PHP var_dump 返回资源(16)

Question

我有以下代码:

function fetch_conversation_messages($conversation_id){
    $conversation_id = (int)$conversation_id;

    $sql = "SELECT
                            `conversations_messages`.`message_date`,
                            `conversations_messages`.`message_date` > `conversations_members`.`conversation_last_view` AS `message_unread`,
                            `conversations_messages`.`message_text`,
                            `users`.`user_name`
                    FROM `conversations_messages`
                    INNER JOIN `users` ON `conversations_messages`.`user_id` = `users`.`user_id`
                    INNER JOIN `conversations_members` ON `conversations_messages`.`conversation_id` = `conversations_members`.`conversation_id`
                    WHERE `conversations_messages`.`conversation_id` = {$conversation_id}
                    AND `conversations_members`.`user_id` = {$_SESSION['user_id']}
                    ORDER BY `conversations_messages`.`message_date` DESC";

    $result = mysql_query($sql);
    var_dump($result);
    $messages = array();

    while (($row = mysql_fetch_assoc($result)) !== false){
            $messages[] = array(
                    'date'          => $row['message_date'],
                    'unread'        => $row['message_unread'],
                    'text'          => $row['message_text'],
                    'user_name'     => $row['user_name'],
            );
    }
    var_dump( $messages );

}

它应该返回类似这样的内容:

Array ( [0] => Array ( [date] => 1322254667 [text] => one [user_name] => bob ) )

但是，它返回

resource(16) of type (mysql result) array(0) { }

我对 PHP 和 MySQL 完全陌生，但我尝试过检查拼写错误、回显 mysql_error 以及在错误时终止脚本，这表明 SQL 中没有错误。

我不知道出了什么问题以及如何修复它。

请帮助我。提前致谢。

Answer 1

var_dump($result); 显示 resource(16) of type (mysql result) 表示您的查询正常。

var_dump($messages); 显示空数组意味着查询结果为空。