我给出了tlb_comments,这个表是tlb_comments表
tlb_comments:
| ID | post_id | user_id | comments | cm_parent |
+----+-------------+---------------+-----------+-------------+
| 11 | 18 | 29 | Nice | 11 |
| 12 | 24 | 30 | Good | 12 |
tlb_avatar表是给定的,这个表是tlb_avatar表
tlb_avatar:
| id | vw_id | image |
+----+-------------+-----------------------------------------+
| 1 | 29 | 1510553363_User_Avatar_2.png |
| 2 | 30 | d968c2fe95629e238b04ee34599c5013.jpg |
tlb_viewer鉴于此表是tlb_viewer表
tlb_viewer:
| id | name |
+----+---------------+
| 1 | john |
| 2 | Quotos |
我的左连接查询输出。
这个Out put是完美的工作,但我希望将tbl_comments id添加到每一行。现在它不会调用tbl_comments id
| id | post_id| user_id | comments| cm_parent| vw_id | image |
+----+----------+----------+---------+-----------+--------+-------+
| 29 | 18 | 29 | Nice | 11 | 29 | 2.png |
| 30 | 24 | 30 | Good | 12 | 30 | 2.png |
它的查询输出很好。但是我想要tlb comment
Id(tbl_comments id)到每一行。
我解释给定的表格。这里我们想要最后一列tbl_comments id(注释ID)。
这张表是我期待的表。如何获得???????
| id | post_id| user_id | comments| cm_parent| vw_id |comment ID |
+----+----------+----------+---------+-----------+--------+------------+
| 29 | 18 | 29 | Nice | 11 | 29 | 11 |
| 30 | 24 | 30 | Good | 12 | 30 | 12 |
MYSQLI QUERY: 那个查询是我的mysqli查询:
$qid = $quotos->id;
$avatar = mysqli_query($conn,"SELECT * FROM tlb_comments LEFT JOIN tlb_avatar ON tlb_avatar.vw_id = tlb_comments.user_id LEFT JOIN tlb_viewer
ON tlb_viewer.id = tlb_comments.user_id WHERE tlb_comments.post_id = '".$qid."'" );
while($comments = mysqli_fetch_object($avatar)){}
答案 0 :(得分:0)
如果使用select *
,则连接表的列列表的输出顺序与其表的连接顺序相同。如果您希望列的顺序不同,则列出列。
select user_id as id,
post_id, user_id, comments, cm_parent,
vw_id, image,
tlb_comments.ID as 'comment ID'
from tlb_comments
left join tlb_avatar
...