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  • MySQL & PHP - Creating Multiple Parent Child Relations

    - by Ashok
    Hi, I'm trying to build a navigation system using categories table with hierarchies. Normally, the table would be defined as follows: id (int) - Primary key name (varchar) - Name of the Category parentid (int) - Parent ID of this Category referenced to same table (Self Join) But the catch is that I require that a category can be child to multiple parent categories.. Just like a Has and Belongs to Many (HABTM) relation. I know that if there are two tables, categories & items, we use a join table categories_items to list the HABTM relations. But here i'm not having two tables but only table but should somehow show HABTM relations to itself. Is this be possible using a single table? If yes, How? If not possible, what rules (table naming, fields) should I follow while creating the additional join table? I'm trying to achieve this using CakePHP, If someone can provide CakePHP solution for this problem, that would be awesome. Even if that's not possible, any solution about creating join table is appreciated. Thanks for your time.

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  • I need help on my C++ assignment using Microsoft Visual C++

    - by krayzwytie
    Ok, so I don't want you to do my homework for me, but I'm a little lost with this final assignment and need all the help I can get. Learning about programming is tough enough, but doing it online is next to impossible for me... Now, to get to the program, I am going to paste what I have so far. This includes mostly //comments and what I have written so far. If you can help me figure out where all the errors are and how to complete the assignment, I will really appreciate it. Like I said, I don't want you to do my homework for me (it's my final), but any constructive criticism is welcome. This is my final assignment for this class and it is due tomorrow (Sunday before midnight, Arizona time). This is the assignment: Examine the following situation: Your company, Datamax, Inc., is in the process of automating its payroll systems. Your manager has asked you to create a program that calculates overtime pay for all employees. Your program must take into account the employee’s salary, total hours worked, and hours worked more than 40 in a week, and then provide an output that is useful and easily understood by company management. Compile your program utilizing the following background information and the code outline in Appendix D (included in the code section). Submit your project as an attachment including the code and the output. Company Background: Three employees: Mark, John, and Mary The end user needs to be prompted for three specific pieces of input—name, hours worked, and hourly wage. Calculate overtime if input is greater than 40 hours per week. Provide six test plans to verify the logic within the program. Plan 1 must display the proper information for employee #1 with overtime pay. Plan 2 must display the proper information for employee #1 with no overtime pay. Plans 3-6 are duplicates of plan 1 and 2 but for the other employees. Program Requirements: Define a base class to use for the entire program. The class holds the function calls and the variables related to the overtime pay calculations. Define one object per employee. Note there will be three employees. Your program must take the objects created and implement calculations based on total salaries, total hours, and the total number of overtime hours. See the Employee Summary Data section of the sample output. Logic Steps to Complete Your Program: Define your base class. Define your objects from your base class. Prompt for user input, updating your object classes for all three users. Implement your overtime pay calculations. Display overtime or regular time pay calculations. See the sample output below. Implement object calculations by summarizing your employee objects and display the summary information in the example below. And this is the code: // Final_Project.cpp : Defines the entry point for the console application. // #include "stdafx.h" #include <iostream> #include <string> #include <iomanip> using namespace std; // //CLASS DECLARATION SECTION // class CEmployee { public: void ImplementCalculations(string EmployeeName, double hours, double wage); void DisplayEmployInformation(void); void Addsomethingup (CEmployee, CEmployee, CEmployee); string EmployeeName ; int hours ; int overtime_hours ; int iTotal_hours ; int iTotal_OvertimeHours ; float wage ; float basepay ; float overtime_pay ; float overtime_extra ; float iTotal_salaries ; float iIndividualSalary ; }; int main() { system("cls"); cout << "Welcome to the Employee Pay Center"; /* Use this section to define your objects. You will have one object per employee. You have only three employees. The format is your class name and your object name. */ std::cout << "Please enter Employee's Name: "; std::cin >> EmployeeName; std::cout << "Please enter Total Hours for (EmployeeName): "; std::cin >> hours; std::cout << "Please enter Base Pay for(EmployeeName): "; std::cin >> basepay; /* Here you will prompt for the first employee’s information. Prompt the employee name, hours worked, and the hourly wage. For each piece of information, you will update the appropriate class member defined above. Example of Prompts Enter the employee name = Enter the hours worked = Enter his or her hourly wage = */ /* Here you will prompt for the second employee’s information. Prompt the employee name, hours worked, and the hourly wage. For each piece of information, you will update the appropriate class member defined above. Enter the employee name = Enter the hours worked = Enter his or her hourly wage = */ /* Here you will prompt for the third employee’s information. Prompt the employee name, hours worked, and the hourly wage. For each piece of information, you will update the appropriate class member defined above. Enter the employee name = Enter the hours worked = Enter his or her hourly wage = */ /* Here you will implement a function call to implement the employ calcuations for each object defined above. You will do this for each of the three employees or objects. The format for this step is the following: [(object name.function name(objectname.name, objectname.hours, objectname.wage)] ; */ /* This section you will send all three objects to a function that will add up the the following information: - Total Employee Salaries - Total Employee Hours - Total Overtime Hours The format for this function is the following: - Define a new object. - Implement function call [objectname.functionname(object name 1, object name 2, object name 3)] /* } //End of Main Function void CEmployee::ImplementCalculations (string EmployeeName, double hours, double wage){ //Initialize overtime variables overtime_hours=0; overtime_pay=0; overtime_extra=0; if (hours > 40) { /* This section is for the basic calculations for calculating overtime pay. - base pay = 40 hours times the hourly wage - overtime hours = hours worked – 40 - overtime pay = hourly wage * 1.5 - overtime extra pay over 40 = overtime hours * overtime pay - salary = overtime money over 40 hours + your base pay */ /* Implement function call to output the employee information. Function is defined below. */ } // if (hours > 40) else { /* Here you are going to calculate the hours less than 40 hours. - Your base pay is = your hours worked times your wage - Salary = your base pay */ /* Implement function call to output the employee information. Function is defined below. */ } // End of the else } //End of Primary Function void CEmployee::DisplayEmployInformation(); { // This function displays all the employee output information. /* This is your cout statements to display the employee information: Employee Name ............. = Base Pay .................. = Hours in Overtime ......... = Overtime Pay Amount........ = Total Pay ................. = */ } // END OF Display Employee Information void CEmployee::Addsomethingup (CEmployee Employ1, CEmployee Employ2) { // Adds two objects of class Employee passed as // function arguments and saves them as the calling object's data member values. /* Add the total hours for objects 1, 2, and 3. Add the salaries for each object. Add the total overtime hours. */ /* Then display the information below. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%% EMPLOYEE SUMMARY DATA%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%% Total Employee Salaries ..... = 576.43 %%%% Total Employee Hours ........ = 108 %%%% Total Overtime Hours......... = 5 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% */ } // End of function

