我怎样才能得到我需要的变量值

Question

我怎样才能得到我需要的变量值，我已经通过MySQL选择了评论和喜欢的值，并且创建了一个类似的按钮，但是当我点击它时，它会添加类似于上次选择/发布的变量值，因此它喜欢上次生成的评论。如何让它像我想要的而不是最后选择的那个？

实施例：

我有ID的评论（1,2,3）时产生其预定他们以该顺序（1,2,3）时，我想用ID评论= 2它喜欢的评论3，因为它是最后生成的，所以在我喜欢评论2时，变量ID值为3.

我没有任何意义，我希望你们能够得到它，请帮助吗？

Answer 1

note.php

//-------- WHILE LOOP FOR GETTING NOTE COMMENTS ------- 

$commentsList = array(); 

$query = mysql_query("SELECT * FROM note_comments WHERE note_id='$note_id'"); 
while (($row = mysql_fetch_assoc($query)) != false) { 
    $commentsList[] = array(
     'comId' => $row['id'], 
     'comment' => $row['comment'], 
     'com_likes' => $row['likes'], 
     'com_likers' => $row['likers'] 
    ); 
} 

note.php（主体）：

 <div id="noteComments"> 
     <h2>Comments</h2> 
     <?php 
     $commentL = $commentsList; 
     if (count($commentL) == 0) { 
      echo 'Sorry, there are no comments.'; 
     } else { 
      echo '<ul style="list-style:none;">'; 
      foreach($commentL as $comment) { 
       echo '<li><p>', $comment['comment'], '</p><p><a style="background-color:#3b3c3b; color:#fff; 
       text-decoration:none; padding: 1 5 1 5; border-radius:5px; border: 1px solid black; " href="#" onclick="like_add(', $comment['comId'] ,');">Like</a><span id="comment_', $comment['comId'] ,'_likes">', $comment['com_likes'] , '</span> like this</p></li>'; 
      } 
      echo '</ul>'; 
     } 
     ?> 
     </div><!-- end noteComments --> 

---

like.js

function like_add(comId) { 
$.post('ajax/like_add.php', {comId:comId}, function(data) { 
    if (data == 'success') { 
     like_get(comId); 
    } else { 
     alert(data); 
    } 
}); 
} 

function like_get(comId) { 
$.post('ajax/like_get.php', {comId:comId}, function(data) { 
    $('#comment_'+comId+'_likes').text(data); 
}); 
} 

---

like_add.php

<?php 
include_once("../scripts/checkuserlog.php"); 
include '../functions/like.php'; 
if (isset($_POST['comId']) && comment_exists($_POST['comId'])) { 
$comId = $_POST['comId']; 
add_like($comId); 
} 
?> 

like_get.php

---

like.php

<?php 
function comment_exists($comId) { 
$comId = (int)$comId; 
return (mysql_result(mysql_query("SELECT COUNT('id') FROM note_comments WHERE id=$comId"), 0) == 0) ? false : true; 
} 

function like_count($comId) { 
$comId = (int)$comId; 
return (int)mysql_result(mysql_query("SELECT likes FROM note_comments WHERE id=$comId"), 0, 'likes'); 
} 
function add_like($comId) { 
$comId = (int)$comId; 
$visitor = $_SESSION['id']; 
$com_likers = ""; 
$query = mysql_query("SELECT likes, likers FROM note_comments WHERE id='$comId'"); 
while (($row = mysql_fetch_array($query)) != false) { 

     $com_likes = $row['likes']; 
     $com_likers = $row['likers']; 
} 
$thisComlikers = explode(",", $com_likers); 
if (!in_array($visitor, $thisComlikers)) { 
    if ($com_likers != "") { $com_likers = "$com_likers,$visitor"; } else { $com_likers = "$visitor"; } 

mysql_query("UPDATE note_comments SET likes = likes + 1 WHERE id=$comId")or die (mysql_error()); 
mysql_query("UPDATE note_comments SET likers =$com_likers WHERE id=$comId")or die (mysql_error()); 
echo 'success'; 
} else { 
    echo 'You have already liked this!'; 
} 
} 

?> 

Answer 2

比方说，我们都在同一个岗位三点意见：

第一个表：评论

id | post_id | comment 
1 | 1  | "something" 
2 | 1  | "something else" 
3 | 1  | "something else entirely" 

二表：comment_likes

id | comment_id 
1 | 2 
2 | 2 
3 | 3 

在这个例子中，注释2将有2个喜欢，并且评论3个。

这可能是显示我们评论的代码。这里的诀窍是为类似链接添加一个自定义属性，以便我们可以在javascript中检测评论ID。

<div class="comments"> 
<?php 
$res = mysql_query('SELECT * FROM `comments` WHERE `post_id` = '.$post_id.';'); 
while ($row = mysql_fetch_array($res)) { 
    $cid = $row['id']; 
    echo '<div class="comment">'; 
    echo $row['comment'].'<br />'; 
    echo '<span class="like" comment-id="'.$cid.'">Like</span>'; 
    $lres = mysql_query('SELECT COUNT(*) FROM `comment_likes` WHERE `comment_id` = '.$cid.';'); 
    $likes = mysql_result($lres,0); 

    echo '<span class="likes">'; 
    if ($likes > 0) { 
     echo '<span class="num_likes">'.$likes.'</span>'; 
     $p = ($likes > 1) ? 'people like' : 'person likes'; 
     echo $likes.' '.$p.' this'; 

    } 
    echo '</span>'; 
    echo '</div>'; 
} 
?> 
</div> 

现在的JavaScript（使用jQuery）：

$(document).ready(function() { 
    $('.like').click(function() { 
     var id = $(this).attr('comment-id'); 
     $.ajax({ 
      url: 'add_like.php', 
      method: 'POST', 
      data: {'id':id}, 
      success: function(res) { 
       if (res === '1') { 
        var likes = $('.like[comment-id='+id+']').find('.num_likes').text(); 
        likes++; 
        var p = likes > 1 ? 'people like' : 'person likes'; 
        var html = '<span class="num_likes">'+likes+'</span> '+p+' this'; 
        $('.like[comment-id='+id+']').find('likes').html(html); 
       } 
       else alert('Error Liking Comment'); 
      }, 
      error: function() { alert('Error liking comment'); } 
     }); 
    }); 
}); 

而PHP（用于AJAX） - add_like.php

<?php 

if (!isset($_POST['id'])) exit; 
$id = $_POST['id']; 
// connect/select db 
$res = mysql_query('INSERT INTO `comment_likes` WHERE `comment_id` = '.$id.';'); 
if ($res) { 
    die(1); // success 
} 
die(0); 

?> 

显然，你会想扩大这是喜欢/不喜欢 - 因为目前用户可以喜欢多次，但你明白了。