Question

我有一些帖子和喜欢的代码按钮：

<a href="likes.php?userid='$userid'&postid='$postid'">Like</a>

现在在likes.php中，我有一些针对userid和postid的编码，以找到哪个用户喜欢哪个帖子并将其存储到数据库中。它运行良好但重新加载页面。它转到likes.php，如果它成功，它会回到主页或我想要的任何其他内容。现在我的问题是如何在不重新加载页面的情况下执行此操作，如果我将likes.php代码包含到页面中的帖子和类似按钮。如果有可能，请使用<a ref=""></a>？或者如果有人有更好的解释，他也可以发布。

Answer 1

是的，因为您使用了代码ahref标签，这就是它应该如何工作。如果您不想重新加载尝试使用AJAX。这不是完整的答案，但您需要通过只有ajax。你会发现很多关于它的视频和演示。

检查此示例： http://www.w3schools.com/ajax/ajax_php.asp

<强>更新

<强> mainfile.php中

<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>

<强> getlike.php

<?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.
?>

Php mysqli喜欢没有页面重新加载

1 个答案: