PDO联合选择结果无

Question

我需要选择name ， surname从users那里username =列following内network_follow表

查询：

"SELECT * FROM network_follow WHERE following = :f UNION SELECT name, surname FROM users where username = (following)";

php代码：

 $f = $conexao_pdo->prepare('SELECT * FROM network_follow WHERE following = :f UNION SELECT name, surname FROM users where username = (following)');
    $f->bindParam(':f', $db_user);
    $f->execute();

    while($values = $f->fetch(PDO::FETCH_ASSOC)){
        $fvalue = $values['follower'];
        $fname = $values['name'];
        echo '<center> 
    <div><a href=""> <img class="rounded-circle"  width="45" src="'.get_user_picture($fvalue).'"></img>&nbsp;@'.$fvalue.'</a>
     <span style="padding-left: 15px;">'.$fname.'<span>
     <div></center>';

}

对我来说$fvalue就好了，但是$fname不好

由于某些原因， 工会选择不起作用，有人可以帮忙吗？

Answer 1

UNION用于从多个表中选择相似的数据。

JOIN用于合并来自多个表的相关数据。

例子：

SELECT `city`, `sights` FROM `places_i_visited`
UNION
SELECT `city`, `sights` FROM `places_i_want_to_visit`

+----------------+--------+
| city           | sights |
+================+========+
| Los Angeles    | 1537   |  -- from `places_i_visited`
| Rio de Janeiro | 829    |  -- from `places_i_visited`
| Moscow         | 1822   |  -- from `places_i_want_to_visit`
+----------------+--------+

SELECT `city`, `sights`, `visits`.`date_visited`, `visits`.`duration`, `visits`.`sights_seen` FROM `places_i_visited`
INNER JOIN `visits` ON `visits`.`city_id` = `places_i_visited`.`city_id`

+----------------+--------+--------------+----------+-------------+
| city           | sights | date_visited | duration | sights_seen |
+================+========+==============+==========+=============+
| Los Angeles    | 1537   | 2017-06-25   | 14       | 25          |
| Rio de Janeiro | 829    | 2018-11-04   | 7        | 12          |
+----------------+--------+--------------+----------+-------------+

在您的情况下，您想从network_following表中获取信息， 并将其与users表中的用户名结合起来 。 JOIN在这里更合适：

SELECT network_follow.*, users.name, users.surname FROM network_follow
INNER JOIN users ON users.username = network_follow.following
WHERE following = :f

我在这里使用INNER JOIN来确保您只在两个表中看到具有结果的条目。

如果要查看数据库中的数据是否不一致，可以改用LEFT JOIN 。 这样，您将看到network_follow表中的所有条目。 如果在users表中没有这样的users ，您仍然会看到这些条目，但name和surname列将为NULL 。

Answer 2

您可以使用以下查询代替使用UNION语句：

SELECT t1.*
    ,t2.name
    ,t2.surname
FROM network_follow AS t1
INNER JOIN users AS t2 ON t1.following = t2.username
WHERE t2.following = ':f'

尽管您需要考虑任何ID的列，但可以使用主键和外键对表进行规范化。