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  • Hiring my first employee

    - by Ady
    A few years ago I moved to a new job having been programming for 2 years using C#, however this new company was mainly using VB6. I made the case for .NET and won, but one of the consessions I had to make was to use VB.NET and not C# (understandable as most of the other developers were already using VB). Three years later it was time to move on, but when applying for jobs I couldn't get past the recruitment agents. I realised that when they were looking at the basic requirements (5 years experience) that they could not add 2 and 3 together to make 5. They were looking for 5 years in VB or C# not across both. Frustrated I decided to combine my skills with a designer friend and start my own company. After two years of hard graft we are now looking for our first employee (a programmer), and this question has hit me again, but now I see the employers perspective. Why take the risk of someone getting up to speed when you have thousands of applicants to choose from. So my question is this, if I define the requirements to be too narrow, I could miss the really great candidates. But if they are too broad it's going to take ages to go through them all. This will be our first 'employee' so the choice needs to be good, I can't afford to make a mistake and employ someone naff. Another option would be to choose a bright university graduate, and train them up (less of a risk because we can pay them less). What have others done in this situation, and what would you recommend I do?

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  • SQL SERVER – Quiz and Video – Introduction to Hierarchical Query using a Recursive CTE

    - by pinaldave
    This blog post is inspired from SQL Queries Joes 2 Pros: SQL Query Techniques For Microsoft SQL Server 2008 – SQL Exam Prep Series 70-433 – Volume 2.[Amazon] | [Flipkart] | [Kindle] | [IndiaPlaza] This is follow up blog post of my earlier blog post on the same subject - SQL SERVER – Introduction to Hierarchical Query using a Recursive CTE – A Primer. In the article we discussed various basics terminology of the CTE. The article further covers following important concepts of common table expression. What is a Common Table Expression (CTE) Building a Recursive CTE Identify the Anchor and Recursive Query Add the Anchor and Recursive query to a CTE Add an expression to track hierarchical level Add a self-referencing INNER JOIN statement Above six are the most important concepts related to CTE and SQL Server.  There are many more things one has to learn but without beginners fundamentals one can’t learn the advanced  concepts. Let us have small quiz and check how many of you get the fundamentals right. Quiz 1) You have an employee table with the following data. EmpID FirstName LastName MgrID 1 David Kennson 11 2 Eric Bender 11 3 Lisa Kendall 4 4 David Lonning 11 5 John Marshbank 4 6 James Newton 3 7 Sally Smith NULL You need to write a recursive CTE that shows the EmpID, FirstName, LastName, MgrID, and employee level. The CEO should be listed at Level 1. All people who work for the CEO will be listed at Level 2. All of the people who work for those people will be listed at Level 3. Which CTE code will achieve this result? WITH EmpList AS (SELECT Boss.EmpID, Boss.FName, Boss.LName, Boss.MgrID, 1 AS Lvl FROM Employee AS Boss WHERE Boss.MgrID IS NULL UNION ALL SELECT E.EmpID, E.FirstName, E.LastName, E.MgrID, EmpList.Lvl + 1 FROM Employee AS E INNER JOIN EmpList ON E.MgrID = EmpList.EmpID) SELECT * FROM EmpList WITH EmpListAS (SELECT EmpID, FirstName, LastName, MgrID, 1 as Lvl FROM Employee WHERE MgrID IS NULL UNION ALL SELECT EmpID, FirstName, LastName, MgrID, 2 as Lvl ) SELECT * FROM BossList WITH EmpList AS (SELECT EmpID, FirstName, LastName, MgrID, 1 as Lvl FROM Employee WHERE MgrID is NOT NULL UNION SELECT EmpID, FirstName, LastName, MgrID, BossList.Lvl + 1 FROM Employee INNER JOIN EmpList BossList ON Employee.MgrID = BossList.EmpID) SELECT * FROM EmpList 2) You have a table named Employee. The EmployeeID of each employee’s manager is in the ManagerID column. You need to write a recursive query that produces a list of employees and their manager. The query must also include the employee’s level in the hierarchy. You write the following code segment: WITH EmployeeList (EmployeeID, FullName, ManagerName, Level) AS ( –PICK ANSWER CODE HERE ) SELECT EmployeeID, FullName, ” AS [ManagerID], 1 AS [Level] FROM Employee WHERE ManagerID IS NULL UNION ALL SELECT emp.EmployeeID, emp.FullName mgr.FullName, 1 + 1 AS [Level] FROM Employee emp JOIN Employee mgr ON emp.ManagerID = mgr.EmployeeId SELECT EmployeeID, FullName, ” AS [ManagerID], 1 AS [Level] FROM Employee WHERE ManagerID IS NULL UNION ALL SELECT emp.EmployeeID, emp.FullName, mgr.FullName, mgr.Level + 1 FROM EmployeeList mgr JOIN Employee emp ON emp.ManagerID = mgr.EmployeeId Now make sure that you write down all the answers on the piece of paper. Watch following video and read earlier article over here. If you want to change the answer you still have chance. Solution 1) 1 2) 2 Now compare let us check the answers and compare your answers to following answers. I am very confident you will get them correct. Available at USA: Amazon India: Flipkart | IndiaPlaza Volume: 1, 2, 3, 4, 5 Please leave your feedback in the comment area for the quiz and video. Did you know all the answers of the quiz? Reference: Pinal Dave (http://blog.sqlauthority.com) Filed under: Joes 2 Pros, PostADay, SQL, SQL Authority, SQL Query, SQL Server, SQL Tips and Tricks, T SQL, Technology

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  • Many-to-many relations in RDBMS databases

    - by Industrial
    What is the best way of handling many-to-many relations in a RDBMS database like mySQL? Have tried using a pivot table to keep track of the relationships, but it leads to either one of the following: Normalization gets left behind Columns that is empty or null What approach have you taken in order to support many-to-many relationships?

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  • Generic relations missing with grappelli

    - by diegueus9
    I'm using the last svn revision of grappelli and rev 11840 of django (before multidatabases and stuff), and i'm trying to use generic relations in the admin, but doesn't work, The model: class AutorProyectoLey(DatedModel): tipo_autor = models.ForeignKey(ContentType) autor_id = models.PositiveIntegerField() content_object = generic.GenericForeignKey('tipo_autor', 'autor_id') proyecto_ley = models.ForeignKey(ProyectoLey) The admin: class AutorInline(GenericInlineModelAdmin): model = AutorProyectoLey allow_add = True ct_field = 'tipo_autor' ct_fk_field = 'autor_id' classes = ('collapse-open',) And i put this model of var inlines in another adminmodel, but the html render is : <!-- Inlines --> <!-- Submit-Row -->

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  • Oracle Query for getting MAximum CTC (Salary) of Each Employee

    - by reply2viveksshah
    i want maximim CTC of each employee following is the design of my table Ecode Implemented Date Salary 7654323 2010-05-20 350000 7654322 2010-05-17 250000 7654321 2003-04-01 350000 7654321 2004-04-01 450000 7654321 2005-04-01 750000 7654321 2007-04-01 650000 i want oracle query for following out put Ecode Salary 7654321 650000 7654322 250000 7654323 350000 thanks in advance Vivek Shah

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  • accepts_nested_attributes_for and has_many :through relations.

    - by antiarchitect
    I want to make a simple examing application on RoR 2.3. The problem area is to make an exam_session in a one form with only one submit action. For the exam session there are selected some number of questions from the question pool in random order. For these questions there are selected some number of alternatives (to check is this a single answer question or multi answer question I use the number of correct alternatives: if only 1 - single, 1 - multi. Radiobuttons or checkboxes in form to answer depends on it). I have models: Questions ---< Alternative and ExamSession. I think there must be has_many :through relations between ExamSession and Questions and has_many :through relation between the intermediate table (for example QuestionsExamSession) and Alternative to point what alternatives are answers of the student on this Question. So the questions are: Is this scheme is too complicated and there is a way to do it simple and clear? Is there any way to organize models in such a way to make the form I want to work?

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  • Kohana 3 simple relations..

    - by matiit
    I'm trying to write a very simple cms (for learning purposes) in kohana 3 web framework. I have my db schemas and i want to map it to ORM but i have problems with relations. Schemas:articles and categories One article has one category. One category might has many articles of course. I think it is has_one relationship in article table.(?) Now php code. I need to create application/classes/models/article.php first, yes? class Model_Article extends ORM { protected // and i am not sure what i suppose to write here }

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  • ACCESS VBA - DAO in VB - problem with creating relations

    - by Justin
    So take the following example: Sub CreateRelation() Dim db As Database Dim rel As Relation Dim fld As Field Set db = CurrentDb Set rel = db.CreateRelation("OrderID", "Orders", "Products") 'refrential integrity rel.Attributes = dbRelationUpdateCascade 'specify the key in the referenced table Set fld = rel.CreateField("OrderID") fld.ForeignName = "OrderID" rel.Fields.Append fld db.Relations.Append rel End Sub I keep getting the error, No unique index found for the referenced field of the primary table. if i include the vb before this sub to create in index on the field, it gives me the error: Index already exists. so i am trying to figure this out. if there are not any primary keys set, will that cause this not to work? i am confused by this, but i really really want to figure this out. So orderID is a FOREIGN KEY in the Products Table please help thanks justin

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  • Symfony 1.4: use relations in fixtures with propel

    - by iggnition
    Hello, I just started to use the PHP symfony framework. Currently I'm trying to create fixture files in YAML to easily insert data into my MySQL database. Now my database has a couple of relations, I have the tables Organisation and Location. Organisation org_id (PK) org_name Location loc_id (PK) org_id (FK) loc_name Now I'm trying too link these tables in my fixture file, but for the life of me I cannot figure out how. Since the org_id is auto-incremented I can't simply use org_id: 1 In the location fixture. How can I fix this?

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  • Zend Framework Relations vs. Table Select

    - by rtmilker
    Hey! I just want to know your guys opinion on using join tables within the zend framework. Of course you can use relations by defining a referenceMap and dependentTables and stuff, or using setIntegrityCheck(false) within a db select(). The setIntegrityCheck version seems a little bit dirty to me, but the other version is not very suitable for big querys and joining many tables... I'm a PHP developer for 5 years now and new to the zend framework and just want get a direction for my first project. Thanks!!!

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  • how to use use case relations - uml

    - by joao alves
    Heys guys! Im have been study UML and im trying to to design the use case diagram of a problem. Lets supose my app consists in this: Two Requesites: - create teams - create players This is the deal: A user can create a team, and after create a team he can create players for that team(not required). But in this app there are multiple users, and a user can create a team and other user can create players. The only constraint is that to create players must exist alreay a team. I research and i end up a little confuse. If i get the concepts of relations on use case diagrams right, i think i should have the folowwing two use cases: [use case - create team] <-------extends---- [use case - create player] I need opinions,Is this the proper solution? or should i have two not related use cases? Thanks in advance, and im sorry my english.

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  • Accessing inter-schema tables and relations in hibernate

    - by nitesh
    There is a typical situation being faced where different tables are scattered through different schemas in Oracle database and they are related to each other (encompassing all different types of relations). How can they be represented in Hibernate using annotations as when a sessionfactory handle is created for one schema, tables in that schema can't access other related tables (foreign key relation to tables in other schema)? For a query like following, exception is thrown - "from table1 as model where model.table2Name.table2column = "+foo Exception comes as - org.hibernate.QueryException: could not resolve property: table2column of: com.test.table1 [from com.test.table1 as model where model.table2Name.table2column = 1] Here table1 and table2 are present in different schemas.

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  • counting employee attendance

    - by jjj
    i am trying to write a statment for counting the employees attendance and execute thier id , name and the days that he has working on the last 3 months by counting the duplicate id on NewTimeAttendance for month 1 , 2 and 3 .. i tried to count : Select COUNT(employeeid) from NewTimeAttendance where employeeid=1 and (month=1 or month =2 or month = 3) This is absolutely working ,but just for one employee... the secound try: SELECT COUNT(NewEmployee.EmployeeID) FROM NewEmployee INNER JOIN NewTimeAttendance ON NewEmployee.EmployeeID = NewTimeAttendance.EmployeeID and (month=1 or month =2 or month = 3) This is working , but it counts all employees .. and i want it to execute each EmployeeId, EmployeeName and number of days as new record last try: (before you see the code ... it is wrong ..but i am trying) for i in 0..27 loop SELECT COUNT(NewEmployee.EmployeeID),NewEmployee.EmployeeId,EmployeeName FROM NewEmployee INNER JOIN NewTimeAttendance ON NewEmployee.EmployeeID(i) = NewTimeAttendance.EmployeeID and (month=1 or month =2 or month = 3) end loop i realy need help...thanks in advance

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  • Setting up relations/mappings for a SQLAlchemy many-to-many database

    - by Brent Ramerth
    I'm new to SQLAlchemy and relational databases, and I'm trying to set up a model for an annotated lexicon. I want to support an arbitrary number of key-value annotations for the words which can be added or removed at runtime. Since there will be a lot of repetition in the names of the keys, I don't want to use this solution directly, although the code is similar. My design has word objects and property objects. The words and properties are stored in separate tables with a property_values table that links the two. Here's the code: from sqlalchemy import Column, Integer, String, Table, create_engine from sqlalchemy import MetaData, ForeignKey from sqlalchemy.orm import relation, mapper, sessionmaker from sqlalchemy.ext.declarative import declarative_base engine = create_engine('sqlite:///test.db', echo=True) meta = MetaData(bind=engine) property_values = Table('property_values', meta, Column('word_id', Integer, ForeignKey('words.id')), Column('property_id', Integer, ForeignKey('properties.id')), Column('value', String(20)) ) words = Table('words', meta, Column('id', Integer, primary_key=True), Column('name', String(20)), Column('freq', Integer) ) properties = Table('properties', meta, Column('id', Integer, primary_key=True), Column('name', String(20), nullable=False, unique=True) ) meta.create_all() class Word(object): def __init__(self, name, freq=1): self.name = name self.freq = freq class Property(object): def __init__(self, name): self.name = name mapper(Property, properties) Now I'd like to be able to do the following: Session = sessionmaker(bind=engine) s = Session() word = Word('foo', 42) word['bar'] = 'yes' # or word.bar = 'yes' ? s.add(word) s.commit() Ideally this should add 1|foo|42 to the words table, add 1|bar to the properties table, and add 1|1|yes to the property_values table. However, I don't have the right mappings and relations in place to make this happen. I get the sense from reading the documentation at http://www.sqlalchemy.org/docs/05/mappers.html#association-pattern that I want to use an association proxy or something of that sort here, but the syntax is unclear to me. I experimented with this: mapper(Word, words, properties={ 'properties': relation(Property, secondary=property_values) }) but this mapper only fills in the foreign key values, and I need to fill in the other value as well. Any assistance would be greatly appreciated.

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  • computes the number of possible orderings of n objects under the relations < and =

    - by hilal
    Here is the problem : Give a algorithm that takes a positive integer n as input, and computes the number of possible orderings of n objects under the relations < and =. For example, if n = 3 the 13 possible orderings are as follows: a = b = c, a = b < c, a < b = c, a < b < c, a < c < b, a = c < b, b < a = c, b < a < c, b < c < a, b = c < a, c < a = b, c < a < b, c < b < a. Your algorithm should run in time polynomial in n. I'm null to this problem. Can you find any solution to this dynamic-programming problem?

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  • Linked List manipulation, issues retrieving data c++

    - by floatfil
    I'm trying to implement some functions to manipulate a linked list. The implementation is a template typename T and the class is 'List' which includes a 'head' pointer and also a struct: struct Node { // the node in a linked list T* data; // pointer to actual data, operations in T Node* next; // pointer to a Node }; Since it is a template, and 'T' can be any data, how do I go about checking the data of a list to see if it matches the data input into the function? The function is called 'retrieve' and takes two parameters, the data and a pointer: bool retrieve(T target, T*& ptr); // This is the prototype we need to use for the project "bool retrieve : similar to remove, but not removed from list. If there are duplicates in the list, the first one encountered is retrieved. Second parameter is unreliable if return value is false. E.g., " Employee target("duck", "donald"); success = company1.retrieve(target, oneEmployee); if (success) { cout << "Found in list: " << *oneEmployee << endl; } And the function is called like this: company4.retrieve(emp3, oneEmployee) So that when you cout *oneEmployee, you'll get the data of that pointer (in this case the data is of type Employee). (Also, this is assuming all data types have the apropriate overloaded operators) I hope this makes sense so far, but my issue is in comparing the data in the parameter and the data while going through the list. (The data types that we use all include overloads for equality operators, so oneData == twoData is valid) This is what I have so far: template <typename T> bool List<T>::retrieve(T target , T*& ptr) { List<T>::Node* dummyPtr = head; // point dummy pointer to what the list's head points to for(;;) { if (*dummyPtr->data == target) { // EDIT: it now compiles, but it breaks here and I get an Access Violation error. ptr = dummyPtr->data; // set the parameter pointer to the dummy pointer return true; // return true } else { dummyPtr = dummyPtr->next; // else, move to the next data node } } return false; } Here is the implementation for the Employee class: //-------------------------- constructor ----------------------------------- Employee::Employee(string last, string first, int id, int sal) { idNumber = (id >= 0 && id <= MAXID? id : -1); salary = (sal >= 0 ? sal : -1); lastName = last; firstName = first; } //-------------------------- destructor ------------------------------------ // Needed so that memory for strings is properly deallocated Employee::~Employee() { } //---------------------- copy constructor ----------------------------------- Employee::Employee(const Employee& E) { lastName = E.lastName; firstName = E.firstName; idNumber = E.idNumber; salary = E.salary; } //-------------------------- operator= --------------------------------------- Employee& Employee::operator=(const Employee& E) { if (&E != this) { idNumber = E.idNumber; salary = E.salary; lastName = E.lastName; firstName = E.firstName; } return *this; } //----------------------------- setData ------------------------------------ // set data from file bool Employee::setData(ifstream& inFile) { inFile >> lastName >> firstName >> idNumber >> salary; return idNumber >= 0 && idNumber <= MAXID && salary >= 0; } //------------------------------- < ---------------------------------------- // < defined by value of name bool Employee::operator<(const Employee& E) const { return lastName < E.lastName || (lastName == E.lastName && firstName < E.firstName); } //------------------------------- <= ---------------------------------------- // < defined by value of inamedNumber bool Employee::operator<=(const Employee& E) const { return *this < E || *this == E; } //------------------------------- > ---------------------------------------- // > defined by value of name bool Employee::operator>(const Employee& E) const { return lastName > E.lastName || (lastName == E.lastName && firstName > E.firstName); } //------------------------------- >= ---------------------------------------- // < defined by value of name bool Employee::operator>=(const Employee& E) const { return *this > E || *this == E; } //----------------- operator == (equality) ---------------- // if name of calling and passed object are equal, // return true, otherwise false // bool Employee::operator==(const Employee& E) const { return lastName == E.lastName && firstName == E.firstName; } //----------------- operator != (inequality) ---------------- // return opposite value of operator== bool Employee::operator!=(const Employee& E) const { return !(*this == E); } //------------------------------- << --------------------------------------- // display Employee object ostream& operator<<(ostream& output, const Employee& E) { output << setw(4) << E.idNumber << setw(7) << E.salary << " " << E.lastName << " " << E.firstName << endl; return output; } I will include a check for NULL pointer but I just want to get this working and will test it on a list that includes the data I am checking. Thanks to whoever can help and as usual, this is for a course so I don't expect or want the answer, but any tips as to what might be going wrong will help immensely!

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  • JPA entity relations are not populated after .persist()

    - by Tomik
    Hello, this is a sample of my two entities: @Entity public class Post implements Serializable { @OneToMany(mappedBy = "post", fetch = javax.persistence.FetchType.EAGER) @OrderBy("revision DESC") public List<PostRevision> revisions; @Entity(name="post_revision") public class PostRevision implements Serializable { @ManyToOne public Post post; private Integer revision; @PrePersist private void prePersist() { List<PostRevision> list = post.revisions; if(list.size() >= 1) revision = list.get(list.size() - 1).revision + 1; else revision = 1; } So, there's a "post" and it can have several revisions. During persisting of the revision, entity takes a look at the list of the existing revisions and finds the next revision number. Problem is that Post.revisions is NULL but I think it should be automatically populated. I guess there's some kind of problem in my source code but I don't know where. Here's my "persistence" code: Post post = new Post(); PostRevision revision = new PostRevision(); revision.post = post; em.persist(post); em.persist(revision); em.flush(); I think that after persisting "post", it becomes "managed" and all the relations should be populated from now on. Thanks for help! (Note: public attributes are just for demonstration)

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  • Umbraco Gold Partner going from strength to strength.

