Question

我有一个LIKE表和一个BOOK表以及我的user_id。

我想只用BOOK NAME和AUTHOR

从BOOK表中拉出来

我的json会：

{book_id:1,bookname:sample,author:sean,user_id:111}

表

-----------BOOK TABLE------------
ID ---- BOOK AUTHOR ---- BOOK NAME

1
2

...................................


USERS TABLE
ID-------NAME
1
2
.............

LIKE TABLE---------------------------
ID-----BOOK ID-------LIKER USER ID---
1
2
.....................................

Answer 1

要获取与喜欢相关的所有数据，您可以运行此sql命令。结果集可以循环并生成JSON。确保在查询中正确地给出了所有列名。

select 
b.book_id,
b.bookname,
b.author,
u.user_id
from 
`like` l
inner join book b on b.book_id = l.book_id
inner join users u on u.user_id = l.user_id

如果您需要过滤数据，只需在最后一次加入后添加where条件

where u.user_id = {your user id}

Answer 2

$stmt = $mysqli->prepare("SELECT book.*, users.*, like.* FROM like INNER JOIN
                          users ON like.liker_user_id = users.id INNER JOIN
                          book ON like.book_id = book.id WHERE
                          users.id = ?
                         ");
$stmt->bind_param( "d", $user_id); 

$stmt->execute();

$stmt->bind_result($col1);
// then fetch and close the statement

Mysql查询，选择连接两个表

2 个答案: