我想要这种出局
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);
它将显示上述结果,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.
所以我将文件ID作为逗号分隔值传递。但我想要这种结果如何使用任何循环创建它。
ID Status
100 Viewed
101 Not
102 Not
103 Viewed
104 Not
105 Viewed
有可能吗?请帮帮我。
答案 0 :(得分:1)
你有一张表存在这些ID吗?那么你可以参加吗?
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,则需要创建临时表或内联表(其中包含这些值)。然后,您可以将该表加入您的数据。
修改强>
内联表的示例。有几种方法可以做到这一点,这只是一种。
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
答案 1 :(得分:0)
假设缺少ID
时“不”,您可以这样做:
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
替换为真实database names
。