大家好。我正在设计一个平台，就像一个社交媒体平台，有帖子和喜欢和评论系统。我使用jquery的喜欢，每当我想要文本它，我想喜欢一个帖子，只有第一个帖子将工作，但其他帖子的喜欢系统不工作。我怎么解决它。这是包含post和jquery代码的index.php。
<!DOCTYPE html>
<html lang="en">
<head>
<meta charset="UTF-8">
<meta http-equiv="X-UA-Compatible" content="IE=edge">
<meta name="viewport" content="width=device-width, initial-scale=1.0">
<title>post</title>
<link rel="stylesheet" href="style.css">
<link rel="stylesheet" href="font/css/all.css">
<link rel="stylesheet" href="https://fonts.googleapis.com/css?family=Poppins">
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.0/jquery.min.js"></script>
<script>
$(document).ready(function(){
$("#likings").click(function(){
var name = $("input").val();
$.post("like.php", {
sugess: name
}, function(data, status){
$("#like").html(data);
});
});
});
</script>
</head>
<?php
include('conn.php');
?>
<body>
<?php
$sql = "SELECT * FROM post";
$result = mysqli_query($conn, $sql);
while($row = mysqli_fetch_assoc($result)){
$id = $row['id'];
$name = $row['name'];
$image = $row['image'];
$action = $row['action'];
$date = $row['date'];
$time = $row['time'];
$sqlcount = "SELECT * FROM likes WHERE postid=$id";
$resultcount = mysqli_query($conn, $sqlcount);
$count = mysqli_num_rows($resultcount);
echo '
<center>
<div class="post">
<div class="up">
<div>
<img class="img" src="img/'.$image.'">
</div>
<div class="uptext"><span class="name">'.$name.' </span><span class="des"> '.$action.'
<br>'.$date.' at '.$time.'</span></div>
<div>
...
</div>
</div>
<img src="img/'.$image.'">
<div class="liking">
<div class="like">
<i class="fa fa-thumbs-up" aria-hidden="true" style="text-align: left; color: navy;"></i><span id="like">'.$count.'</span>
</div>
<div>
<p>23 comments</p>
</div>
</div>
<div class="likenow">
<div>
<input name="id" value="'.$id.'" hidden>
<i class="fa fa-thumbs-up" id="likings" aria-hidden="true" style="text-align: left; color: navy;"></i>like
</div>
<div>
<i class="fa-solid fa-message"></i> comment
</div>
</div>
<p style="text-align: left; margin-left: 20px; font-size: 10pt;">view more comments</p>
<div class="comments">
<p>itz celeb <br> very nice</p>
<p>titi kosi <br> so cute</p>
</div>
<div class="entercomment">
<div>
<img class="img" src="img/'.$image.'">
</div>
<div>
<input placeholder="Write a comment..."><i class="fa-solid fa-message" style="margin-left: 20px;"></i>
</div>
</div>
</div>
</center>';
}
?>
</body>
</html>
这是喜欢的php代码。
<?php
include("conn.php");
$name = $_POST['sugess'];
$sql = "INSERT INTO likes(postid, likes) VALUES('$name', '1')";
$result = mysqli_query($conn, $sql);
$sqlcount = "SELECT * FROM likes WHERE postid=$name";
$resultcount = mysqli_query($conn, $sqlcount);
$count = mysqli_num_rows($resultcount);
echo $count;
?>
那我该怎么解决呢
like应该适用于每个岗位。我使用了while循环，但我仍然有同样的问题。
你的问题是在选择器中，它是一个ID选择器，使用ID选择器你只能选择一个元素尝试将其更改为类选择器
<i class="fa fa-thumbs-up" id="likings" aria-hidden="true" style="text-align: left; color: navy;"></i>
<i class="fa fa-thumbs-up likings" aria-hidden="true" style="text-align: left; color: navy;"></i>
应该可以工作
<script>
$(document).ready(function(){
$(".likings").click(function(){
var name = $("input").val();
$.post("like.php", {
sugess: name
}, function(data, status){
$("#like").html(data);
});
});
});
</script>