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nChars = recv(sock, (char *) buffPos, readSize, MSG_NOSIGNAL); readlen += nChars; if (nChars <= 0) { return -1; } if (buffPos[nChars - 1] == '\n') { complete = true; buffPos[nChars - 1] = '\0'; } buffPos += nChars; }

Note that each record in the junction table links one category with one product. You can have as many records as you like in the junction table, linking any category to any product. The linking table contains two fields, each one referencing the primary key of one of the two linked tables. In our case, the junction table will contain two fields: a category_id field and a product_id field. Each record in the junction table will consist of a (product_id, category_id) pair, which will be used to associate a particular product with a particular category. By adding more records to the product_category table, you can associate a product with more categories or a category with more products, effectively implementing the Many-to-Many relationship. Because the Many-to-Many relationship is implemented using a third table that makes the connection between the linked tables, there is no need to add additional fields to the related tables in the way that we added the department_id to the category table for implementing the One-to-Many relationship. There s no definitive naming convention to use for the junction table. Most of the time it s okay to just join the names of the two linked tables in this case, the junction table is named product_category.

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Summary

return readlen; }

delimiter. Since we can have at most two operands for a command, after it finds the second space it places the remaining text from the buffer into the second operand. This works perfectly for our commands such as PMSG, where the second operand can contain spaces. It also works well enough for the MSG command, which has only one operand. If the message contains spaces, we just concatenate the two operands, in the msg_command() function.

Relationships between tables can be physically enforced in the database using FOREIGN KEY constraints, or simply foreign keys. You learned in the previous chapter about the PRIMARY KEY and UNIQUE constraints. We covered them there because they apply to the table as an individual entity. Foreign keys, on the other hand, occur between two tables: the table in which the foreign key is defined (the referencing table) and the table the foreign key references (the referenced table).

cmd_t decodeCommand(const char *buffer) { struct cmd_t ret_cmd; int state; state = 0; for (int x = 0; x < strlen(buffer); x++) { if (buffer[x] == ' ' && state < 2) { state++; } else {

Actually, the referencing table and the referenced table can be one and the same. This isn t seen too Tip

switch (state) { case 0: case 1: default: } } } return ret_cmd; } ret_cmd.command += toupper(buffer[x]); break; ret_cmd.op1 += buffer[x]; break; ret_cmd.op2 += buffer[x];

often in practice, but it s not unusual either. For example, you can have a table with employees, where each employee references the employee that is his or her boss (in this scenario the big boss would probably reference itself).

int join_command(const cmd_t &cmd, string &msg) { int retval; map<string, client_t>::iterator client_iter;

Now that you have analyzed the domain model of an OOP application, you are ready to transform the design into an actual implementation. The next part of this book will introduce you to the Visual Basic language. You will look at the .NET Framework and see how Visual Basic applications are built on top of the framework. You will be introduced to working in the Visual Studio IDE and become familiar with the syntax of the Visual Basic language. The next section will also demonstrate the process of implementing OOP constructs such as class structures, object instantiation, inheritance, and polymorphism in the Visual Basic .NET language. You will revisit the case study introduced in this chapter in 10, at which time you will look at transforming the application design into actual implementation code.

First, we do a preliminary check on the nickname. If the nickname has a space in it, then it will have been separated into op1 and op2. Therefore, if op2 is not an empty string, then the nickname is invalid.

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