Ajax PHP關注腳本 - 沒有存儲在數據庫中

Question

我最近在ajax上爲初學者發現了一個樹屋博客http://blog.teamtreehouse.com/beginners-guide-to-ajax-development-with-php我一直在尋找一段時間的關注腳本，並且我已經陷入死衚衕。目前，按鈕按鈕應該會消失，但到目前爲止還沒有值存儲在數據庫中。

Profile.php（遵循按鈕）：

  <div id="followbtncontainer" class="btncontainer"><a href="#" id="followbtn" class="bigblue">Follow</a></div> 

Ajax.js

$(function(){ 
    $('#followbtn').on('click', function(e){ 
    e.preventDefault(); 
    $('#followbtn').fadeOut(300); 

    $.ajax({ 
     url: '../ajax-follow.php', 
     type: 'post', 
     data: {'action': 'follow'}, 
     success: function(data, status) { 
     if(data == "ok") { 
      $('#followbtncontainer').html('<p><em>Following!</em></p>'); 
      var numfollowers = parseInt($('#followercnt').html()) + 1; 
      $('#followercnt').html(numfollowers); 
     } 
     }, 
     error: function(xhr, desc, err) { 
     console.log(xhr); 
     console.log("Details: " + desc + "\nError:" + err); 
     } 
    }); // end ajax call 
    }); 

    $('body').on('click', '#morefllwrs', function(e){ 
    e.preventDefault(); 
    var container = $('#loadmorefollowers'); 

    $(container).html('<img src="images/loader.gif">'); 
    var newhtml = ''; 

    $.ajax({ 
     url: 'ajax-followers.php', 
     type: 'post', 
     data: {'page': $(this).attr('href')}, 
     cache: false, 
     success: function(json) { 
     $.each(json, function(i, item) { 
      if(typeof item == 'object') { 
      newhtml += '<div class="user"> <a href="#" class="clearfix"> <img src="'+item.profile_pic+'" class="avi"> <h4>'+item.username+'</h4></a></div>'; 
      } 
      else { 
      return false; 
      } 
     }) // end $.each() loop 

     if(json.nextpage != 'end') { 
      // if the nextpage is any other value other than end, we add the next page link 
      $(container).html('<a href="'+json.nextpage+'" id="morefllwrs" class="bigblue thinblue">Load more followers</a>'); 
     } else { 
      $(container).html('<p></p>'); 
     } 

     $('#followers').append(newhtml); 
     }, 
     error: function(xhr, desc, err) { 
     console.log(xhr + "\n" + err); 
     } 
    }); // end ajax call 
    }); 
}); 

ajax.php

<?php require 'database.php' //<?php include 'session-check-index.php' ?> 
<?php include 'authentication.php' ?> 
<?php 
    session_start(); 
$follower=$_SESSION['id']; 

    $sql = "SELECT * FROM users WHERE username='$username'"; 
    $result = mysqli_query($database,$sql); 
    $rws = mysqli_fetch_array($result); 
$following=$rws['id']; 

/** 
* this script will auto-follow the user and update their followers count 
* check out your POST data with var_dump($_POST) 
**/ 



if($_POST['action'] == "follow") { 

$sql=" INSERT INTO `user_follow` (`follower`, `following`, `subscribed`) VALUES ('$follower', '$following', CURRENT_TIMESTAMP);" 
    /** 
    * we can pass any action like block, follow, unfollow, send PM.... 
    * if we get a 'follow' action then we could take the user ID and create a SQL command 
    * but with no database, we can simply assume the follow action has been completed and return 'ok' 
    **/ 
    mysqli_query($database,$sql) or die(mysqli_error($database)); 

} 

?> 

我不知道，如果實際$以下和$ follower值導致問題，並且不傳遞任何數據。任何幫助將不勝感激，謝謝！

Answer 1

嘗試在ajax.js

$(function(){ 
    $('#followbtn').on('click', function(e){ 
     e.preventDefault(); 
     $('#followbtn').fadeOut(300); 

     $.ajax({ 
      url: '../ajax-follow.php', 
      ... 

更改URL參數：

url: 'ajax-follow.php', 

看它是否會工作方式