Question

我有3张桌子：

用户

id
username

图像

id
user_id
image

user_follow

id
user_id
follow_id

我尝试进行查询，以显示我朋友的所有朋友和朋友。我需要通过他们的id（来自表“users”）获取他们的用户名

我的查询是：

$sql = "SELECT f.user_id, username, i.image, i.date FROM users u
      INNER JOIN user_follow f on (u.id = f.follow_id)
      LEFT OUTER JOIN images i on (f.follow_id = i.user_id)
      WHERE
      f.user_id = 3 OR f.user_id IN
      (
      select f2.follow_id
      from user_follow f2
      where f2.user_id = 3
      )
    ";

并且这个：

[0] => Array
    (
        [user_id] => 6
        [username] => 7777
        [image] => 
        [date] => 
    )

[1] => Array
    (
        [user_id] => 6
        [username] => 8888
        [image] => 86o99fyyocscfvbzlco0_380.jpg
        [date] => 2012-01-23 15:03:26
    )

[2] => Array
    (
        [user_id] => 3
        [username] => 6666
        [image] => my.jpg
        [date] => 2012-01-23 16:40:28
    )

[user_id] =＆gt; 6 =“用户”表中的名称是6666 [user_id] =＆gt; 7 =“用户”表中的名称是7777

此查询显示除user_id名称之外的所有内容。我需要对查询进行更改，它还会显示user_id（来自“user_follow”表） - ＆gt;用户名（在“用户表”中）

Answer 1

使用join添加users user_id，并在匹配的行中显示username？

<强>更新

SELECT f.user_id, fn.username, i.image, i.date
FROM
  users u
  INNER JOIN user_follow f on (u.id = f.follow_id)
  LEFT OUTER JOIN images i on (f.follow_id = i.user_id)
  JOIN users AS fn ON (f.user_id=fn.id)
WHERE
  f.user_id = 3 OR f.user_id IN
  (
  select f2.follow_id
  from user_follow f2
  where f2.user_id = 3
  )

按用户标识SQL查询返回用户名

1 个答案: