Question

所以我在页面上有一个项目列表。我试图添加功能，以便用户可以喜欢/不喜欢这些项目。当用户点击;收藏夹＆＃39;链接它应该将moodmemo_id和user_id发布到数据库中的收藏表，但是它不起作用。

HTML：

<a href="javascript:;" id="button" name="button0" class="ajaxFavourite">

Javascript：

$(function(){
  $(document).on('click','.ajaxFavourite',function(){

    // Configure those variables as appropriate

    var divid = 'status';
    var favid= $(this).attr('id');
    var url = 'addremove.php';


    // The XMLHttpRequest object

    var xmlHttp;
    try{
    xmlHttp=new XMLHttpRequest(); // Firefox, Opera 8.0+, Safari
    }
    catch (e){
    try{
    xmlHttp=new ActiveXObject("Msxml2.XMLHTTP"); // Internet Explorer
    }
    catch (e){
    try{
    xmlHttp=new ActiveXObject("Microsoft.XMLHTTP");
    }
    catch (e){
    alert("Your browser does not support AJAX.");
    return false;
    }
    }
    }

    // Generate timestamp for preventing IE caching the GET request

    fetch_unix_timestamp = function()
    {
    return parseInt(new Date().getTime().toString().substring(0, 10))
    }

    var timestamp = fetch_unix_timestamp();
    var nocacheurl = url+"?t="+timestamp;


    // This code sends the variables through AJAX and gets the response

    xmlHttp.onreadystatechange=function(){
    if(xmlHttp.readyState!=4){
    document.getElementById(divid).innerHTML='<img src="images/spinner.gif">  Wait...';
    }
    if(xmlHttp.readyState==4){
    document.getElementById(divid).innerHTML=xmlHttp.responseText;
    }
    }
    xmlHttp.open("GET",nocacheurl+"&favid="+favid,true);
    xmlHttp.send(favid);


    // Finally, some code for button toggle

    var button = document.getElementById('button');

    switch(button.name)
    {
    case 'button0':
      button.text = 'Unfavourite';
      button.name = 'button1';
      break;
    case 'button1':
      button.text = 'Favourite';
      button.name = 'button0';
      break;
    }

  });
});

addremove.php

$favid = mysql_real_escape_string($_GET['favid']);

// Firstly, check if article is favourite or not

$query = mysqli_query($mysqli, "SELECT * FROM tbl_favourites WHERE user_id='$loggedInUser->user_id' AND moodmemo_id=$favid");
$matches = mysqli_num_rows($query);

// If it is not favourited, add as favourite

if($matches == '0'){
mysqli_query($mysqli,"INSERT INTO tbl_favourites (user_id, moodmemo_id) VALUES ($loggedInUser->user_id, '$favid')");

echo "<div class=\"green\">This is a favourite</div>";
}

// Instead, if it is favourited, then remove from favourites

if($matches != '0'){
mysqli_query($mysqli,"DELETE FROM tbl_favourites WHERE user_id='$loggedInUser->user_id' AND moodmemo_id=$favid");

echo "<div class=\"red\">This is NOT a favourite</div>";
}

addremove.php中的查询失败。与...有关：我认为$query = mysqli_query($mysqli, "SELECT * FROM tbl_favourites WHERE user_id='$loggedInUser->user_id' AND moodmemo_id=$favid");。我得到警告：mysqli_num_rows（）要求参数1为mysqli_result，布尔值为错误。

我可以帮忙吗？感谢

无法使用ajax / php添加/删除收藏夹数据

0 个答案: