Javascript 为什么当我点击';比如';形象?
我有一个类似按钮的图片,当点击时,应该会将分数更新1。 我正在使用ajax来执行此操作。 但是,当我单击图像时,什么也没有发生。当我将鼠标悬停在图像上时,指针甚至不会变成手 index.phpJavascript 为什么当我点击';比如';形象?,javascript,php,jquery,ajax,Javascript,Php,Jquery,Ajax,我有一个类似按钮的图片,当点击时,应该会将分数更新1。 我正在使用ajax来执行此操作。 但是,当我单击图像时,什么也没有发生。当我将鼠标悬停在图像上时,指针甚至不会变成手 index.php <!DOCTYPE html> <html> <head> <link rel="stylesheet" type="text/css" href="css/style.css"></link> <s
<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" type="text/css" href="css/style.css"></link>
<script src="js/jquery-1.11.3.min.js"></script>
<script src="js/script.js"></script>
</head>
<body>
Welcome: <?php
// Start the session
session_start();
//Use session variable created in login.php
echo $_SESSION['username'];
?>
<div class="friend">
<?php
while($row = $res->fetch_assoc())
{
echo "<image class='img' src='images/" .$row['image'] . "'></image>";
echo "<div class='info'>";
echo $row['name'] . "<br>";
echo $row['surname'] . "<br>";
echo "SCORE: " . ($row['likes'] - $row['dislikes']);
echo "</div>";
}
?>
<div class='friend_actions'>
<image class='button' id='likes' src='images/like.jpg'></image><br>
<image class='button' id='dislikes' src='images/dislike.jpg'></image>
</div>
</body>
<?php
require "php/conn.php";
$db->query("UPDATE friends_list SET likes = likes + 1 WHERE id = 1");
?>
likes.php
<!DOCTYPE html>
<html>
<head>
<link rel="stylesheet" type="text/css" href="css/style.css"></link>
<script src="js/jquery-1.11.3.min.js"></script>
<script src="js/script.js"></script>
</head>
<body>
Welcome: <?php
// Start the session
session_start();
//Use session variable created in login.php
echo $_SESSION['username'];
?>
<div class="friend">
<?php
while($row = $res->fetch_assoc())
{
echo "<image class='img' src='images/" .$row['image'] . "'></image>";
echo "<div class='info'>";
echo $row['name'] . "<br>";
echo $row['surname'] . "<br>";
echo "SCORE: " . ($row['likes'] - $row['dislikes']);
echo "</div>";
}
?>
<div class='friend_actions'>
<image class='button' id='likes' src='images/like.jpg'></image><br>
<image class='button' id='dislikes' src='images/dislike.jpg'></image>
</div>
</body>
<?php
require "php/conn.php";
$db->query("UPDATE friends_list SET likes = likes + 1 WHERE id = 1");
?>
手部问题是因为它是一个图像,而不是一个链接(a
-tag)。您可以添加:#likes:hover{cursor:pointer;}
来解决这个问题。对于手动指针,您应该使用#likes:hover{cursor:pointer;}
当您单击图像时,在开发人员控制台中得到了什么?您是否可以确认likes.php
正在从控制台中获得调用?您的控制台是否显示任何错误?@MagnusEriksson未能加载资源:服务器以404状态响应(未找到)