relational-database

问题 When I researched horizontal scaling for relational databases on the internet, I got the impression that the only option which includes write scaling as well as read scaling is sharding, which seems to be a manual design process that involves complex application specific configurations and is hard to maintain if you need to change your sharding structure. On the other hand, NoSQL seems to be natively supporting horizontal scaling but it has the drawback of not supporting transactions, ACID

问题 i have 3 tables (Categories, Articles and ArticleCategories). The ArticleCategories has a combined PK and concists of: ArticleID (PK, FK, int) CategoryID (PK, FK, int) Now when I'm using the subsonic paging methods I'm running into some very strange behaviour. It returns double the amount of records then I expect (the 2 PK's?). The issue is resolved by either removing the paging completely or by adding a PK field called ArticleCategoryID. SubSonic.Query qry = new SubSonic.Query(DAL

问题 I am starting to learn SQL using MySQL and I got my hands on book "SQL for MySQL Developers: A Comprehensive Tutorial and Reference". Why the author doesn't place semicolons ; at the end of the statements? Can you form legal statement in MySQL without using semicolon? Here is an example pulled right from the book (p.45) CREATE DATABASE TENNIS . Of course, MySQL shell responds by -> . 回答1: Many flavors and applications of SQL require a semi-colon at the end. A few do not. 来源： https:/

问题 I have a few tables which I am trying to join and fetch the results for a list Interviews Table +--------------+-----------+ | interview_id | Candidate | +--------------+-----------+ | 1 | Ram | | 2 | Rahim | | 3 | Joseph | +--------------+-----------+ Participant Ratings Table +--------------+-----------+-------+ | interview_id | Rater Type|Rating | +--------------+-----------+-------+ | 1 | Candidate | 4 | | 2 | Candidate | 4 | | 1 | Recruiter | 5 | +--------------+-----------+-------+

问题 I have the following table: products (id, name, price, category_id) categories (id, name, description) now because each category's products may have different attributes than other category's product I want to use EVA but here is my problem: I want to make sure that any product attribute's value is being added belong to the category of that product and not allowing the addition of arrtibutes that does not belong to it. would you please tell me how to do this? 回答1: I want to make sure that any

问题 Is it bad to have two tables with the same attributes but with different meanings and relationships? I.e. table1 's attributes = table2 's attributes. However table1 has a relationship with table3 and table2 has a relationship with table4 . Is this considered bad practice? Or should table1 and table2 be collated into one table? Although if it were to be collated into one table wouldn't it follow that table1 would also have a relationship with table4 ? 回答1: It is ok to have tables with the

问题 I'm trying to make a quiz that supports multiple languages. I have made the database schema for the quiz and user results but not sure how to implement support for different languages in a good way. This what it looks like right now (feel free to point out design mistakes): * User: - id PK - name - email - password * Quiz - id PK - slug - title * Question - id PK - quiz_id FK - question - image - message * Option - id PK - question_id FK - option - is_right Relation between user and quiz to

问题 I have queuries with a LOT of these situations Sel TB1.C1 TB2.C2, TB3.C4 Tb5.C5 where < Join conditions involving all tables TB1 through TB4 . Most are inner some are LOJ > where TB2.C2 NOT In ( List ) OR TB3.C5 <> 'string' OR Tb5.C8 NOT IN ( another long list ) Is there a better way to rewrite the filter conditions NOT IN ( List ) , Col <> and IN LIST Optimization some of the cols are selected some are not which are there in the sel Vs the filter condition. 来源： https://stackoverflow.com

问题 I have multiple Raspberry Pi's collecting wifi beacons from mobile devices and save them in a mySQL DB. I have created a view in the DB. Each entry in the DB has the mobile device mac address, the pi id, rssi, location and a timestamp I have created a view from multiple tables that looks like this. +----------+--------------+------+-----+---------+-------+ | Field | Type | Null | Key | Default | Extra | +----------+--------------+------+-----+---------+-------+ | pi_id | varchar(64) | NO | |

问题 Let R(A,B,C,D,E) be a relation schema and F = {A→C, B→D, C→E, E→A}, Find all candidate keys. I believe that there exists no CK's in this set due to not being able to map. B or D to any other relation besides B -> D . Does this mean that that there are no Candiate Keys? Although I am able to map A to all other entities besides B and D. 回答1: There are three candidate keys. B doesn't appear on the right-hand side of any functional dependency. That means B must be part of every candidate key. I