Introduction to Entity-relationship modelling
For example, LEFT-POINTING TRIANGLE's unicode code point is Browser to render the HTML Entity on iPads/iPhones, but I managed to find. has been reduced to those codes which have a one-to-one relationship with code pages / (This means .. Differential vector operator (down triangle). –. ▾ ENTITY drarr SDATA "[drarr ]" -- Downward right-pointing arrow. image of Unicode Character 'BLACK RIGHT-POINTING TRIANGLE' (U+25B6); Browser Test Page · Outline (as SVG file) · Fonts that support U+25B6.
They represent items such as fields, tables, types and keys. ERD physical symbols are the building material for the database.
Notation lines are used to illustrate the relationships. There are three main components of any ER diagram: Basing on these three components, one can define other, less used elements, such as weak entity or relationship, derived attribute, recursive relationship etc.
This is the set of graphic elements of ERD Chen's notation. It involves the set of geometric forms: Connections are displayed with arrowed lines. It is known that the Chen's ERD notation is used to show a detailed view of entities and relationships.
Using special symbols described in ERD notations you can depict a database of any complexity. The vector graphic ER diagrams produced when using ConceptDraw ERD solution can be used in whitepapers, presentations, datasheets, posters, or any technical materials. An entity-relationship ER diagram is used to show the structure of a business database.
▶ | black right-pointing triangle (U+25B6) @ Graphemica
ERD represents data as objects entities that are connected with standard relationships symbols which Illustrate an association between entities. ERD, there is a wide range of ERD notations used by data bases architects for reflecting the relationships between the data entities.
Not all joins are about recovering relations from an ER diagram. Also, I said earlier that entity T should not have an attribute that was another entity of type S; instead, we should create a relationship R between T and S. If S was at all a candidate for an attribute, each T would be related to at most one S and so this would have cardinality constraint TNRS.
Then, when we did the above conversion, in step four we would add S's key to T with a foreign key constraint referring to S.
But suppose we did add S as an entity attribute to T. Then we would end up with the same situation: So in the end we get the same thing. Invoice How shall we model invoices?XML Chapter 4 - XML Entities
An invoice is a collection of parts ordered, each with a quantity. An invoice is thus all the items to the same customer with the same date. For a given c and d there might be multiple parts p that were part of the invoice. Ternary relationships tend to be inefficient.
Even if we do this, we have another issue: That is, we declare that orders are "things" rather than relationships.
ERD Symbols and Meanings
This is an instance of a rather general strategy that might be called the synthetic-key trick: In this case the synthetic key has a natural interpretation: For now we are going to stop this part of the analysis since the purpose of this example is to demonstrate what ER modelling is all about.
The ER modelling process is not something for which a set of steps can be given and then performed. The process contains almost as much art as science. Some steps are performed many times and many decisions are re-visited and revised. Given these conditions, a broad outline can be given: Determine what entity types are involved. Determine which entity types are related. Refine the definition of the relationships.
Understand now that there are several methods for representing ER models graphically. Notice what has happened with this situation. Initially we had a text description of the problem. After analysing it and making some necessary assumptions, we created an ER diagram that reflects the situation accurately and makes explicit the relationship among the entity types. This is why we perform ER modelling. It is quite a straight-forward step to go from this ER model to an implemented database.
Remember why we are doing all this: We are finding out all we need to know to create a database that will hold our data. And a well-defined database can be a very useful tool for solving business problemsand it is also in high demand by recruiters. You will learn how to perform the steps necessary to create such a database in later chapters. In this section I present more detail on some of the basic concepts. In the example in an earlier section, we saw that divisions are directly associated with departments and departments are directly associated with employees.
No direct association between division and employee was given. This does not mean that there is no relationship between division and employee.
In fact, the ER diagram tells us that there is a relationship between the two: Given any one division, there can be many employees managing departments within that division. Certainly, this is not earth shattering news. But it is in the ER diagram. The above fact is not represented as a separate relationship between division and employee because it can be inferred from existing relationships.
An ER diagram should contain the minimum number of relationships necessary to reflect the situation. For relationships between two entity types, there are three basic cardinalities. Each of the following descriptions are given in terms of a relationship between entity type X and entity type Y.
One entity of type Y can be associated with, at most, one entity of type X. A car has only one steering wheel and a steering wheel can only be installed in one car. M one-to-many One entity of type X can be associated with many entities of type Y. A building can have many rooms but a room can be in, at most, one building.
M many-to-many One entity of type X can be associated with many entities of type Y.
One entity of type Y can be associated with many entities of type X. A car can have many options and an option can be installed on many cars. Determine the cardinality of the relationships between the following four pairs of entity types.
For each relationship you have to answer two questions: Answering these two questions gives you the answer to the following questions. For example, if you answered M to the first question and 1 for the second question, then this relationship between entity types X and Y is of cardinality M: Patient under care of primary care physician Physician performs operation Doctors have speciality in disease Needle injected into patient It would seem that at any particular time a patient can only have one primary care physician and that any physician can have many patients M: One physician can perform many operations and one operation can be performed by many physicians M: One doctor can have specialities in many diseases and one disease can be the speciality of many doctors M: In this section we examine the minimum number of entities in a relationship.
Existence is given as optional, mandatory, or unknown. This is best clarified with an example. Consider again the example discussed in Section 2.
Specifically, focus on the manage relationship between department and employee. We know the cardinality is 1: This tells us that at most one department is managed by an employee and an employee can manage, at most, one department. Be sure you understand the distinction between these two phrases. The existence of this relationship tells us the fewest number of departments that can be managed by an employee and the fewest number of employees that can manage a department.
Only one of the following can be true: Similarly, only one of the following may be true: For each set of three above, which ones would you choose? It is not entirely clear from the situation description which of the above are true. I make the relatively standard assumptions that a department must have at least one manager and that an employee need not be the manager of any department.
Thus, the existence of this relationship is mandatory in one direction and optional in the other. Going back to the definition of existence, we can also look at this situation in this way: Given any randomly chosen department, there must be an employee on the other side of the manage relationship. Thus, the relationship is mandatory in this direction.
Given any randomly chosen employee, there need not be any department on the other side of the manage relationship.
Thus, the relationship is optional in this direction. I assume that the contains relationship is mandatory in both directions. Given this information, the ER diagram is modified in the following manner: This diagram is beginning to look a little complicated but remember the following pieces of information and it gets a little easier: The marks on the lines tell you the minimum number in a relationship. A dash on the line looks like a 1; it tells you the minimum number in the relationship is one so the existence is mandatory.
A circle on the line looks line a 0; it tells you the minimum number in the relationship is zero so the existence is optional. If there are not any marks on the line, then the existence is unknown. The marks below the line tell you the maximum number in a relationship.
An M tells you the cardinality is many while a 1 tells you it is one. Look at the manages relationship again. Cover up the line connecting the manages diamond with the department rectangle.
Now all you see are the department entity type, the employee entity type, and the right side of the line with a dash on it and a one below it. This tells us that a department is managed by at least the dash one employee and at most the 1 one employee. A department is managed by one and only one employee. Cover up the line connecting the manages diamond with the employee rectangle. Now all you see are the department entity type, the employee entity type, and the left side of the line with a circle on it and a 1 below it.
This tells us that an employee does not have to manage any departments the circle and may manage at most the 1 one department. An employee may manage no more than one department. For each of the relationships listed in Problem 5: