php - 绑定(bind)变量错误

Question

我在编写的准备好的语句中遇到错误...

 Number of elements in type definition string doesn't match number of bind variables on line 33. Create Topic Insert bind_param() failed:

我唯一能想到的是因为 NOW() 字段，但我认为不需要为此设置变量？

我是否明显遗漏了什么或者是什么导致了这个错误？

//Prepared INSERT stmt for the forum topic
    $stmt = $con->prepare("INSERT INTO forum_topics (`category_id`, `topic_title`, `topic_creator`, `topic_date`, `topic_reply_date`) 
    VALUES(?, ?, ?, NOW(), NOW())");
    if ( !$stmt || $con->error ) {
        die('Create Topic Insert prepare() failed: ' . htmlspecialchars($con->error));
    }
    if(!$stmt->bind_param('sssii', $cid, $title, $creator)) {
        die('Create Topic Insert bind_param() failed: ' . htmlspecialchars($stmt->error));
    }
    if(!$stmt->execute()) {
        die('Create Topic Insert execute() failed: ' . htmlspecialchars($stmt->error));
    }

Answer 1

mysqli_stmt::bind_param -- mysqli_stmt_bind_param — Binds variables to a prepared statement as parameters

因此在您的 bind_param 中，您需要插入 5 个参数。