使用一个MySQL查询从两个表中获取数据？

我正在构建一个社交网络页面，希望在一个提要页面上获取用户的所有活动

我有一个评论表：

user_id    comment                date
1          "Hi...."               24 March 2013
2          "Hello..."             25 March 2013
1          "Another comment..."   26 March 2013

还有一张桌子：

user_id    id_of_liked_item     date
1          101                  24 March 2013 
1          145                  25 March 2013

是否可以执行一个查询，将这些内容按日期排序，放入用户所有活动的数组中？如果是这样的话，这个问题会是什么样子？我的阵列需要如下所示：

$activity = array(
    array(
        "type"=>"comment",
        "content"=>"Comment goes here"
    ),
    array(
        "type"=>"like",
        "item_id"=>345,
    )
)

Type    Content            Item_ID Date
------- ------------------ ------- -------------
comment Hi....             null    24 March 2013
comment Another comment... null    26 March 2013
like    null               101     24 March 2013
like    null               145     25 March 2013

---

我非常擅长处理php数据，因此我可能不需要太多帮助来组织数组，但我的问题主要是首先如何从数据库中获取数据。谢谢

最简单的方法可能是使用工会：

SELECT
    'comment' AS type,
    comment AS content,
FROM comments
UNION
SELECT
    'like' AS type,
    id_of_liked_item AS item_id

---

当然，您也可以在联合中为用户组织的每个查询添加

WHERE

条件和附加列。

最简单的方法可能是使用联合：

SELECT
    'comment' AS type,
    comment AS content,
FROM comments
UNION
SELECT
    'like' AS type,
    id_of_liked_item AS item_id

---

当然，您也可以在联合中为用户组织的每个查询添加

WHERE

条件和附加列。

是的，联合是一种方法：

SELECT 'comment' AS Type, Comment AS Content, NULL as Item_ID, Date
  FROM Comments
  WHERE User_Id = 1
UNION ALL SELECT 'like', NULL, id_of_liked_item, Date
  FROM Likes
  WHERE User_Id = 1

这将为您提供一个如下所示的结果集：

$activity = array(
    array(
        "type"=>"comment",
        "content"=>"Comment goes here"
    ),
    array(
        "type"=>"like",
        "item_id"=>345,
    )
)

Type    Content            Item_ID Date
------- ------------------ ------- -------------
comment Hi....             null    24 March 2013
comment Another comment... null    26 March 2013
like    null               101     24 March 2013
like    null               145     25 March 2013

---

你可以从那里得到它。

是的，工会是前进的道路：

SELECT 'comment' AS Type, Comment AS Content, NULL as Item_ID, Date
  FROM Comments
  WHERE User_Id = 1
UNION ALL SELECT 'like', NULL, id_of_liked_item, Date
  FROM Likes
  WHERE User_Id = 1

这将为您提供一个如下所示的结果集：

$activity = array(
    array(
        "type"=>"comment",
        "content"=>"Comment goes here"
    ),
    array(
        "type"=>"like",
        "item_id"=>345,
    )
)

Type    Content            Item_ID Date
------- ------------------ ------- -------------
comment Hi....             null    24 March 2013
comment Another comment... null    26 March 2013
like    null               101     24 March 2013
like    null               145     25 March 2013

您可以从那里开始使用。

尝试使用一个接头：

SELECT * FROM 
(SELECT `user_id`, `comment` AS activity, `date` FROM comments

UNION

SELECT `user_id`, `id_of_liked_item` AS activity, `date` FROM likes) 

AS user_activity

ORDER BY user_activity.`date`

尝试使用联合：

SELECT * FROM 
(SELECT `user_id`, `comment` AS activity, `date` FROM comments

UNION

SELECT `user_id`, `id_of_liked_item` AS activity, `date` FROM likes) 

AS user_activity

ORDER BY user_activity.`date`

看一看，看一看