PHP var_dump返回资源（16）

I have the following code:

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 );

}

It should return something like this:

它应该返回这样的东西：

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

However, it returns

然而，它返回

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

I am completely new to PHP and MySQL, but I have tried checking for typos, echoing mysql_error, and killing the script at an error, which shows that there is no error in the SQL.

我是PHP和MySQL的新手，但我已经尝试检查拼写错误，回显mysql_error，并在出错时终止脚本，这表明SQL中没有错误。

I do not know what has gone wrong and how to fix it.

我不知道出了什么问题以及如何解决它。

Please help me. Thanks in advance.

请帮帮我。提前致谢。

3 个解决方案

#1

var_dump($result); shows resource(16) of type (mysql result) means your query is ok.

后续代码var_dump（$结果）;显示类型的资源（16）（mysql结果）表示您的查询正常。

var_dump($messages); shows empty array means the result of your query is empty.

的var_dump（$消息）;显示空数组表示查询结果为空。

#2

For SELECT, SHOW, DESCRIBE, EXPLAIN and other statements returning resultset, mysql_query() returns a resource on success, or FALSE on error.

对于SELECT，SHOW，DESCRIBE，EXPLAIN和其他返回结果集的语句，mysql_query（）在成功时返回资源，如果出错则返回FALSE。

For other type of SQL statements, INSERT, UPDATE, DELETE, DROP, etc, mysql_query() returns TRUE on success or FALSE on error.

对于其他类型的SQL语句，INSERT，UPDATE，DELETE，DROP等，mysql_query（）在成功时返回TRUE，在出错时返回FALSE。

#3

mysql_query function is expected to return resource when your query is valid.

当查询有效时，mysql_query函数应返回资源。

For you to see the actual returned result of your query, use row fetching functions like mysql_fetch_assoc, mysql_fetch_object, and the like. In your case, you put the fetched values in your $row variable, so you may use var_dump on it.

为了查看查询的实际返回结果，请使用行提取函数，如mysql_fetch_assoc，mysql_fetch_object等。在您的情况下，您将获取的值放在$ row变量中，因此您可以在其上使用var_dump。

while ($row = mysql_fetch_assoc($result)){
        var_dump($row);
}

Try to check this reference.

尝试检查此参考。

#1