    - by Vizioz Limited
    It is amazing how time flies when you are having fun and we certainly have been having an interesting year at Vizioz, I thought it was about time I wrote another blog post and shared some of our news.Over the last 6 months we have:a) Had the pleasure of working with some great clients from the USA, Ireland and the UKb) Built some interesting and complex sites for Multi-national brands (under NDA's, you'd be impressed if you knew!)c) Converted the Umbraco Users Manual to a free iBook for all those iPad owning Umbraco users.d) We have hired three new employees (Sam, Pearl & Zaara)e) We have given our notice on our current office (see below)And on the horizon:a) We have submitted Allen for Umbraco to the Apple App store for approval (hopefully this will be available very soon!)b) We are about to sign a new office lease for a new office that is twice the size of our current office, so we will have room for a meeting room, a chill out room and some more employees!So it's exciting times at Vizioz, thank you to all our fantastic clients for making this possible, we look forward to working with you all over the years to come.One thing we don't shout about as much as we probably should, we also renewed our Umbraco Gold Partner status for the second year, showing our commitment to the Umbraco CMS, if you are looking for a great Umbraco partner with experienced developers to build your new site, or to take over the on-going maintenance of an existing site, then pick up the phone and give us a call, we would love to add you to our list of happy customers!

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  • When should complexity be removed?

    - by ElGringoGrande
    Prematurely introducing complexity by implementing design patterns before they are needed is not good practice. But if you follow all (or even most of) the SOLID principles and use common design patterns you will introduce some complexity as features and requirements are added or changed to keep your design as maintainable and flexible as needed. However once that complexity is introduced and working like a champ when do you removed it? Example. I have an application written for a client. When originally created there where several ways to give raises to employees. I used the strategy pattern and factory to keep the whole process nice and clean. Over time certain raise methods where added or removed by the application owner. Time passes and new owner takes over. This new owner is hard nosed, keeps everything simple and only has one single way to give a raise. The complexity needed by the strategy pattern is no longer needed. If I where to code this from the requirements as they are now I would not introduce this extra complexity (but make sure I could introduce it with little or no work should the need arise). So do I remove the strategy implementation now? I don't think this new owner will ever change how raises are given. But the application itself has demonstrated that this could happen. Of course this is just one example in an application where a new owner takes over and has simplified many processes. I could remove dozens of classes, interfaces and factories and make the whole application much more simple. Note that the current implementation does works just fine and the owner is happy with it (and surprised and even happier that I was able to implement her changes so quickly because of the discussed complexity). I admit that a small part of this doubt is because it is highly likely the new owner isn't going to use me any longer. I don't really care that somebody else will take this over since it has not been a big income generator. But I do care about 2 (related) things I care a bit that the new maintainer will have to think a bit harder when trying to understand the code. Complexity is complexity and I don't want to anger the psycho maniac coming after me. But even more I worry about a competitor seeing this complexity and thinking I just implement design patterns to pad my hours on jobs. Then spreading this rumor to hurt my other business. (I have heard this mentioned.) So... In general should previously needed complexity be removed even though it works and there has been a historically demonstrated need for the complexity but you have no indication that it will be needed in the future? Even if the question above is generally answered "no" is it wise to remove this "un-needed" complexity if handing off the project to a competitor (or stranger)?

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  • Tester that doesn't test

    - by George
    What should I do about a tester that does not test? We have a complicated dry run scenario, that takes a lot of time to execute. Mostly this tester will execute it's tests in very slow way...checking emails, internet, etc. He reports just a few bugs, but! Whenever the official dry-run begins (these are logged with testlink) the tester starts to open new bugs that where not discovered before. Is he not doing his job correctly? Or am I just overlooking how tests work? I'm not his supervisor, but he is testing code that I wrote.

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