Php 从5个表中获取数据的MySQL查询？

什么是正确的MySQL查询？以下是我想要实现的目标

我想显示所有状态更新条目、发布日期和 海报用户名、性别、城市和标签

我想显示用户为_id='1'的成员的所有状态更新条目，状态更新发布的日期，海报用户名、性别、城市以及该成员在其个人资料上的标签

我如何在一行中以“，”分隔显示用户配置文件上的所有标记？我之前试过group_concat，但没能成功。请帮忙

这是我的桌子

STATUS_UPDATE Table
+------------------------------------------------------+
| status_id | user_id |      body       |  postdate    |
+-----------+---------+-----------------+--------------+
| 1         |   1     | hello world     | Aug 12, 2012 |
+-----------+---------+-----------------+--------------+
| 2         |   1     | i miss you      | Aug 13, 2012 |
+-----------+---------+-----------------+--------------+
| 3         |   2     | lorem ipsum     | Aug 14, 2012 |
+-----------+---------+-----------------+--------------+
| 4         |   2     | why me? why?    | Aug 14, 2012 |
+-----------+---------+-----------------+--------------+

MEMBERS Table    //Primary data of members
+---------------------------------------------------+
| user_id | username |  password  |   email         |
+---------+----------+------------+-----------------+
|    1    | john_doe |  qwerty    | john@doe.com    |
+---------+----------+------------+-----------------+
|    2    | maryjane |  pass123   | mary@jane.com   |
+---------+----------+------------+-----------------+

MEMBERS_DATA Table   //Profile Fields
+----------------------------------------------+
| user_id | gender |  description  |   city    |
+---------+--------+---------------+-----------+
|    1    | male   |  i am simple  | chicago   |
+---------+--------+---------------+-----------+
|    2    | female |  i am flirty  | newyork   |
+---------+--------+---------------+-----------+

MEMBERS_TAGS Table   //tags that members added to their profiles
+-------------------+
| user_id | item_id |
+---------+---------+
|  1      |  555    |
+---------+---------+
|  1      |  666    |
+---------+---------+
|  1      |  7777   |
+---------+---------+
|  2      |  8888   |
+---------+---------+


TAGS Table    //the info of the tags
+------------------------------+
| item_id | cat_id |  name     |
+---------+--------+-----------+
| 555     |   5    |  sexy     |
+---------+--------+-----------+
| 666     |   5    |  beauty   |
+---------+--------+-----------+
| 7777    |   6    |  music    |
+---------+--------+-----------+
| 8888    |   6    |  movies   |
+---------+--------+-----------+

CATEGORY Table   //the category of the tags
+-------------------------+
| cat_id  | category_name |
+---------+---------------+
| 5       |  appearance   |
+---------+---------------+
| 6       |  hobbies      |
+---------+---------------+

---

顺便说一句，我用PHP编写代码，我想提前感谢您的帮助。

选择b.*，a.status\u id，a.body，a.postdate，b.username，b.email，c.gender，c.city，c.description，e.name FROM status_更新b.user_id=a.user_id LEFT-OUTER-JOIN members_数据c ON c.user_id=b.user_id LEFT-OUTER-JOIN members_标签d ON d.user_id=b.user_id LEFT-OUTER-JOIN标签e ON e.item_id=d.item_id.“（$user_id？），其中a.user_id=$user_id=$user_id:“：”）ORDER BY a.$orderby$orderdir LIMIT$LIMIT

为什么不使用代替on？两者都可以使用，但这是我的做法

SELECT 
      STATUS_UPDATE.* ,
      MEMBERS.username ,
      MEMBERS_DATA.gender ,
      MEMBERS_DATA.city
      TAGS.*
FROM  UPDATE
LEFT JOIN MEMBERS ON MEMBERS.user_id = STATUS_UPDATE.user_id
LEFT JOIN MEMBERS_DATA ON MEMBERS_DATA.user_id = STATUS_UPDATE.user_id
LEFT JOIN MEMBERS_TAGS ON MEMBERS_TAGS.user_id = STATUS_UPDATE.user_id
LEFT JOIN TAGS ON TAGS.item_id = MEMBERS_TAGS.item_id