Question

我正在开发像twitter这样的关注/取消关注系统。我在过去的两个晚上尝试了几件事，但我无法弄清楚为什么元素的ID没有通过JQUERY传递给PHP。

请帮我纠正我的代码：

HTML：

<a class="btn btn-sm btn-default follow_user" id="<?php echo $row['user_id'];?>" title="Follow"><i class="glyphicon glyphicon-plus"></i> Follow</a>

JQUERY：

$(document).ready(function(){
 $(document).on('click', '.follow_user', function(){
  if($(this).attr('title') == 'Follow'){
    $that = $(this);
    var ID=$that.attr('id');
    var action="follow";
    $.post('include/common/follow_user.php?user_id='+ID+'&action='+action,function(){
    $that.html('<i class="glyphicon glyphicon-check"></i> Following');
    $that.attr('title','Unfollow');
    $that.removeClass('btn btn-sm btn-default');
    $that.addClass('btn btn-sm btn-success');
   });
  }

PHP代码：

<?php session_start(); ob_start();?>
<?php $user_id=$_SESSION['user_id'];?>

<!-- Follow OR Unfollow User 
=============================================== -->
<?php
        include "../db.php";
        echo $follow_user=strip_tags(stripslashes(trim(mysqli_real_escape_string($con, $_POST['user_id']))));
        echo $followee=$user_id;
        echo $action=strip_tags(stripslashes(trim(mysqli_real_escape_string($con, $_POST['action']))));
        if ($action=='follow'){
            //follow
            $sql="insert into follow(follower_id,following_id,follow_date)values('$followee','$follow_user',UTC_TIMESTAMP())";
            if($con,$sql){}else{echo '<script type="text/javascript">toastr.error("We encountered a problem in doing that operation. Please try again after sometime.");</script>';}
        }else{}
?>

Answer 1

IEnumerator downloadImage() { string url = "http://wallpaper-gallery.net/images/hq-images-wallpapers/hq-images-wallpapers-12.jpg"; UnityWebRequest www = UnityWebRequestTexture.GetTexture(url); DownloadHandler handle = www.downloadHandler; //Send Request and wait yield return www.SendWebRequest(); if (www.isHttpError || www.isNetworkError) { UnityEngine.Debug.Log("Error while Receiving: " + www.error); } else { UnityEngine.Debug.Log("Success"); //Load Image Texture2D texture2d = DownloadHandlerTexture.GetContent(www); Sprite sprite = null; sprite = Sprite.Create(texture2d, new Rect(0, 0, texture2d.width, texture2d.height), Vector2.zero); if (sprite != null) { imageToUpdate.sprite = sprite; } } }应该像这样发送给post方法

Data

问题是您在url中将传递数据作为get方法。但是在尝试从post方法获取的服务器上。使用get方法。

$.post('include/common/follow_user.php',
       {user_id:ID,action:ac‌tion},
       function(data,‌status){ alert(data); 
      });

从jquery到php点击LINK的ID

1 个答案: