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Determine the Three Views of Data
We know that the similar thing, if viewed from dissimilar angles produces difference sights. Similarly, the database that we have formed already can have different aspects to reveal if seen from dissimilar levels of abstraction. The term Abstraction is very significant here. Generally it means the amount of detail you want to hide. Any entity can be seen from dissimilar perspectives and levels of complexity to make it a reveal its current amount of abstraction.
In the same manner, the database can also be viewed from dissimilar levels of abstraction to reveal dissimilar levels of details.
These are interactive interfaces, graphical that hide the difficulty of the DBMS and query language behind an intuitive, simple to understand, and convenient interface. This permit
File Organisation And Its Types Just as arrays, trees, lists and other data structures are used to execute data Organisation in main memory, a number of planning's are used to
What is the difference between WHERE and HAVING clauses?. Predicates in the WHERE clause applies to every tuple while the predicate in the HAVING clause applies to groups.
Example : Create a sequence named SEQSS that starts at 105, has a step of 1 and can take maximum value as 2000. CREATE SEQUENCE SEQSS START WITH 105 INCREMENT BY 1 MAX VA
Optimistic Concurrency Control Is locking the only way to stop concurrency related problems? There exist some other ways too. One such way is known as an Optimistic Concurrency
Explain the difference between the three storage types - volatile, non volatile and stable. Ans: Volatile storage: if storage media loses data while power goes off,
Question 1 List and explain various Normal Forms. How BCNF differs from the Third Normal Form and 4th Normal forms? Question 2 Describe the concepts of Structural Semantic Da
What is Normalization? Database normalization is a data design and organization method applied to data structures based on rules that help building relational databases. In rel
it can be possible to create any table without primary key
Prove that a relation which is 4NF must be BCNF. Ans Let R be in 4NF. Suppose it is not in BCNF. Hence, there exists an FD X→Y in R such that x is not a super key. Although
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