Question

我正在尝试显示已登录用户关注的艺术家的状态更新列表。

到目前为止，我有这个：

#Get the list of artists that the user has liked
$q = "SELECT * FROM artist_likes WHERE user_id = '1' ";
$r = mysqli_query($dbc,$q);
while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) {

  #Now grab the statuses for each artist
  $status_query = "SELECT * FROM status_updates WHERE artist_id = '".$row['artist_id']."' ";
  $status_result = mysqli_query($dbc,$status_query)

}

但我不知道如何循环并显示返回的状态更新？

这不是我的强项，所以任何指针都会非常感激！

Answer 1

是什么阻止了您执行类似于第一次查询的操作？如下所示：

#Get the list of artists that the user has liked
$q = "SELECT * FROM artist_likes WHERE user_id = '1' ";
$r = mysqli_query($dbc,$q);
while ($row = mysqli_fetch_array($r, MYSQLI_ASSOC)) {

  #Now grab the statuses for each artist
  $status_query = "SELECT * FROM status_updates WHERE artist_id = '".$row['artist_id']."' ";
  $status_result = mysqli_query($dbc,$status_query)

  while($status_result_row = mysqli_fetch_assoc($status_result)) {
    echo $status_result_row['mycol']; // This is where you know better than us
  }
}

或者如果这两个表artist_likes和status_updates共有artist_id，那么您可以使用一个带有连接的查询。（但不知道你是否要求这样做。）

Answer 2

为避免多次查询，您可以使用以下一个查询：

SELECT l.*, s.* 
from artist_likes l, status_updates s
WHERE
l.artist_id = s.artist_id and
l.user_id = '1'

或

SELECT l.*, s.* 
from artist_likes l
JOIN status_updates s on (l.artist_id = s.artist_id)
WHERE
l.user_id = '1'

循环通过mysqli结果

2 个答案: