Question

我有这样的表

用户

id |  name 
1  |  john 
2  |  alicia 
3  |  alex

follow_table

id |  follow  | follower
1  |  john    | alex
2  |  alicia  | alex

帖子

id |  title         | content | name
1  |  title john    | ....... | john  
2  |  title alicia  | ....... | alicia 
3  |  title alex    | ....... | alex

示例;

$id = 'alex';

$select_one = select * from follow_table where follower = '$id';
    $query_one = mysqli_query($mysqli, $select_one) or die (mysqli_error($mysqli));
                 while ($follow = mysqli_fetch_array($query_one)) { 
                 $follow = $follow['follow']; 

$select_two = select * from posts where name = '$follow';
    $query_two = mysqli_query($mysqli, $select_two) or die (mysqli_error($mysqli));
                 while ($posts = mysqli_fetch_array($query_two)) {  

                 echo $posts['title'];

                  }
                 }

结果将是这样的：

id  | title
1   | title john    
2   | title alicia

上面的实际sql是正确的，但我只是想在$ follow中添加一个查询。

我的问题是，如何将 user.id 3（alex）纳入字段 $ follow ？我想要这样的东西

id  | title
1   | title john    
2   | title alicia  
3   | title alex

感谢您已经愿意提供的帮助

Answer 1

将$select_one替换为：

SELECT name
FROM user
JOIN follow_table ON (
       follow_table.follower = user.name 
    OR follow_table.follow = user.name
)
WHERE follower = '$id';

这将检索每个用户，然后是alex，和 alex本身。

Answer 2

您可以使用CONCAT()或CONCAT_WS()作为派生字段来连接两列的结果。

SELECT
  CONCAT_WS(' ', title, name) as full_name
FROM
 //...

Answer 3

在这种情况下你必须使用别名，你必须实现你的选择查询，我已经尝试过这个并且效果很好，

<?php 
$select_one = select * from follow_table where id = '$id';
    $query_one = mysqli_query($mysqli, $select_one) or die (mysqli_error($mysqli));
                 while ($follow = mysqli_fetch_array($query_one)) { 
                 $follow = $follow['follow']; 

$select_two = select posts.name as posted_name,posts.* from posts where name = '$follow';
    $query_two = mysqli_query($mysqli, $select_two) or die (mysqli_error($mysqli));
                 while ($posts = mysqli_fetch_array($query_two)) {  

                 echo $posts['title'];
                 echo $posts['posted_name'];
                 }
                 }

在php的同时加入两列同一列

3 个答案: