如何使用静态值选择选择?
-
24-10-2019 - |
题
我想为Select提供的值添加一个值。
给定名为foo的表:
+----+
| id |
+----+
| 1 |
| 2 |
| 3 |
+----+
我想用:
+----+------+
| id | type |
+----+------+
| 1 | R |
| 2 | R |
| 3 | R |
+----+------+
我想这看起来像这样:
SELECT foo.id, type INTO bar FROM foo LEFT JOIN 'R' AS type;
...不幸的是,这无效。谁能告诉我该怎么做
我同时使用MySQL和MSSQL
解决方案
SELECT foo.id, 'R' AS type INTO bar FROM foo;
在MySQL中,通常会完成:
懒惰没有索引
CREATE TABLE bar SELECT id, 'R' AS type FROM foo;
更好的方法(假设您已经创建了表格 bar
已经)
INSERT INTO bar SELECT id, 'R' AS type FROM foo;
其他提示
SELECT foo.id
, 'R'
INTO bar (id, type)
FROM foo
用于MySQL:
INSERT INTO bar (id, type)
SELECT foo.id
, 'R'
FROM foo
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