Php-Ajax喜欢/不喜欢按钮_Php_Ajax_Social Media Like

Php-Ajax喜欢/不喜欢按钮

php ajax

Php-Ajax喜欢/不喜欢按钮,php,ajax,social-media-like,Php,Ajax,Social Media Like,我通过对脚本使用“喜欢/不喜欢”按钮从internet搜索中找到了此脚本当用于单个帖子时，一切都很好，但是我想对来自while cycle的所有文档使用此分级脚本例如，我有一个新闻脚本，列出了所有的新闻，并且在标题附近可以看到喜欢/不喜欢按钮。（比如stackoverflow网页）我想对所有这些内容分别打分。在这个脚本中，我无法在同一页中分隔文章id 您能帮我一下吗？首先，您需要为每个按钮指定文章的目的，您可以使用data-attribute轻松完成 <button class="

我通过对脚本使用“喜欢/不喜欢”按钮从internet搜索中找到了此脚本

当用于单个帖子时，一切都很好，但是

我想对来自while cycle的所有文档使用此分级脚本

例如，我有一个新闻脚本，列出了所有的新闻，并且在标题附近可以看到喜欢/不喜欢按钮。（比如stackoverflow网页）我想对所有这些内容分别打分。在这个脚本中，我无法在同一页中分隔文章id

您能帮我一下吗？

首先，您需要为每个按钮指定文章的目的，您可以使用data-attribute轻松完成

<button class="votebutton" data-article="1" data-vote="1">Up vote</button>
<button class="votebutton" data-article="1" data-vote="-1">Down vote</button>
<span class="votes" data-article="1"><?=//print the current number of votes of this article//?>

现在您需要rateArticle（）函数，该函数如下所示：

function rateArticle(vote, article){ 
  var data = 'vote='+rating+'&articleID='+article;
  $.ajax({
     type: 'POST',
     url: 'rateArticle.php', 
     data: data,
     success: function(votes){
         $('.vote[data-article='+article+']').html(votes); 
     }
   });
}

$vote = $_POST['vote'];
$articleID = $_POST['articleID'];
$currentVotes = ; //fill it with the votes that the article currently have

if($vote != "-1" || vote != "1"){
  //goddamn script kids messing with us again, let's put them back where they belong
  header("Location: http://www.lemonparty.com"); //probably not this, but you can think of something
}

if(userHasAlreadyVoted()){ //if user has already voted we just return the current number of votes
  echo $currentVotes;
}else{
  $newVotes = $currentVotes + $vote;
  //save newVotes to db...
  echo $newVotes;
}

在服务器端，您将发布到的rateArticle.php文件如下所示：

function rateArticle(vote, article){ 
  var data = 'vote='+rating+'&articleID='+article;
  $.ajax({
     type: 'POST',
     url: 'rateArticle.php', 
     data: data,
     success: function(votes){
         $('.vote[data-article='+article+']').html(votes); 
     }
   });
}

$vote = $_POST['vote'];
$articleID = $_POST['articleID'];
$currentVotes = ; //fill it with the votes that the article currently have

if($vote != "-1" || vote != "1"){
  //goddamn script kids messing with us again, let's put them back where they belong
  header("Location: http://www.lemonparty.com"); //probably not this, but you can think of something
}

if(userHasAlreadyVoted()){ //if user has already voted we just return the current number of votes
  echo $currentVotes;
}else{
  $newVotes = $currentVotes + $vote;
  //save newVotes to db...
  echo $newVotes;
}

这就是问题所在。

首先，您需要为每个按钮指定文章的目的，您可以使用data-attribute轻松完成

<button class="votebutton" data-article="1" data-vote="1">Up vote</button>
<button class="votebutton" data-article="1" data-vote="-1">Down vote</button>
<span class="votes" data-article="1"><?=//print the current number of votes of this article//?>

现在您需要rateArticle（）函数，该函数如下所示：

function rateArticle(vote, article){ 
  var data = 'vote='+rating+'&articleID='+article;
  $.ajax({
     type: 'POST',
     url: 'rateArticle.php', 
     data: data,
     success: function(votes){
         $('.vote[data-article='+article+']').html(votes); 
     }
   });
}

