如何从2个不同的表中获取数据？

Question

我有2个表User_Posts

CREATE TABLE IF NOT EXISTS `User_Posts` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `user_id` int(12) NOT NULL,
  `wall_id` int(12) NOT NULL,
  `text` text COLLATE utf8_bin NOT NULL,
  `img` varchar(255) COLLATE utf8_bin NOT NULL,
  `time` date NOT NULL
  PRIMARY KEY (`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=utf8 COLLATE=utf8_bin AUTO_INCREMENT=7 ;

User_Activity

CREATE TABLE IF NOT EXISTS `User_Activity` (
  `id` int(11) NOT NULL AUTO_INCREMENT,
  `content` varchar(55) COLLATE utf8_bin NOT NULL,
  `activity_id` int(12) NOT NULL,
  `activity` varchar(88) COLLATE utf8_bin NOT NULL,
  `user_id` int(12) NOT NULL,
  `time` date NOT NULL,
  `value` varchar(12) COLLATE utf8_bin NOT NULL,
  PRIMARY KEY (`id`)
) ENGINE=MyISAM  DEFAULT CHARSET=utf8 COLLATE=utf8_bin AUTO_INCREMENT=131 ;

我想获取所有活动和帖子，其中user_id =吗？

试图做到这一点

$act = $this->app->db->rawQuery("SELECT p.*, a.*
                                 FROM User_Posts p 
                                 JOIN User_Activity a
                                 ON a.user_id = ? OR p.user_id = ?", array($id,$id));

我正在获取这样的数据

Array
(
    [0] => Array
        (
            [id] => 130
            [user_id] => 16
            [wall_id] => 0
            [text] => 'Text'
            [img] => 0
            [time] => 2014-10-31
            [deleted] => 0
            [content] => movie
            [activity_id] => 5
            [activity] => favorite
            [value] => 0
        )

    [1] => Array
        (
            [id] => 130
            [user_id] => 16
            [wall_id] => 0
            [text] => 'Text'
            [img] => /posts/ceee9b387dcf72bcb19d1c1c73147ef3.jpg
            [time] => 2014-10-31
            [deleted] => 0
            [content] => movie
            [activity_id] => 5
            [activity] => favorite
            [value] => 0
        )

)

但是我需要

Array
(
    [0] => Array
        (
            [id] => 1
            [user_id] => 16
            [wall_id] => 0
            [text] => 'Text'
            [img] => 0
            [time] => 2014-10-31
        )

    [1] => Array
        (
            [id] => 2
            [user_id] => 16
            [wall_id] => 0
            [text] => 'Text'
            [img] => /posts/ceee9b387dcf72bcb19d1c1c73147ef3.jpg
            [time] => 2014-10-31
        )

      [2] => Array
        (
            [id] => 130
            [content] => movie
            [activity_id] => 5
            [activity] => favorite
            [value] => 0
        )

)

回复是带有附加活动的帖子，但是我每个人都需要不同的结果

怎么做？

Answer 1

只是一个小的修改

$act = $this->app->db->rawQuery("SELECT p.*, a.*
                             FROM User_Posts p, User_Activity a
                             WHERE a.user_id = p.user_id
                             AND p.user_id = ?", array($id));

Answer 2

您不需要JOIN关键字。

SELECT *
FROM User_Posts, User_Activity 
WHERE `User_Activity`.`User_id` = `User_Posts`.`user_id`