Question

我将这个表单与ajax脚本连接，当我提交长文本时结果不保存在数据库中（在posts表上，我有post_content列为＆＃34; longtext＆＃34; type）。当我提交简单的东西时（比如＆＃34;你好世界＆＃34;）结果正确保存。当我以前使用mysqli时，我用这行代码解决了这个问题（使用mysqli_real_escape_string）：

//$insert_posts = "INSERT INTO posts (post_content,post_date,post_topic,post_by_uid) VALUES ('".mysqli_real_escape_string($conn,$_POST['val'])."',NOW(),$id,'".$_SESSION['uid']."')";

但是现在有了PDO，我无法做到。我试过这个：$conn->quote($reply)但它没有用。

＆＃13;

<script>
function saveEditorTrigger()
{
  for ( instance in CKEDITOR.instances ) CKEDITOR.instances[instance].updateElement();
}
</script>
 <script>
	  $(function () {
		
        $('form').on('submit', function (e) {
		saveEditorTrigger();
		var str = CKEDITOR.instances['reply-post'].getData();
         var id = <?php echo $_GET['id']; ?>;
		e.preventDefault();

          $.ajax({
            type: 'post',
            url: 'post.php',
            data:{ val : str, id : id },
            success: function (data) {
             alert('Your answer is submitted');
            location.replace("topic.php?id=<?php echo $_GET['id']; ?>");
			}
          });

        });

      });
    </script>

＆＃13;

<form id="form" >
				  <br>
				  <textarea name="reply-post"></textarea>
						<script>
						CKEDITOR.replace( "reply-post" );
            </script>
					<input type="submit" name="submit" value="submit">  
				</form>

＆＃13;

post.php文件：

＆＃13;

<?php
include 'dbh.php'; 
session_start();
$reply = $_POST['val'];
$id = $_POST['id'] ;
$insert_posts = $conn -> prepare("INSERT INTO posts (post_content,post_date,post_topic,post_by_uid) VALUES ('".$reply."',NOW(),$id,'".$_SESSION['uid']."')");
//$insert_posts = "INSERT INTO posts (post_content,post_date,post_topic,post_by_uid) VALUES ('".mysqli_real_escape_string($conn,$_POST['val'])."',NOW(),$id,'".$_SESSION['uid']."')";
$insert_posts -> execute();

?>

＆＃13;

我的问题是我做错了什么？有什么想法吗？

Answer 1

<?php
include 'dbh.php'; 
session_start();
$reply = $_POST['val'];
$id = $_POST['id'] ;
$userId=$_SESSION['uid'];
$insert_posts = $conn -> prepare("INSERT INTO posts 
                (post_content,post_date,post_topic,post_by_uid) 
                VALUES 
                (:post_content,:post_date,:post_topic,:post_by_uid)");
            $insert_posts->bindParam(':post_content', $reply, PDO::PARAM_STR);
            $insert_posts->bindParam(':post_date', NOW(), PDO::PARAM_STR);
            $insert_posts->bindParam(':post_topic', $id, PDO::PARAM_STR);
            $insert_posts->bindParam(':post_by_uid', $userId, PDO::PARAM_STR);
            $insert_posts -> execute();
?>

https://www.phpro.org/tutorials/Introduction-to-PHP-PDO.html可能是一个很好的学习资源

Answer 2

更正此代码，我让它成功。

<?php
include 'dbh.php'; 
session_start();
$reply = $_POST['val'];
$id = $_POST['id'] ;
if (isset($_SESSION['uid'])){
$user=$_SESSION['uid'];

$insert_posts = $conn -> prepare("INSERT INTO posts (post_content,post_date,post_topic,post_by_uid) VALUES (:post_content,NOW(),:post_topic,:post_by_uid)");
            $insert_posts->bindParam(':post_content', $reply);
            $insert_posts->bindParam(':post_topic', $id);
            $insert_posts->bindParam(':post_by_uid', $user);
            $insert_posts->execute();
}
?>

PDO插入不起作用

2 个答案: