Question

我试图将注释插入数据库，但是每当尝试提交它时我都会收到一个Object对象警报，并且它永远不会进入数据库。这些文件都在同一个文件夹中。

提前感谢任何帮助。

的index.php

 <?php require_once('include/header.php'); require_once('include/browser.php');   ?>
<div class="content">
    <section>

        <div class="article">
    <?php


        $userNameQuery = "SELECT * FROM (SELECT * FROM `tbl_posts` ORDER BY 'post_id' DESC) 
                        t ORDER BY `post_id` DESC
                        LIMIT 3";

        $result = mysqli_query($connection, $userNameQuery)
                  or die("Error in query: ". mysqli_error($connection));


        while ($row = mysqli_fetch_assoc($result)){
            $post_id = $row ['post_id'];
    ?>

    <div class="wrapper">
        <div class="titlecontainer">
        <h1><?php echo $row['post_title']; ?></h1>
        </div>
        <div class="textcontainer">
        <?php echo $row['post_content']?>
        </div>
<?php
        if (!empty($row['imagePath'])) //This will check if there is an path in the textfield
        {
?>  
            <div class="imagecontainer">
            <img src="<?php echo $row['imagePath']; ?>" alt="Article Image">
            </div>
<?php
        }
?>
        <div class="timestampcontainer">
        <b>Date posted :</b><?php echo $row['post_timestamp']; ?>
        <b>Author :</b> Admin 
        </div>
<?php
        //Selecting comments which correspond to the post
        $selectCommentQuery = "SELECT * FROM `tbl_comments` 
                                LEFT JOIN `tbl_users` 
                                ON tbl_comments.tbl_comments_users_id = tbl_users.id 
                                WHERE tbl_comments.tbl_comments_post_id ='$post_id'";

        $commentResult = mysqli_query($connection,$selectCommentQuery)
            or die ("Error in the query: ". mysqli_error($connection));

        //showing the comments

        echo '<div class="comment-block_' . $post_id .'">';

        while ($commentRow = mysqli_fetch_assoc($commentResult)) 
        {
?>
            <div class="commentcontainer">
            <div class="commentusername"><h1>Comment by: <?php echo $commentRow['username']?></h1></div>
            <div class="commentcontent"><?php echo $commentRow['comment_content']?></div>
            <div class="commenttimestamp"><?php echo $commentRow['comment_timestamp']?></div>
            </div>
<?php
        }
?>
        </div>
<?php           
if (!empty($_SESSION['cleanUsername']) ) 
        {
?>
            <form method="POST" class="post-form" action="index.php" >
            <label>New Comment</label>
            <textarea name="comment" class="comment"></textarea>
            <input type="hidden" name="postid" value="<?php echo $post_id ?>">
            <input type="submit" name ="submit" class="submitComment"/>
            </form>
<?php
        }
        echo "</div>";
        echo "<br /> <br /><br />"; 
    }



require_once('include/footer.php'); ?>

comments.js

     $(document).ready(function(){

   $(document).on('click','.submitComment',function(e) {

        e.preventDefault();
        //send ajax request
        var form = $(this).closest('form');
        var comment  = $('.comment',form);
        if (comment.val().length > 1) 
        {
            $.ajax({
                url: 'postComments.php',
                type: 'POST',
                cache: false,
                dataType: 'json',
                data: $(form).serialize(), //form serialize data so we can enter into database
                beforeSend: function(){
                    //Changing submit button value text and disabling it
                    $(this).val('Submiting ....').attr('disabled', 'disabled');
                },
                success: function(data)
                {
                    var item = $(data.html).hide().fadeIn(800);
                    $('.comment-block_' + data.id).append(item);

                    // reset form and button
                    $(form).trigger('reset');
                    $(this).val('Submit').removeAttr('disabled');
                },
                error: function(e)
                {
                    alert(e);
                }
            });
        }
        else
        {
            alert("Please fill the field!");
        }
    });
});

postComments.php

 <?php
if (isset($_SERVER['HTTP_X_REQUESTED_WITH'])):

    session_start();
    require_once('ajaxconnection.php');
    $connection2 = connectToMySQL();

        $userId = $_SESSION['userID'];
        $username = $_SESSION['cleanUsername'];
        $comment = $_POST['comment'];
        $postId = $_POST['post_id'];
        $date_format = " Y-m-d  g : i : s";
        $time = date ($date_format);


          $insertCommentQuery = "INSERT INTO `tbl_comments` 
                                 (`comment_content`,`tbl_comments_users_id`,`tbl_comments_post_id`) 
                                VALUES ('$comment', $userId, $postId)";
          $result = mysqli_query($connection,$insertCommentQuery);

$obj = array();

$obj['id'] = $postId;
$obj['html'] = '<div class="commentcontainer">
                    <div class="commentusername"><h1> Username :'.$username.'</h1></div>
                    <div class="commentcontent">'.$comment.'</div>
                    <div class="commenttimestamp">'.$time.'</div>
               </div>';
echo json_encode($obj);

    connectToMySQL(0);
   endif?>

ajaxconnection.php

<?php
    function connectToMySQL()
    {
         $connection2 = mysqli_connect("localhost","root","","ascaniobajada2ed8s")
            or die('Error connecting to the database');

            return $connection2;
    }
?>

Answer 1

JQuery .Ajax返回这样的错误（jqXHR，textStatus，errorThrown）。因此，像这样使用它来查看真正的错误而不仅仅是对象对象

error: function (jqXHR, textStatus, errorThrown) {
   alert(textStatus);
}

索引页面注释未使用AJAX插入数据库

1 个答案: