我想在帖子下面制作一个不喜欢的按钮。注册用户可以给予喜欢。我已经制作了按钮，但我不知道如何在用户点击按钮时发送记录。我想我需要like表，所以下面是表格和我所拥有的字段：
user：id_user，username
posting：id_post，id_user，content
like：id_like，id_user，id_post
发布页面和类似按钮脚本：
<?php
include "database_connection.php";
$query=$dbc->query("select user.username, posting.content FROM posting inner join user on user.id_user = posting.id_user where id_post='$_GET[id]'");
$array= $query->fetch_array()
?>
<!doctype html>
<html>
<head>
<script type="text/javascript" src="post.js"></script>
</head>
<body>
<?php echo $array['username'];?>
<?php echo $array['content'];?>
<!--THIS IS LIKE BUTTON-->
<a class="like-button" href="#"><i class="fa fa-thumbs-up"></i></a>
<!--LIKE BUTTON END-->
</body>
</html>
post.js
$(function() {
$('.like-button').click(function(){
var obj = $(this);
if( obj.data('liked') ){
obj.data('liked', false);
obj.html('<i class="fa fa-thumbs-up"></i>');
}
else{
obj.data('liked', true);
obj.html('<i class="fa fa-thumbs-down"></i>');
}
});
});
答案 0 :(得分：1)
好的，所以我花了很多时间为你创建一个基本的工作示例。
为了简单起见，我在内联脚本中包含了post.js的工作方式而不是单独的文件，并在脚本中包含了一个PHP变量。
您的HTML页面
<?php
include "database-connection.php";
$user = 1;// get your accessing user ID (not user id of poster)
$post = $_GET['id'];
// query checks whether user has liked the post or not and returns it as well
$query=$dbc->prepare("
SELECT `user`.`username`, `posting`.`content`, IFNULL(`like`.`id_like`,0) AS `id_like`
FROM `posting`
INNER JOIN `user` ON `user`.`id_user` = `posting`.`id_user`
LEFT JOIN `like` ON `like`.`id_user` = ? AND `like`.`id_post` = ?
WHERE `posting`.`id_post`=?");
// bind the parameters to avoid injection
$query->execute(array($user, $post, $post));
$array= $query->fetch(PDO::FETCH_ASSOC);
?>
<!doctype html>
<html>
<head>
<script type="text/javascript" src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.0/jquery.min.js"></script>
<script type="text/javascript">
// previously post.js
$(function() {
$('.like-button').click(function(e){
e.preventDefault();
var obj = $(this);
// ajax query that returns a JSON object with the result of the request
$.getJSON('likes.php',{post:obj.data('post'), user: <?php echo $user; ?>}, function(data){
//console.log(data); // uncomment for debugging
if(data.error){
// query returned error, handle it however you want
} else {
if (data.like == 1){ // user now likes the post
obj.html('<i class="fa fa-thumbs-up"></i>');
} else { // user now doesn't like the post
obj.html('<i class="fa fa-thumbs-down"></i>');
}
}
});
});
});
</script>
</head>
<body>
<?php echo $array['username'];?>
<?php echo $array['content'];?>
<!--THIS IS LIKE BUTTON-->
<?php
if ($array['id_like']==0){
// user hasn't liked the post, show thumbs down
echo '<a class="like-button" href="#" data-post="'.$post.'"><i class="fa fa-thumbs-down"></i></a>';
} else {
// user has liked the post, show thumbs up
echo '<a class="like-button" href="#" data-post="'.$post.'"><i class="fa fa-thumbs-up"></i></a>';
}
?>
<!--LIKE BUTTON END-->
</body>
</html>
likes.php（PHP脚本）
<?php
include "database-connection.php";
$post = $_GET['post'];
$user = $_GET['user'];
$result = (object) ['like'=>null, 'post'=>$post, 'user'=>$user];
$q = $dbc->prepare("SELECT id_like FROM `like` WHERE id_post=? AND id_user=?");
$q->execute(array($post, $user));
$r = $q->fetch(PDO::FETCH_OBJ);
if ($q->rowCount() > 0){
$like = $r->id_like;
} else {
$like = 0;
}
if ($like == 1){
// user likes post, so we unlike it by setting id_like to 0 (for false)
$like = 0;
$u = $dbc->prepare("UPDATE `like` SET id_like = 0 WHERE id_post=? AND id_user=?");
} elseif ($q->rowCount()>0) {
// update because the record exists
$like = 1;
$u = $dbc->prepare("UPDATE `like` SET id_like = 1 WHERE id_post=? AND id_user=?");
} else {
// create the record because it doesn't exist yet
$like = 1;
$u = $dbc->prepare("INSERT INTO `like` (id_like, id_post, id_user) VALUES(1, ?, ?)");
}
if($u->execute(array($post, $user))){
// update succeeded
$result->like = $like;
} else{
// there was an error
$result->error = 'failed to execute in database';
}
// return the json object to your page
echo json_encode($result);
同样，这只是它如何工作的基础知识。您必须自己研究登录，会话和安全性，以管理正在访问，发布和喜欢内容的用户。但我希望这有帮助！
答案 1 :(得分：0)
点击like按钮将请求发送到PHP页面并在那里处理以更新数据库。
答案 2 :(得分：-1)
您需要向服务器发送Ajax请求，然后在PHP脚本中处理。
这是一种方法。
post.js：
$(function() {
$('.like-button').click(function(){
var obj = $(this);
if( obj.data('liked') ){
obj.data('liked', false);
obj.html('<i class="fa fa-thumbs-up"></i>');
}
else{
obj.data('liked', true);
obj.html('<i class="fa fa-thumbs-down"></i>');
}
$.post('url/to/your_script.php', {
action: 'updateLikeStatus',
status: obj.data('liked'),
post_id: obj.data('id') // ID of the object that user "liked"
});
});
});
您可以阅读有关jQuery.post() here的更多信息。关于更一般的jQuery.ajax()方法的here's文档。
your_script.php（处理Ajax请求的脚本）可能如下所示：
<?php
include "database_connection.php";
if (isset($_POST['action']) && $_POST['action'] === 'updateLikeStatus') {
$id_user = $_SESSION['user_id'];
$id_post = $_POST['post_id'];
if ($_POST['status'] === true) {
// adding "like"
$query = $dbc->query("
INSERT INTO like
(id_user, id_post)
VALUES ({$id_user}, {$id_post});
");
$query->query();
} else {
// removing "like"
$query = $dbc->query("
DELETE FROM like
WHERE id_user = {$id_user}
AND id_post = {$id_post};
");
$query->query();
}
}
请注意，此代码只是一个示例，您不应该直接在生产中使用它。首先，您不能将$_POST中的变量直接放入MySQL查询中，因为它会创建一个SQL注入类型漏洞，允许人们在您的服务器上执行任意查询。避免它的一种方法是使用prepared statements。
另一个问题是您需要处理用户身份验证和授权。我在我的示例中使用了$_SESSION['user_id']，但除非您初始化会话并首先填充user_id值，否则它将无效。会话是必需的，以便一个用户不能代表另一个用户喜欢帖子。你可以read more about sessions here。