php 如何 link 评论 sql 博客 sql

Question

第一次来，希望大家多多帮助。

我创建了一个博客系统，供我的员工通过管理员登录系统使用。还有我的 customers/viewers 使用的登录系统和注册系统。所以现在我为博客系统创建了一个评论部分，人们可以在其中发表评论。问题是每个博客上的评论都是一样的。如何才能link评论到每个blogID？

    <?php
     ob_start();
     require('config.php');

     session_start();
     $id = $_SESSION['user'];

     // if session is not set this will redirect to login page
    if( !isset($_SESSION['user']) ) {
    $buttonlog = 'Login';   
    $smglog = 'Log nu snel in!';
    }if(isset($_SESSION['user'])) {

        $smglog = ' Logout';
        $buttonlog = ' Mijn account';

    }
     // select loggedin users detail
     $sql = mysql_query("SELECT * FROM `userlogin` WHERE `id` = '$id'");
     $userRow = mysql_fetch_array($sql);

     $sql1 = mysql_query("SELECT * FROM `comments`");
     $userRow1 = mysql_fetch_array($sql1);
    ?>


    <?php require('includes/config.php'); 

    $stmt = $db->prepare('SELECT postID, postTitle, postCont, postDate, imgBackground, imgNewspost FROM blog_posts WHERE postID = :postID');
    $stmt->execute(array(':postID' => $_GET['id']));
    $row = $stmt->fetch();

    //if post does not exists redirect user.
    if($row['postID'] == ''){
        header('Location: ./');
        exit;
    }

    ?>
<?php


function setComments(){
    if(isset($_POST['commentSubmit'])){
        $id = $_POST['id'];
        $username = $_POST['username'];
        $date = $_POST['date'];
        $message = $_POST['message'];

        mysql_query("INSERT INTO `comments` (id, username, date, message) VALUES ('$id','$username', '$date', '$message')");

        }

}
?>
<?php
function deleteComments(){
    if(isset($_POST['DeleteComment'])){
        $id = $_POST['id'];

        mysql_query("DELETE FROM `comments` WHERE `comments`.`id` = '$id'");

    }

} 
?>      
<?php
if (isset($_SESSION['user'])){

echo            "<form method='POST' action='".setComments()."'>
                <input type='hidden' name='id' value=''>
                <input type='hidden' name='username' value='".$_SESSION['user']."'>
                <input type='hidden' name='date' value='".date('Y-m-d H:i:s')."'>
                <textarea name='message'> </textarea>
                <br>
                <input style='' type='submit' name='commentSubmit' value='REAGEER'>
                </form>";
}else{
    echo " Log in om te reageren";
}
?>

</div>


    <div class="reacties"><br></div>

<?php   

 $sql = mysql_query("SELECT * FROM `comments`");
 while ($row = mysql_fetch_assoc($sql)){
        $id= $row['username'];
        $sql2 = mysql_query("SELECT * FROM `userlogin` WHERE `id` = '$id'");
        if ($row2 = mysql_fetch_assoc($sql2)){
                echo "<div class='commentsDiv'>";
                echo "<img style='margin: 5px; left: -25px; width: 100px; height: 100px; border: 1px black solid;position: relative; display: inline;'src='".$row2['image']."'>";
                echo '<p style="color: #b73a09;font-size: 16px; font-weight: bold; top: 120px; left: 0px; display: inline; position: absolute;"> '.$row2['uname'].'</p>';
                echo '<p style="color: #44444; font-size: 10px; left: 5px; top: -10px;  display: inline; position: absolute;"> '.$row['date'].'</p>';
                echo '<p style="font-family: Poppins, sans-serif; color: #44444; text-align: justify;  padding-left: 150px; top: -125px; position: relative;"> '.nl2br($row['message']);
            echo "";
            if(isset($_SESSION['user'])){
                if($_SESSION['user'] == $row2['id']){

                        echo "<form class='commentDelete' method='POST' action='".deleteComments()."'> 
                        <input type='hidden' name='id' value='".$row['id']."'>
                        <button name='DeleteComment'>Delete</button>
                        </form>
                        <form class='commentEdit' method='POST' action='editComments.php'> 
                            <input type='hidden' name='id' value='".$row['id']."'>
                            <input type='hidden' name='username' value='".$row['username']."'>
                            <input type='hidden' name='date' value='".$row['date']."'>
                            <input type='hidden' name='message' value='".$row['message']."'>
                        <button>Edit</button>
                        </form>";
            }else { 
                    echo "<form class='commentEdit' method='POST' action='replycomment.php'> 
                        <input type='hidden' name='id' value='".$row['id']."'>
                        <button type=''>Reply</button>
                        </form>";

                    }
            } else{
                echo "<p class='commentMessage'> Je moet inglogd zijn om te reageren! </p>";
            }
                        echo "</div>";
        }
 }



?>

Answer 1

试试这个查询：

<?php   
 $id = $_SESSION['user'];
 $sql2 = mysql_query("SELECT * FROM `userlogin` WHERE `id` = '$id'");
?>

您正在获取所有评论并根据您显示评论的 ID，因此根据会话用户 ID 获取评论。

Answer 2

当您在 database.Send 博客的评论 table 中输入用户评论时，-post-id（每个博客 post 唯一）以及表格的其余部分。你可以使用像这样的隐藏输入来完成

<input type="hidden" name="blogpost_id" id="blog_entry" value="<?php echo $postId; ?>"/>

并且当您在博客 post 的页面 select 上加载来自评论 table 的评论时，并使用该博客 -post-id 对它们进行排序。像这样

$query = "SELECT * FROM blog_comments WHERE blogpost_id='{$postId}';
$result = mysqli_query($connection,$query);

php 如何 link 评论 sql 博客 sql

php How to link comment sql to the blog sql

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