问题
Two points I don’t understand about RDBMS being CA in CAP Theorem :
1) It says RDBMS is not Partition Tolerant but how is RDBMS any less Partition Tolerant than other technologies like MongoDB or Cassandra? Is there a RDBMS setup where we give up CA to make it AP or CP?
2) How is it CAP-Available? Is it through master-slave setup? As in when the master dies, slave takes over writes?
I’m a novice at DB architecture and CAP theorem so please bear with me.
回答1:
A lot of databases now actually have different configurations and depending on the settings you set, it can be either CA, CP, AP, etc but can not achieve all three at the same time. Some databases actually make an effort to support all three but still prioritizes them in a certain way.
For example, MySQL can be CP and CA depending on the configurations. By default, it is CA because it follows a master slave paradigm which data is replicated to the slaves. Partition tolerance is sacrificed in the event that a set of the slaves loses the connection to the master and therefore decides to elect a new master creating two masters with their own set of slaves.
However, MySQL also has another configuration which is a clustered configuration. It prioritizes CP over availability eg. the cluster will shutdown if there are not enough live nodes to serve all the data.
There are probably more configurations for MySQL that makes it satisfy other CAP theorem combinations but overall, I just wanted say that it depends on what your system requires. Sometimes databases are better for one configuration vs another so its best to see what kinds of problems that may also occur in using a certain configuration.
As for implementing the CAP theorem, I would advise taking a further look into different databases and how they implement the priorities for the CAP theorem. There are just too many different ways of implementing them eg. generally, the master slave model is used for CA systems, the hash ring for AP systems, etc.
回答2:
I agree that RDBMS can have all the properties of CAP. I have started studying noSQL DBs and had prior experience with IBM DB2.
Here is how IBM DB2 satisfies all the 3 CAP properties
C : Consistency : Every relational database satisfies this due to the transactional nature of RDBMS.
A : Availability : Availability means that when a query is made for a data that exists, it should be returned. Again, a relational database is designed to do this easily.
P : Partition Tolerance : This is the most interesting one. From DB2 stand point, in the application that I was working on, we had 2 databases spread across different data centres. One was the primary and communicated with the secondary via heartbeats. Each of these primary and secondary databases, had 12 physical instances where data was distributed on the basis of some predefined logic. If the primary goes down, the secondary detects this and takes the place of primary. Since the primary and secondary were always maintained in sync, data remains consistent as well.
This is how I think that RDBMS satisfies all 3 properties of CAP Theorem.
I may be wrong, and open to discussion on this.
回答3:
CAP theorem is problematic and it applies only to distributed database systems. When you have distributed databases then network partition and node crashes can happen. And when network partition happens you must have partition tolerance (the P of your CAP).
So to answer your question number 1) It’s either CP or AP. It can be configured as Will mentioned.
More about why partition tolerance is a must: https://codahale.com/you-cant-sacrifice-partition-tolerance/
More about problems around CAP theorem: https://martin.kleppmann.com/2015/05/11/please-stop-calling-databases-cp-or-ap.html
来源：https://stackoverflow.com/questions/36404765/why-isnt-rdbms-partition-tolerant-in-cap-theorem-and-why-is-it-available