Php mysqli喜欢不重新加载页面
我有一些带有代码的post和like按钮：Php mysqli喜欢不重新加载页面,php,jquery,ajax,Php,Jquery,Ajax,我有一些带有代码的post和like按钮： <a href="likes.php?userid='$userid'&postid='$postid'">Like</a> 现在在likes.php中，我对userid和postid进行了一些Get编码，以找到哪个用户喜欢哪个帖子，并将其存储到数据库中。它工作良好，但它可以重新加载页面。它转到likes.php，如果成功，它将返回主页或我想要的任何其他页面。现在我的问题是，如果我应该在页面中包含likes.php
<a href="likes.php?userid='$userid'&postid='$postid'">Like</a>
现在在likes.php中，我对userid和postid进行了一些Get编码，以找到哪个用户喜欢哪个帖子，并将其存储到数据库中。它工作良好，但它可以重新加载页面。它转到likes.php，如果成功，它将返回主页或我想要的任何其他页面。现在我的问题是，如果我应该在页面中包含likes.php代码，其中有posts和like按钮，我如何在不重新加载页面的情况下完成这项工作。如果可能的话，使用
ahref<html>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.0/jquery.min.js"> </script>
<?php
session_start();
$userid=$_SESSION['userid'];
?>
<div id="like">
<a href="#" id="postid" onclick="likeclick(this);">Like</a>
<!-- here id should be different for every post .I would prefer using post id to this ahref id because i would use that to detect what post it is actually. -->
</div>
</html>
<script type="text/javascript">
function likeclick(element)
{
var postid=element.id;
var userid=<?php echo json_encode($userid);?>;
//You can use different methods to pass variable to javascript.I used this one because it is easy to implement
//it has some cons to it .Do check for that on google also.
//http://stackoverflow.com/questions/23740548/how-to-pass-variables-and-data-from-php-to-javascript
//Check this link for more.
$.ajax({
type:'POST',
url:'getlike.php',
data:{"userid":userid,"postid":postid},
success : function(content)
{
$('#like').html(content);
//This gets the html content from the getlike page and displays in the div on this page.
//Note:I have used .html which replaces any previous content inside the 'like' div.
})
}
</script>
<?php
require 'connect.inc.php'; //This make a connection to the database
$userid=$_POST['userid'];
$postid=$_POST['postid'];
$statement=$mysqli->prepare("select `likes` from `posts` where `postid`=?");
$statement->bind_param("s",$postid);
$statement->execute();
$result=$statement->get_result();
while($row=$result->fetch_assoc())
{
$likes_on_this_post=$row['likes'];
}
$likes_on_this_post=$likes_on_this_post+1; //Added one like more.
$statement=$mysqli->prepare("update `posts` set `likes`=?");
$statement->bind_param("s",$likes_on_this_post);
$statement->execute();
echo "+".$likes_on_this_post;
//This echo is actually the main thing.When this page runs through ajax the response is given back to the calling object
//What goes with response are the html contents on this page as well as whatever i echo on this page.
//Considering this example i have only echoed the no of likes and it doesn't contain any sort of html content so only the echoed element goes.
?>
函数likeclick（元素）
{
var postid=element.id；
var userid=；
//您可以使用不同的方法将变量传递给javascript
//它有一些缺点。请在谷歌上检查一下。
//http://stackoverflow.com/questions/23740548/how-to-pass-variables-and-data-from-php-to-javascript
//查看此链接了解更多信息。
$.ajax({
类型：'POST'，
url:'getlike.php'，
数据：{“userid”：userid，“postid”：postid}，
成功：功能（内容）
{
$（'#like'）.html（内容）；
//这将从getlike页面获取html内容，并显示在此页面的div中。
//注意：我使用了.html来替换“like”div中以前的任何内容。
})
}
<html>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/1.12.0/jquery.min.js"> </script>
<?php
session_start();
$userid=$_SESSION['userid'];
?>
<div id="like">
<a href="#" id="postid" onclick="likeclick(this);">Like</a>
<!-- here id should be different for every post .I would prefer using post id to this ahref id because i would use that to detect what post it is actually. -->
</div>
</html>
<script type="text/javascript">
function likeclick(element)
{
var postid=element.id;
var userid=<?php echo json_encode($userid);?>;
//You can use different methods to pass variable to javascript.I used this one because it is easy to implement
//it has some cons to it .Do check for that on google also.
//http://stackoverflow.com/questions/23740548/how-to-pass-variables-and-data-from-php-to-javascript
//Check this link for more.
$.ajax({
type:'POST',
url:'getlike.php',
data:{"userid":userid,"postid":postid},
success : function(content)
{
$('#like').html(content);
//This gets the html content from the getlike page and displays in the div on this page.
//Note:I have used .html which replaces any previous content inside the 'like' div.
})
}
</script>
<?php
require 'connect.inc.php'; //This make a connection to the database
$userid=$_POST['userid'];
$postid=$_POST['postid'];
$statement=$mysqli->prepare("select `likes` from `posts` where `postid`=?");
$statement->bind_param("s",$postid);
$statement->execute();
$result=$statement->get_result();
while($row=$result->fetch_assoc())
{
$likes_on_this_post=$row['likes'];
}
$likes_on_this_post=$likes_on_this_post+1; //Added one like more.
$statement=$mysqli->prepare("update `posts` set `likes`=?");
$statement->bind_param("s",$likes_on_this_post);
$statement->execute();
echo "+".$likes_on_this_post;
//This echo is actually the main thing.When this page runs through ajax the response is given back to the calling object
//What goes with response are the html contents on this page as well as whatever i echo on this page.
//Considering this example i have only echoed the no of likes and it doesn't contain any sort of html content so only the echoed element goes.
?>
听说过ajax吗？然后阅读文档。我相信你会找到有用的演示和代码。很抱歉，我不明白你在说什么……你有没有为它编写过ajax？@Noob查看这篇文章：没有，我没有。只有php和mysqliCan我可以使用ajax get var with ref？这可能是个愚蠢的问题，但当我在iframe内使用iframe with it load page时。ref可以像页面上的任何地方一样使用iframe，然后使用ajax get var从ref获取用户ID吗？说真的，我甚至没有理解你刚才说的，但是你可以用PHP和ajax做任何事情。我的意思是，你的问题的真正答案是使用PHP和ajax。相信我，去看一些视频和阅读一些文档。从长远来看，这肯定会对你有所帮助。：）I从phpacademy找到了一些东西这不完全是我需要的，但它会帮助很多。无论如何，谢谢你，我可以给你整个工作代码，但你不会得到任何东西。给我ofc我不会使用它，但它更容易尝试了解如何工作，然后制作我自己的，然后粘贴在代码中…如果你可以。。