php - 使用 AJAX 切换 Like 按钮颜色

Question

点击帖子下方的“赞”按钮。如果点击它，它会变成橙色，否则它会变成绿色。我无法切换颜色。如何使用 AJAX 请求从 MySQL 数据库查询数据来执行此操作？

我尝试使用 jQuery 来使用切换属性修改 css 文件。我还尝试查询数据库以查看用户是否喜欢该帖子，并将结果设置为要在 AJAX 函数(数据)中使用的变量。

注意。为了演示简单起见，删除了防止黑客攻击的代码。

索引 php 文件:

        <?php
            $userid = session_id();
            $query = "SELECT * FROM posts";
            $result = mysqli_query($con,$query);
            while($row = mysqli_fetch_array($result)){
                $postid = $row['id'];
                $title = $row['title'];
                $content = $row['content'];

                // Checking user status
                $status_query = "SELECT count(*) as type FROM likes WHERE userid='".$userid. "'" .  "and postid=".$postid;
                $status_result = mysqli_query($con,$status_query);
                $status_row = mysqli_fetch_array($status_result);
                $type = $status_row['type'];

                // Count post total likes and unlikes
                $like_query = "SELECT COUNT(*) AS cntLikes FROM likes WHERE postid=".$postid;
                $like_result = mysqli_query($con,$like_query);
                $like_row = mysqli_fetch_array($like_result);
                $total_likes = $like_row['cntLikes'];

        ?>

                <div class="post">
                    <h1><?php echo $title; ?></h1>
                    <div class="post-text">
                        <?php echo $content; ?>
                    </div>
                    <div class="post-action">

                        <input type="button" value="Like" id="like_<?php echo $postid . "_" . $userid; ?>" class="like" style="<?php if($type == 1){ echo "color: #ffa449;"; } ?>" />&nbsp;(<span id="likes_<?php echo $postid . "_" . $userid; ?>"><?php echo $total_likes; ?></span>)&nbsp;

                    </div>
                </div>
        <?php
            }
        ?>

    </div>
</body>

Ajax jQuery:

$(".like").click(function(){
    var id = this.id;   // Getting Button id
    var split_id = id.split("_");

    var postid = split_id[1]; 
    var userid = split_id[2];

    // AJAX Request
    $.ajax({
        url: 'likeunlike.php',
        type: 'post',
        data: {postid:postid,userid:userid},
        dataType: 'json',
        success: function(data){
            var likes = data['likes'];
            var type = data['type'];

            $("#like_" + postid + "_" + userid).text(likes);

            if(type == 1){
                $("#likes_" + postid + "_" + userid).css("color","#ffa449");
            }

            if(type == 0){
                $("#likes_" + postid + "_" + userid).css("color","lightseagreen");
            }
        }
    });

});

调用php文件:

$postid = $_POST['postid'];
$userid = $_POST['userid'];

// Check entry within table
$query = "SELECT COUNT(*) AS cntpost FROM likes WHERE postid=".$postid." and userid='".$userid . "'";

$result = mysqli_query($con,$query);
$fetchdata = mysqli_fetch_array($result);
$count = $fetchdata['cntpost'];

if($count == 0){
    $insertquery = "INSERT INTO likes(userid,postid) values('".$userid."',".$postid.")";
    mysqli_query($con,$insertquery);
}else {
    $updatequery = "DELETE FROM likes where userid='" . $userid . "'" . " and postid=" . $postid;
    mysqli_query($con,$updatequery);
}

// count numbers of likes in post
$query = "SELECT COUNT(*) AS cntLike FROM likes WHERE postid=".$postid;
$result = mysqli_query($con,$query);
$fetchlikes = mysqli_fetch_array($result);
$totalLikes = $fetchlikes['cntLike'];

$return_arr = array("likes"=>$totalLikes,"type"=>$count);

echo json_encode($return_arr);

Answer 1

修改了 jQuery 文件如下。 jQuery 之前没有命中正确的标签。我还使用 Safari，需要清除缓存才能查看页面的修改后的更改。哎呀。新手。

$("#like_" + postid + "_" + userid).text(likes);

  if(type == 1){
    $("#likes_" + postid + "_" + userid).css("color","#ffa449");
            }

   if(type == 0){
     $("#likes_" + postid + "_" + userid).css("color","lightseagreen");
          }

至:

$("#likes_" + postid + "_" + userid).text(likes);

  if(type == 1){
    $("#like_" + postid + "_" + userid).css("color","lightseagreen");

                }

if(type == 0){
  $("#like_" + postid + "_" + userid).css("color","#ffa449"); 
                }