谁禁言/忽略了用户

社区 数据与报告
1

大家好！:smiley:

使用数据探索器，是否可以在表格中显示哪些用户被静音和忽略，以及是由谁操作的？

如果有人能提供帮助，我将不胜感激 :heart:
谢谢

2

以前从未见过此功能，我怀疑它不会被添加，因为它可能会危及安全的隐私实践。

3

尝试：

SELECT id AS user_id, muted_users, ignored_users
FROM (
  SELECT u.id,
    ARRAY_AGG(m.muted_user_id)
      FILTER (WHERE m.muted_user_id IS NOT NULL) AS muted_users,
    ARRAY_AGG(i.ignored_user_id)
      FILTER (WHERE i.ignored_user_id IS NOT NULL) AS ignored_users
  FROM users u
  LEFT JOIN ignored_users i ON u.id = i.user_id
  LEFT JOIN muted_users m ON u.id = m.user_id
  GROUP BY u.id
) data
WHERE (muted_users IS NOT NULL OR ignored_users IS NOT NULL)

或者，如果您想要用户名列表而不是 ID，可以尝试：

SELECT id AS user_id, muted_users, ignored_users FROM (
  SELECT u.id,
    ARRAY_AGG(muteds.username)
      FILTER (WHERE muteds.username IS NOT NULL)  AS muted_users,
    ARRAY_AGG(ignores.username)
      FILTER (WHERE ignores.username IS NOT NULL) AS ignored_users
  FROM users u
  LEFT JOIN (
    SELECT i.user_id AS id, u1.username
    FROM users u1
    INNER JOIN ignored_users i ON u1.id = i.ignored_user_id) ignores
  ON u.id = ignores.id
  LEFT JOIN (
    SELECT m.user_id AS id, u1.username
    FROM users u1
    INNER JOIN muted_users m ON u1.id = m.muted_user_id) muteds
  ON u.id = muteds.id
  GROUP BY u.id
) data
WHERE (muted_users IS NOT NULL OR ignored_users IS NOT NULL)
4

感谢 Robert 创建这个查询！用户名那个对我们来说更好用。
祝您有美好的一天！