KEY = parseInt($(this).attr("data")); 
    var ID = $(this).attr("id"); 
    if (KEY == '1') { 
     var sid = ID.split("likes"); 
    } else { 
     var sid = ID.split("like"); 
    } 


    var New_ID = sid[1]; 
    var REL = $(this).attr("rel"); 
    var URL = $.base_url + 'post_like_ajax.php'; 
    var dataString = 'post_id=' + New_ID + '&rel=' + REL; 
    $.ajax({ 
     type: "POST", 
     url: URL, 
     data: dataString, 
     cache: false, 
     success: function (html) { 
      if (html) { 
       if (REL == 'Like') { 
        $("#elikes" + New_ID).show('fast').prepend("<span id='you" + New_ID + "'><a href='#'>You</a> like this.</span>"); 
        $("#likes" + New_ID).prepend("<span id='you" + New_ID + "'><a href='#'>You</a>, </span>"); 
        $('#' + ID).html('Unlike').attr('rel', 'Unlike').attr('title', 'Unlike'); 
       } else { 
        $("#elikes" + New_ID).hide('slow'); 
        $("#you" + New_ID).remove(); 
        $('#' + ID).attr('rel', 'Like').attr('title', 'Like').html('Like'); 
       } 
      } 

     } 
    }); 

    return false; 
}); 

PHP代码：

<?php 
if($login) 
{ 
?> 
<a href='#' class='like like_button icontext' id='like<?php echo $post_id;?>' title='<?php echo $like_status;?>' rel='<?php echo $like_status;?>' data=''><?php echo $like_status;?></a> 
<a href='#' class='commentopen commentopen_button icontext' id='<?php echo $post_id;?>' rel='<?php echo $post_id;?>' title='Comment'>Yorum yap </a> 
<?php if($uid != $post_id) { ?> 

<?php } } else { ?> 
<a href='<?php echo $index_url; ?>' class='like icontext' >Like</a> 
<a href='<?php echo $index_url; ?>' class='commentopen icontext' title='Comment'>Comment</a> 
<a href='<?php echo $index_url; ?>' class='share icontext' title='Share'>Share</a>  
<?php 
} 
?> 


<?php if($post_like_count>0) 
{ 
$likesuserdata=$POLL->post_Like_Users($post_id); 
if($likesuserdata) 
{ 
echo '<div class="likes" id="likes'.$post_id.'">'; 
$i=1; 
$j=count($likesuserdata); 
foreach($likesuserdata as $likesdata) 
{ 
$you="likeuser".$post_id; 
$likeusername=$likesdata['username']; 
if($likeusername == $session_username) 
{ 
$likeusername='You'; 
$you="you".$post_id; 
} 

echo '<a href="#" id="'.$you.'">'.$Wall->UserFullName($likeusername).'</a>'; 
if($j!=$i) 
{ 
echo ', '; 
} 
$i=$i+1; 
} 

if($post_like_count>3) 
{ 
$post_like_count=$post_like_count-3; 
echo ' and <span id="like_count'.$post_id.'" class="numcount">'.$post_like_count.'</span> others like this.'; 
} 
else 
{ 
echo ' like this.'; 
} 

echo '</div>'; 
} 
} 
else 
{ 
echo '<div class="likes" id="elikes'.$post_id.'" style="display:none"></div>'; 
} 
?> 

post_like_ajax.php

<?php 
include_once 'includes.php'; 

if(isSet($_POST['post_id']) && isSet($_POST['rel'])) 
{ 
$haber_id=$_POST['post_id']; 
$rel=$_POST['rel']; 
if($rel=='Like') 
{ 
$cdata=$POLL->POST_Like($post_id,$uid); 
} 
else 
{ 
$cdata=$POLL->POST_Unlike($post_id,$uid); 
} 
echo $cdata; 
} 
?> 

Answer 1

我想你忘记了刚刚以显示您正在前面加上因为开始您添加显示的DIV：无，echo '<div class="likes" id="elikes'.$post_id.'" style="display:none"></div>';

尝试改变这一行：

$("#likes" + New_ID).prepend("<span id='you" + New_ID + "'><a href='#'>You</a>, </span>"); 

这样：

$("#likes" + New_ID).show().prepend("<span id='you" + New_ID + "'><a href='#'>You</a>, </span>");