如何根据where条件值创建mysql查询结果 - How create mysql query result according to where condition values?

How create mysql query result according to where condition values?

我想要这种输出

ID      Status
100     Viewed
103     Not Viewed
105     Viewed

这是我的sql:

select id, status from status_table where ID in (100, 101,102,103,104,105);

它将显示上面的结果，因为在状态表中其他id没有任何条目。ID is foreign key of another table named as table_file table. It contains in another database. So I cannot join the table due to some performance issue.所以我以逗号分隔的值传递文件ID。但我想要这样的结果，我怎么能在不使用任何循环的情况下创建它呢。

ID    Status
100   Viewed
101   Not
102   Not
103   Viewed
104   Not
105   Viewed

有可能吗？请帮帮我。

您有一个表中确实存在这些ID吗？这样你就可以加入了？

SELECT
  source.ID,
  status.value
FROM
  source
LEFT JOIN
  status
    ON status.id = source.id
WHERE
  source.ID in (100, 101,102,103,104,105);

如果没有，则需要创建一个临时表或内联表（其中包含这些值）。然后您可以将该表连接到您的数据中。

编辑

内联表的示例。有几种方法可以做到这一点，这只是其中之一。

SELECT
  source.ID,
  status.value
FROM
  (
    SELECT 100 AS id UNION ALL
    SELECT 101 AS id UNION ALL
    SELECT 102 AS id UNION ALL
    SELECT 103 AS id UNION ALL
    SELECT 104 AS id UNION ALL
    SELECT 105 AS id
  )
  AS source
LEFT JOIN
  status
    ON status.id = source.id

当ID丢失时，假设"不是"，可以这样做：

SELECT
  so.ID,
  CASE WHEN st.value IS NULL THEN 'Not' ELSE st.value END
FROM
  databasename1..source so
LEFT JOIN
  databasename2..status st
    ON st.id = so.id
WHERE
  so.ID in (100, 101,102,103,104,105)

将databasename1和databasename2替换为实database names。