php - 有没有一种方法可以将 "load more"代码实现到我的原始代码中？

Question

好吧，我正在尝试创建一个网站，如您所见，但我在使用“加载更多”按钮时遇到了困难，我的设想是，每当有人单击此按钮时，它最好会通过 php告诉数据库再加载 5 个帖子的代码。

show_posts.php

<?php

    $query = $con->query("SELECT * FROM `posts` ORDER BY `id` DESC LIMIT 5");

    if( isset( $_POST['load'] ) ){
        //code goes here
    }

    while($result = mysqli_fetch_assoc($query)){
        // if video is empty then echo this line
        if($result["video"] == ""){
            $account_assoc = $result["account_assoc"];

            $result2 = mysqli_fetch_assoc($con->query("SELECT * FROM `Accounts` WHERE username='$account_assoc' OR email='$account_assoc'"));

            if($result2["username"] == ""){
                $identifier = $result2["firstname"];
            }else{
                $identifier = $result2["username"];
            }

            if($result2['image'] == ""){
                $image = "http://jnvbaghmara.nic.in/images/staff/Blank.png";
            }else{
                $image = 'data:image/jpeg;base64,'.base64_encode($result2['image']);
            }

            if($result['dislikes'] > 1){
                $dislike = "<label style='color: red;'>".format_num($result['dislikes'])."</label> Dislikes";
            }else{
                $dislike = "<label style='color: red;'>".format_num($result['dislikes'])."</label> Dislike";
            }

            if($result['likes'] > 1){
                $like = "<label style='color: #0096f3;'>".format_num($result['likes'])."</label> likes";
            }else{
                $like = "<label style='color: #0096f3;'>".format_num($result['likes'])."</label> like";
            }

            echo '<li>
            <h4><img src="'.$image.'"/> <label>'.$identifier.'</label></h4>
            <div class="all-content">
            <label style="color: #777;"> '.$like.' </label><label style="color: #777;"> | '.$dislike.'</label><label style="color: #777;"> | 2 comment</label>
            <p>'.$result["text"].'</p>
            </div></li>';
        }
        // if video is not empty then echo this line
        elseif($result["video"] != ""){
            $account_assoc = $result["account_assoc"];

            $result2 = mysqli_fetch_assoc($con->query("SELECT * FROM `Accounts` WHERE username='$account_assoc' OR email='$account_assoc'"));

            if($result2["username"] == ""){
                $identifier = $result2["firstname"];
            }else{
                $identifier = $result2["username"];
            }

            if($result2['image'] == ""){
                $image = "http://jnvbaghmara.nic.in/images/staff/Blank.png";
            }else{
                $image = 'data:image/jpeg;base64,'.base64_encode($result2['image']);
            }

            if($result['dislikes'] > 1){
                $dislike = "<label style='color: red;'>".format_num($result['dislikes'])."</label> Dislikes";
            }else{
                $dislike = "<label style='color: red;'>".format_num($result['dislikes'])."</label> Dislike";
            }

            if($result['likes'] > 1){
                $like = "<label style='color: #0096f3;'>".format_num($result['likes'])."</label> likes";
            }else{
                $like = "<label style='color: #0096f3;'>".format_num($result['likes'])."</label> like";
            }

            echo '<li>
            <h4><img src="'.$image.'"/> <label>'.$identifier.'</label></h4>
            <div class="all-content">
            <iframe width="90%" height="90%" src="https://www.youtube.com/embed/'.$result["video"].'" frameborder="0" allowfullscreen></iframe>
            <p>'.$result["text"].'</p>
            <label style="color: #777;"> '.$like.' </label><label style="color: #777;"> | '.$dislike.'</label><label style="color: #777;"> | 2 comment</label>
            </div>
            </li>';
        }
    }

    echo '<li>
    <div class="all-content">
    <form method="post" action="#">
    <input type="submit" name="load" value="LOAD MORE..." class="load_more_button"/>
    </form>
    </div>
    </li>';
?>

配置文件.php

<?php include('include/header.php'); ?>
<?php
    $error = "";

    if(isset($_POST['post'])){          
        $video = substr($_POST["video"], 17);
        $text = $_POST['paragraph'];
        $account_assoc = "";

        if($_SESSION['username'] != ""){
            $account_assoc = $_SESSION['username'];
        }elseif($_SESSION['email'] != ""){
            $account_assoc = $_SESSION['email'];
        }

        $con->query("INSERT INTO `posts` (`id`, `image`, `video`, `text`, `account_assoc`, `likes`, `dislikes`) VALUES ('', '$image', '$video', '$text', '$account_assoc', '0', '0')");

    }

?>
<div class="content">
    <div class="actual_content">
        <div class="right-content">
            <div style="padding: 10px;">
                <div id="post">
                    <h3>Latest News</h3>
                    <ul class="inner-post">
                        <?php

                        $sql = $con->query("SELECT * FROM `news` ORDER BY `id` DESC LIMIT 5");

                        function myTruncate($string, $limit, $break=" ", $pad="..."){
                        if(strlen($string) <= $limit){
                        return $string;
                        }

                        $string = substr($string, 0, $limit);
                        if(false !== ($breakpoint = strrpos($string, $break))) {
                        $string = substr($string, 0, $breakpoint);
                        }

                        return $string . $pad;
                        }

                        while($result = mysqli_fetch_array($sql)){
                        $header = $result['header'];
                        $statement = $result['statement'];
                        $image = $result['image'];
                        $shortdesc = myTruncate($statement, 300);

                        echo '<li><h5><img src="#" /><label>'.$header.'</label></h5><p>'.$shortdesc.'</p></li>';
                        }
                        ?>
                    </ul>
                </div>
            </div>
        </div>
        <div class="left-content">
            <div class="post-holder" width="70%" style="padding: 10px;">
                <form class="post" method="post" action="#">
                    <textarea name="paragraph" placeholder="Hey there! Share a game highlight with a photo or video... <?php echo $error; ?>" maxlength="250"></textarea>
                    <input type="submit" name="post" class="button_post" value="POST" />
                    <label class="video"><input type="text" name="video" placeholder="Use a YouTube link in order to post video..."/></label><br />
                </form>
            </div>
            <div class="new-posts" style="padding: 10px;">
                <ul>
                    <?php include('show_posts.php')?>
                </ul>
            </div>
        </div>
    </div>
</div>
<div class="footer">

</div>
</div>
</body>
</html>

此外，如果有任何方法可以使我的代码更简单，那么我愿意接受建议。

*我希望代码是 php，因为我比 js、jquery 等更了解 php。

Answer 1

$postsNum = $_GET['num'];

$query = $con->query("SELECT * FROM `posts` ORDER BY `id` DESC LIMIT $postsNum");

带有链接的“加载更多”按钮:YOURURL?num=n

n 可以通过你的 php 代码改变。