I have a special case,
for example in table ta
in database A
, it stores all the products I buy
table ta( id, name, price )
in table tb
in database B
, it contain all the product that people can buy
table tb( id, name, price .... )
Can I create a view in database A
to list all the products that I haven`t bought?
A view that combines data from multiple tables enables you to show relevant information in multiple tables together. You can create a view that combines data from two or more tables by naming more than one table in the FROM clause.
Multiple Databases on One Server Instance For example, you can have a MySQL server installation on machine X that hosts both the customers and orders database. It is possible to use SQL to write one query that combines the information from many databases.
It appears as though you cannot have multiple databases running under one instance. Without OPS or RAC, you must have one instance for each database if you want to access data in that database. However, it is possible to access another database's data through an instance by using Database Links.
Yes you can - the t-sql syntax is the same as within any other cross database call (within a stored procedure for example).
To reference your tables in the second database you simply need:
[DatabaseName].[Schema].[TableName]
So you would end up with something like
CREATE VIEW [dbo].[YourView] as select a.ID, a.SomeInfo, b.SomeOtherInfo from TableInA a join DatabaseB.dbo.TableInB b on -- your join logic goes here
Note that this will only work on the same server - if your databases are on different servers them you will need to create a linked server.
As the other answers indicate, you can use the {LINKED_SERVER.}DATABASE.SCHEMA.OBJECT notation.
You should also be aware that cross-database ownership chaining is disabled by default.
So within a database, granting SELECT on a view allows a user who may not have SELECT on the underlying tables to still SELECT from the view. This may not work across to another database where the user does not have permissions on the underlying table.
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