用php和jquery实现一个喜欢/不喜欢的计数器 [英] Implementing a like/dislike counter with php and jquery

问题描述

我有一个漫画网站，我想跟踪有多少人对一个漫画ID表示喜欢和不喜欢点击...我想将值存储在数据库中，并报告给我进入浏览器.

I have a comics website, and I'd like to keep track of how many people click like and dislike for a single comic id... I'd like to store the value in my database, and also report it back out to the browser.

这是我无法使用的解决方案:

This is my non-working solution:

viewcomic.php:

    <script type="text/javascript">
    function likeCounter(choice) {    
       var site = $("#site").val();
       var imgid = $("#imgid").val();
       $.get("./scripts/likecounter.php", {_choice : choice, _site : site, _id : imgid},
            function(returned_data) {
                $("#choice").html(returned_data);
            }
        );
}
</script>
       //Changed to try and work with buttons
   <button id="like" onClick="likeCounter(this.id)">Like</button>
   <button id="dislike" onClick="likeCounter(this.id)">Dislike</button>
   <input id="site" type="hidden" value="<?php echo $site; ?>">
   <input id="imgid" type="hidden" value="<?php echo $imgid; ?>">

<br />
Likes: <span id="choice"></span>

likecounter.php

<?php 
include 'dbconnect.php';

$site = $_GET['_site'];
$imgid = $_GET['_id'];
$input = $_GET['_choice'];

if ($site == "artwork") {
   $table = "comics";
}
else {
   $table = "artwork";
}

$likes = $mysqli->query("SELECT like FROM $table WHERE id = $imgid");
$dislikes = $mysqli->query("SELECT dislike FROM $table WHERE id = $imgid");


if ($input == "like") {
   $sql = "UPDATE $table SET like = like + 1 WHERE id = $imgid";
   $mysqli->query($sql);
   $likes++;

else if ($input == "dislike") {
   $sql = "UPDATE $table SET like = dislike + 1 WHERE id = $imgid";
   $mysqli->query($sql);
   $dislikes++;
}   
mysqli_close($mysqli);

echo "Likes: " . $likes . ", Dislikes: " . $dislikes;

?>

它没有增加数据库值，也没有向浏览器报告一个值.有什么建议吗?

It's not incrementing the database value, and it's not reporting out a value to the browser. Any suggestions?

谢谢！

推荐答案

技术问题是这一行:

$.get("./scripts/likecounter.php", {_choice : choice, _site : site, _id : id}...

由于未定义网站和ID，您会收到JavaScript错误ReferenceError: site is not defined

as site and id are undefined you'll be getting a javascript error ReferenceError: site is not defined

尝试:

$.get("./scripts/likecounter.php", {_choice : choice, _site : "<?php echo $site; ?>", _id : "<?php echo $imgid; ?>"}...

您不需要隐藏的输入，onClick也应为onclick.

You don't need the hidden inputs, also onClick should be onclick.

其他快速建议:

确保$ imgid实际上是一个整数:

Make sure $imgid is actually an integer:

$imgid = intval($_GET['_id']);
if($imgid == 0) exit('Invalid id');

以下内容无需分开:

$likes = $mysqli->query("SELECT like FROM $table WHERE id = $imgid");
$dislikes = $mysqli->query("SELECT dislike FROM $table WHERE id = $imgid");

$result = $mysqli->query("SELECT like_count, dislike_count FROM $table WHERE id = $imgid");

$mysqli->query将仅返回mysqli_result，这意味着您不能只将其回显，而必须进行提取，并且因为您知道只得到一个结果，所以可以使用list($likes, $dislikes) = $result->fetch_array(MYSQLI_NUM);

$mysqli->query will only return a mysqli_result, meaning you can't just echo it out, you'll have to do a fetch, and because you know you are only getting a single result you can just use list($likes, $dislikes) = $result->fetch_array(MYSQLI_NUM);

已阅读文档以帮助您理解我到底要怎么做.

Have a read of the docs to help you understand what the heck I'm going on about.

正如您提到的那样，人们可以一直喜欢/不喜欢他们想要的东西，您需要使用某种IP日志来限制它.例如，使用以下字段设置一个新的数据库表； ip，表格，imgid.然后，当有人喜欢"登录其IP时:

As you mentioned people can just keep hitting like/dislike all they want, you'll need to limit this with an IP log of some sort. For example setup a new database table with the following fields; ip, table, imgid. Then when someone "likes" log their IP:

$sql = "INSERT INTO xxx (ip, table_name, imgid) VALUES(\"".$_SERVER['REMOTE_ADDR']."\", \"$table\", $imgid)";
$mysqli->query($sql);

然后在添加新的类似项之前检查他们是否有记录:

Then check if they have a record before adding a new like:

$sql = "SELECT ip FROM xxx WHERE ip = \"".$_SERVER['REMOTE_ADDR']."\" AND table_name = "$table" AND imgid = $imgid";
$result = $mysqli->query($sql);
if($result->num_rows == 0) { ...if($input ==... }

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