无法使用 AJAX 将数据插入 mysql 数据库

debugcn 发表于 Dev

维吉特艾尔

我正在制作一个类似项目的时间表，允许对帖子发表评论。但是我无法插入数据。评论按预期显示。表单值似乎没有传递到 save_comments.php 页面。

下面是代码

配置文件

<?php

   $con=new PDO("mysql:host=localhost;dbname=posts&comments","root","");
    $con->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

?>

索引.php

<?php 
    include 'config.php';
?>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.0/jquery.min.js"></script>




<?php


$sql=$con->prepare("SELECT * FROM posts");
$sql->execute();
$results=$sql->fetchAll(PDO::FETCH_ASSOC);
echo "<hr>";
foreach($results as $post){
    echo "<h2>".$post['post_title']."</h2><br>";
    echo $post['post_content']."<br>";
    ?>
<br><b>Comments</b>
<form id=comment_form_<?php echo $post['post_id']; ?> action="">
    <input type=hidden name=pid value=<?php echo $post['post_id']; ?> >
    <p><input type=text name=cbody id="comment_body_<?php echo $post['post_id']; ?>" placeholder="Write comment" ></p>
    <p><input type=text name= cauth id="comment_author_<?php echo $post['post_id']; ?>" placeholder="Your Name" ></p>
    <p><button class="submit_comment" post_id="<?php echo $post['post_id']; ?>" >Submit</button> </p>

</form>
<p><button class="show_comments" post_id="<?php echo $post['post_id']; ?>">Show comments</button></p>
<div id="comments_<?php echo $post['post_id']; ?>">


</div>
    <?php
   echo "<hr>"; 
}
?>
<script>
    $(document).ready(function() {
       $(".show_comments").click(function() {
            var post_id=$(this).attr('post_id'); 
            $("#comments_"+post_id).load("comments.php", {
                post_id:post_id
            }); 
    });
        $(".submit_comment").on("click",function() {
            var post_id=$(this).attr('post_id');
            $.ajax({
                data:$('#comment_form_'+post_id).serialize(),
                url:'save_comments.php',
                async : true,
                success: function(){
                    $("#comments_"+post_id).load("comments.php", {
                        post_id:post_id
                     }); 
                }
            })
        });
    });

</script>

评论.php

<?php

include 'config.php';
$post_id=$_POST['post_id'];
$pid=$post_id;
$sql1=$con->prepare("SELECT * FROM comments where post_id=:pid");
$sql1->bindParam(':pid',$pid);
$sql1->execute();
$results1=$sql1->fetchAll(PDO::FETCH_ASSOC);
foreach($results1 as $comment){
echo "<br><b><cite>".$comment['comment_author']."</cite></b><br>";
echo $comment['comment_body']."<br>";
}

save_comments.php

    <?php

include 'config.php';

$comment_body=$_POST['cbody'];
$comment_author=$_POST['cauth'];
$post_id=$_POST['pid'];   
$sql2=$con->prepare("INSERT INTO comments Values(NULL,:pid,:cauth,:cbody)");
$sql2->bindParam(':pid',$post_id);
$sql2->bindParam(':cauth',$comment_author);
$sql2->bindParam(':cbody',$comment_body);
$sql2->execute();
?>

PS：我还是 AJAX 和 jquery 的初学者

用户2341406

您没有将 ajax 请求定义为 a POST，因此没有发送 POST 变量。

$.ajax({
  data:$('#comment_form_'+post_id).serialize(),
  url:'save_comments.php',
  async : true,
  method: 'POST', <---- ADD THIS
  success: function(){
      $("#comments_"+post_id).load("comments.php", {
          post_id:post_id
       }); 
  }
})

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