$vote = $_POST['vote'];
$articleID = $_POST['articleID'];
$currentVotes = ; //fill it with the votes that the article currently have

if($vote != "-1" || vote != "1"){
  //goddamn script kids messing with us again, let's put them back where they belong
  header("Location: http://www.lemonparty.com"); //probably not this, but you can think of something
}

if(userHasAlreadyVoted()){ //if user has already voted we just return the current number of votes
  echo $currentVotes;
}else{
  $newVotes = $currentVotes + $vote;
  //save newVotes to db...
  echo $newVotes;
}

在服务器端，您将发布到的rateArticle.php文件如下所示：

function rateArticle(vote, article){ 
  var data = 'vote='+rating+'&articleID='+article;
  $.ajax({
     type: 'POST',
     url: 'rateArticle.php', 
     data: data,
     success: function(votes){
         $('.vote[data-article='+article+']').html(votes); 
     }
   });
}

$vote = $_POST['vote'];
$articleID = $_POST['articleID'];
$currentVotes = ; //fill it with the votes that the article currently have

if($vote != "-1" || vote != "1"){
  //goddamn script kids messing with us again, let's put them back where they belong
  header("Location: http://www.lemonparty.com"); //probably not this, but you can think of something
}

if(userHasAlreadyVoted()){ //if user has already voted we just return the current number of votes
  echo $currentVotes;
}else{
  $newVotes = $currentVotes + $vote;
  //save newVotes to db...
  echo $newVotes;
}

这就是问题所在……

我已经编写了这段代码，它对于在p.h.p-code中使用ajax的人来说效果很好：）

我将它分为三个页面，分别命名为test1.php、test1.js、test1ppp.php

    <?php
    mysql_connect("localhost","root","");
    mysql_select_db("school");
    $id = $_GET['status_id'];
    $q = mysql_query("SELECT * from s_user WHERE u_id = $id");
    $r = mysql_fetch_array($q);
    $fetch_status = $r['u_status'];
    if($fetch_status == 1){
        mysql_query("UPDATE s_user SET u_status = 0 WHERE u_id = '$id'");
    }else{
        mysql_query("UPDATE s_user SET u_status = 1 WHERE u_id = '$id'");
    }
?>
            <table id="status_id">
            <?php
                $qry = mysql_query("SELECT * from s_user");
                while($row = mysql_fetch_array($qry)){
                    $uid = $row['u_id'];
                    $status = $row['u_status'];
                    $uname = $row['u_uname'] . "<br />";
            ?>
                <tr>
                    <td><?php echo $uname; ?></td>
                    <td><div id=""><?php echo $status; ?></div></td>
                    <td><input type="button" onclick="return change_status(this.value);" value="<?php echo $uid; ?>"></td>
                <?php } ?>
                </tr>
            </table>

test1.php

<?php
    mysql_connect("localhost","root","");
    mysql_select_db("school");
?>
<html>
    <head>
        <script type="text/javascript" src="test1.js"></script>
    </head>
    <body>
        <form action="" method="post" enctype="">
            <table id="status_id">
            <?php
                $qry = mysql_query("SELECT * from s_user");
                while($row = mysql_fetch_array($qry)){
                    $uid = $row['u_id'];
                    $status = $row['u_status'];
                    $uname = $row['u_uname'] . "<br />";
            ?>
                <tr>
                    <td><?php echo $uname; ?></td>
                    <td><div id=""><?php echo $status; ?></div></td>
                    <td><input type="button" onclick="return change_status(this.value);" value="<?php echo $uid; ?>"></td>
                <?php } ?>
                </tr>
            </table>                    
        </form>
    </body>
</html>

test1ppp.php

    <?php
    mysql_connect("localhost","root","");
    mysql_select_db("school");
    $id = $_GET['status_id'];
    $q = mysql_query("SELECT * from s_user WHERE u_id = $id");
    $r = mysql_fetch_array($q);
    $fetch_status = $r['u_status'];
    if($fetch_status == 1){
        mysql_query("UPDATE s_user SET u_status = 0 WHERE u_id = '$id'");
    }else{
        mysql_query("UPDATE s_user SET u_status = 1 WHERE u_id = '$id'");
    }
?>
            <table id="status_id">
            <?php
                $qry = mysql_query("SELECT * from s_user");
                while($row = mysql_fetch_array($qry)){
                    $uid = $row['u_id'];
                    $status = $row['u_status'];
                    $uname = $row['u_uname'] . "<br />";
            ?>
                <tr>
                    <td><?php echo $uname; ?></td>
                    <td><div id=""><?php echo $status; ?></div></td>
                    <td><input type="button" onclick="return change_status(this.value);" value="<?php echo $uid; ?>"></td>
                <?php } ?>
                </tr>
            </table>

