Question

$users_id = $this->db->query('SELECT users_id FROM users ORDER BY daily_points desc limit 1');
$text = 'Congratulations! You are the winner for today!';
$this->game->send_message( '0', $users_id, $text, 'yes' );

我想问一下$users_id我的错误是什么？如果我通过MySQL执行查询，我会得到所请求的信息但是如果我运行脚本php，它不会返回任何值吗？那是为什么？

Answer 1

在查询之后，您需要先获取结果：

$result = $this->db->query('
    SELECT users_id FROM users 
    ORDER BY daily_points desc limit 1')
->result_array(); // you need to fetch it either result()/result_array()

$text = 'Congratulations! You are the winner for today!';
$this->game->send_message( '0', $result[0]['users_id'], $text, 'yes' );

Answer 2

您的代码不清楚，确定发生的事情有很多可能性。尝试对你的$ users_id变量进行重新整理，看看它是什么。

$users_id = $this->db->query('SELECT users_id FROM users ORDER BY daily_points desc limit 1');
var_dump($users_id);
$text = 'Congratulations! You are the winner for today!';
$this->game->send_message( '0', $users_id, $text, 'yes' );

如果那是一个资源，那么你需要获取结果，看看@Ghost说的那样。

Answer 3

query（） - 方法返回一个对象。获得所需值有多种可能性，例如：通过这个：

$users_id = $this->db->query('
    SELECT users_id FROM users ORDER BY daily_points desc limit 1
')->row()->users_id;

有关参考，请参阅CodeIgniter Guide

php中的MySQL查询问题

3 个答案: