试图对从数据库中获取的“帖子”发表评论，但它只对一篇帖子发表评论

Question

我在我的数据库中有一个用于帖子的表，另一个用于 cmets 的表，我想做的是从数据库中获取所有帖子，并且对于每个帖子，我都提供了一个表单来提交 cmets。当我尝试使用带有 ajax 的此表单发布 cmets 时，它仅适用于一个帖子而不适用于其他帖子。（在这里我想提一下，由于每个帖子都会自动填充此表单，因此其文本区域具有相同的 ID ，可能这可能会导致错误）但我不知道如何解决这个问题？任何帮助或建议将不胜感激。

（注意：我不是专家，只是学习者。）

这是我的代码：

    <?php
echo '<div class="'.$row['post_id'].'">';            

$connect = mysqli_connect("localhost", "root", "", "testing") or die("<p><span style=\"color: red;\">Unable to select database</span></p>");

$post_id = $row['post_id'];

$resultt = mysqli_query($connect,"SELECT * FROM tbl_comment WHERE parent_comment_id = '0' AND post_id = '$post_id' ORDER BY comment_id DESC ");
while($rowss = mysqli_fetch_array($resultt, MYSQLI_BOTH))
{
if ($rowss>0){
echo '
 <div>
  <div>'.$rowss["comment"].'</div>
  <div style="font-size: 12px; font-family: serif;">By <b>'.$rowss["comment_sender_name"].'</b> on '.$rowss["date"].'</div>
  <div align="right" style="margin-bottom: 19px;"><a href="#" class="reply" id="'.$rowss["comment_id"].'">Reply</a></div>
 </div>
 ';
} else { echo ''; } }

echo '</div>
<form method="POST" class="comment_form">
<div class="form-group"><textarea name="comment_content" class="form-control comment_content" post_id ="'.$row['post_id'].'" placeholder="Enter Comment" rows="2"></textarea></div>
    <div class="form-group">
     <input type="submit" name="submit" class="btn btn-info" value="Publish" onClick="adcomnt()" />
    </div>
   </form>

 <script>

                    function adcomnt() {
                    var post_id = $(".comment_content").attr("post_id");
                    var comment_name = '.$_SESSION['username'].';   
                    var comment_content = $(".comment_content").val();
                    var comment_id = 0;

                    $.ajax({
                        type: "POST",
                        url: "add_comment.php",
                        dataType: "json",
                        data: "post_id=" +post_id+ "&comment_content=" + comment_content+ "&comment_id=" + comment_id+ "&comment_name=" + comment_name, 
                        success: function(data) {
                if(data.status == "error"){
                alert("Oops, Comment not inserted!");
                } 
                else if(data.status == "success"){
                alert("Thank you! comment inserted!");


                }},
                    });


}

</script>

            </div>';
            ?>

Answer 1

我不确定，但我认为问题在于 javascript 变量“post_id”和“comment_content”，它们看起来没有链接到 textarea。

所以你可以尝试这样的事情：

    <?php
    echo '<div class="'.$row['post_id'].'">';            

    $connect = mysqli_connect("localhost", "root", "", "testing") or die("<p><span style=\"color: red;\">Unable to select database</span></p>");

    $post_id = $row['post_id'];

    $resultt = mysqli_query($connect,"SELECT * FROM tbl_comment WHERE parent_comment_id = '0' AND post_id = '$post_id' ORDER BY comment_id DESC ");
    while($rowss = mysqli_fetch_array($resultt, MYSQLI_BOTH))
    {
    if ($rowss>0){
    echo '
     <div>
      <div>'.$rowss["comment"].'</div>
      <div style="font-size: 12px; font-family: serif;">By <b>'.$rowss["comment_sender_name"].'</b> on '.$rowss["date"].'</div>
      <div align="right" style="margin-bottom: 19px;"><a href="#" class="reply" id="'.$rowss["comment_id"].'">Reply</a></div>
     </div>
     ';
    } else { echo ''; } }

    echo '</div>

    <div class="form-group"><textarea name="comment_content" class="form-control comment_content" id="content_'.$row['post_id'].'" placeholder="Enter Comment" rows="2"></textarea></div>
        <div class="form-group">
         <input type="submit" id="submit_comment" post_id="'.$row['post_id'].'" name="submit" class="btn btn-info" value="Publish" onClick="adcomnt()" />
        </div>


     <script>
   $(document).ready(function()
{
    $( "#submit_comment" ).click(function(e) { 

    e.preventDefault(); 

    var post_id = $(this).attr("post_id");
    var comment_name = '.$_SESSION['username'].';   
    var comment_content = $("#content_"+post_id).val();
    var comment_id = 0;

    $.ajax({
        type: "POST",
        url: "add_comment.php",
        dataType: "json",
        data: "post_id=" +post_id+ "&comment_content=" + comment_content+ "&comment_id=" + comment_id+ "&comment_name=" + comment_name, 
        success: function(data) {
            if(data.status == "error"){
            alert("Oops, Comment not inserted!");
            } 
            else if(data.status == "success"){
            alert("Thank you! comment inserted!");


            }},
        });

    });

});
    </script>

                </div>';
                ?>

Answer 2

问题已得到解决，通过进行一些更改，因为它是循环内的表单，为每个元素提供了唯一的 ID：

这是更新后的代码：

    <form method="POST" class="comment_form" id="f'.$post_id.'">
        <div class="form-group"><textarea name="comment_content" class="form-control comment_content" id="content_'.$post_id.'" placeholder="Enter Comment" rows="2"></textarea></div>
            <div class="form-group">
              <input type="submit" id="submit_comment'.$post_id.'" post_id="'.$post_id.'" name="submit" class="btn btn-info" value="Publish" />
            </div></form>

<script>
       $(document).ready(function(){
        $( "#submit_comment'.$post_id.'" ).click(function(e) { 
        e.preventDefault()

现在它按预期工作了。