我已经编写了这段代码，它在p.h.p-code中使用ajax时效果很好：）
我将它分为三个页面，分别命名为test1.php、test1.js、test1ppp.php
    <?php
    mysql_connect("localhost","root","");
    mysql_select_db("school");
    $id = $_GET['status_id'];
    $q = mysql_query("SELECT * from s_user WHERE u_id = $id");
    $r = mysql_fetch_array($q);
    $fetch_status = $r['u_status'];
    if($fetch_status == 1){
        mysql_query("UPDATE s_user SET u_status = 0 WHERE u_id = '$id'");
    }else{
        mysql_query("UPDATE s_user SET u_status = 1 WHERE u_id = '$id'");
    }
?>
            <table id="status_id">
            <?php
                $qry = mysql_query("SELECT * from s_user");
                while($row = mysql_fetch_array($qry)){
                    $uid = $row['u_id'];
                    $status = $row['u_status'];
                    $uname = $row['u_uname'] . "<br />";
            ?>
                <tr>
                    <td><?php echo $uname; ?></td>
                    <td><div id=""><?php echo $status; ?></div></td>
                    <td><input type="button" onclick="return change_status(this.value);" value="<?php echo $uid; ?>"></td>
                <?php } ?>
                </tr>
            </table> 

test1.php
<?php
    mysql_connect("localhost","root","");
    mysql_select_db("school");
?>
<html>
    <head>
        <script type="text/javascript" src="test1.js"></script>
    </head>
    <body>
        <form action="" method="post" enctype="">
            <table id="status_id">
            <?php
                $qry = mysql_query("SELECT * from s_user");
                while($row = mysql_fetch_array($qry)){
                    $uid = $row['u_id'];
                    $status = $row['u_status'];
                    $uname = $row['u_uname'] . "<br />";
            ?>
                <tr>
                    <td><?php echo $uname; ?></td>
                    <td><div id=""><?php echo $status; ?></div></td>
                    <td><input type="button" onclick="return change_status(this.value);" value="<?php echo $uid; ?>"></td>
                <?php } ?>
                </tr>
            </table>                    
        </form>
    </body>
</html> 

test1ppp.php
    <?php
    mysql_connect("localhost","root","");
    mysql_select_db("school");
    $id = $_GET['status_id'];
    $q = mysql_query("SELECT * from s_user WHERE u_id = $id");
    $r = mysql_fetch_array($q);
    $fetch_status = $r['u_status'];
    if($fetch_status == 1){
        mysql_query("UPDATE s_user SET u_status = 0 WHERE u_id = '$id'");
    }else{
        mysql_query("UPDATE s_user SET u_status = 1 WHERE u_id = '$id'");
    }
?>
            <table id="status_id">
            <?php
                $qry = mysql_query("SELECT * from s_user");
                while($row = mysql_fetch_array($qry)){
                    $uid = $row['u_id'];
                    $status = $row['u_status'];
                    $uname = $row['u_uname'] . "<br />";
            ?>
                <tr>
                    <td><?php echo $uname; ?></td>
                    <td><div id=""><?php echo $status; ?></div></td>
                    <td><input type="button" onclick="return change_status(this.value);" value="<?php echo $uid; ?>"></td>
                <?php } ?>
                </tr>
            </table> 

当然可以。你有什么，你尝试过什么，什么对你不起作用？我成功地给一篇文章打分，我没有单独给一页中列出的文章打分。剧本只是第一篇文章。我也要像stackoverflow一样思考。页面中列出了帖子和答案，每个人都可以通过点击喜欢或不喜欢按钮来给每个帖子打分。你有什么，你尝试过什么，什么对你不起作用？我成功地给一篇文章打分，我没有单独给一页中列出的文章打分。剧本只是第一篇文章。我也要像stackoverflow一样思考。页面中列出了帖子和答案，每个人都可以通过点击喜欢或不喜欢按钮对每个帖子进行评分