Question

这是我想要实现的一个非常简单的例子：

USERS
 id      |    name          
 12      |    Phil      

ACTIONS
 sender  |  recipient
 12      |  Alice

$table = query("SELECT id, sender FROM users WHERE recipient = Alice");
foreach ($table as $row) {
$sender = $row['sender'];

echo '$sender has sent a gift to Alice';
}

这将输出：

12 sent a gift to Alice

如何获得以下输出？

**Phil** sent a gift to Alice

我应该加入这两张桌子吗？

Answer 1

您需要加入：

SELECT u.name, sender
FROM actions a join
     users u
     on a.sender = u.id
WHERE recipient = 'Alice';

如何从不同的表中回显值？

1 个答案: