这是家庭作业,所以如果你能帮忙,谢谢你,如果不能,我明白了。
一切都按预期工作。但我觉得我可以更好地实现事情或拥有更清晰的代码。我希望按钮只调用一个读取 JSON 文件的方法,运行查询并发出正确的显示,而无需像我一样复制代码。
using System;
using System.Collections.Generic;
using System.Data;
using System.IO;
using System.Linq;
using System.Windows.Forms;
using Newtonsoft.Json;
using Newtonsoft.Json.Linq;
namespace EmployeePayDataWk4
{
public partial class Employee_Pay_Form : Form
{
public Employee_Pay_Form()
{
InitializeComponent();
}
private void Employee_Pay_Form_Load(object sender, EventArgs e)
{
EmployeeDataGridView.ColumnCount = 8;
EmployeeDataGridView.Columns[0].Name = "Employee Name";
EmployeeDataGridView.Columns[1].Name = "Zip Code";
EmployeeDataGridView.Columns[2].Name = "Age";
EmployeeDataGridView.Columns[3].Name = "Monthly Gross Pay";
EmployeeDataGridView.Columns[4].Name = "Department ID";
EmployeeDataGridView.Columns[5].Name = "Developer Type";
EmployeeDataGridView.Columns[6].Name = "Annual Taxes";
EmployeeDataGridView.Columns[7].Name = "Annual Net Pay";
}
private void LoadAllButton_Click(object sender, EventArgs e)
{
EmployeeDataGridView.Rows.Clear();
//Read from JSON file
string JSONstring = File.ReadAllText("JSON.json");
List<Employee> employees = JsonConvert.DeserializeObject<List<Employee>>(JSONstring);
//Display into DataGridView
foreach (Employee emp in employees)
{
string[] row = { emp.Name, emp.Zip, emp.Age.ToString(), string.Format("{0:C}", emp.Pay),
emp.DepartmentId.ToString(), SetDevType(emp.DepartmentId),
string.Format("{0:C}", emp.CalculateTax(emp.Pay)),
string.Format("{0:C}", AnnualPay(emp.Pay) - emp.CalculateTax(emp.Pay))};
EmployeeDataGridView.Rows.Add(row);
}
}
private void FTEmployeeButton_Click(object sender, EventArgs e)
{
EmployeeDataGridView.Rows.Clear();
//Read from JSON file
string JSONstring = File.ReadAllText("JSON.json");
List<Employee> employees = JsonConvert.DeserializeObject<List<Employee>>(JSONstring);
//LINQ Query for FT Employees
var FTEmp = from emp in employees
where emp.GetTaxForm == "W2"
select emp;
//Display into DataGridView
foreach (Employee emp in FTEmp)
{
string[] row = { emp.Name, emp.Zip, emp.Age.ToString(), string.Format("{0:C}", emp.Pay),
emp.DepartmentId.ToString(), SetDevType(emp.DepartmentId),
string.Format("{0:C}", emp.CalculateTax(emp.Pay)),
string.Format("{0:C}", AnnualPay(emp.Pay) - emp.CalculateTax(emp.Pay))};
EmployeeDataGridView.Rows.Add(row);
}
}
private void ContractEmployeeButton_Click(object sender, EventArgs e)
{
EmployeeDataGridView.Rows.Clear();
//Read from JSON file
string JSONstring = File.ReadAllText("JSON.json");
List<Employee> employees = JsonConvert.DeserializeObject<List<Employee>>(JSONstring);
//LINQ Query for Contract Employees
var contractEmp = from emp in employees
where emp.GetTaxForm == "1099"
select emp;
//Display into DataGridView
foreach (Employee emp in contractEmp)
{
string[] row = { emp.Name, emp.Zip, emp.Age.ToString(), string.Format("{0:C}", emp.Pay),
emp.DepartmentId.ToString(), SetDevType(emp.DepartmentId),
string.Format("{0:C}", emp.CalculateTax(emp.Pay)),
string.Format("{0:C}", AnnualPay(emp.Pay) - emp.CalculateTax(emp.Pay))};
EmployeeDataGridView.Rows.Add(row);
}
}
//Method to determine developer type
string typeName;
public string SetDevType(int id)
{
if (id == 1)
{
typeName = "Object-Oriented";
}
else if (id == 2)
{
typeName = "Scripts";
}
else { typeName = "Unknown"; }
return typeName;
}
public double AnnualPay(double amount) => 12 * amount;
}
class Employee : IFilingStatus
{
public Employee() { }
public string Name { get; set; }
public string Zip { get; set; }
public int Age { get; set; }
public double Pay { get; set; }
public int DepartmentId { get; set; }
public string GetTaxForm { get; set; }
public double CalculateTax(double basis)
{
double monthlyTax;
if ((GetTaxForm == "W2") || (GetTaxForm == "w2"))
{
monthlyTax = .07 * basis;
}
else
{
monthlyTax = 0;
}
return 12 * monthlyTax;
}
public double AnnualPay(double amount) => 12 * amount;
}
public interface IFilingStatus
{
double CalculateTax(double basis);
}
}
最佳答案
我会说这一行:
List<Employee> employees = JsonConvert.DeserializeObject<List<Employee>>(JSONstring);
应该是
IList<Employee> employees = JsonConvert.DeserializeObject<List<Employee>>(JSONstring);
根据 StyleCop,局部变量、私有(private)字段和参数应为小写驼峰式:
string jsonString = File.ReadAllText("JSON.json");
另一件事是不要重复自己 (DRY)。这几行可以重构为一个单独的方法:
string JSONstring = File.ReadAllText("JSON.json");
List<Employee> employees = JsonConvert.DeserializeObject<List<Employee>>(JSONstring);
//LINQ Query for Contract Employees
var contractEmp = from emp in employees
where emp.GetTaxForm == "1099"
select emp;
在SetDevType
中完全可以使用switch/case,性能更好。
string typeName; // why do we need this?
private string SetDevType(int id)
{
string typeName = string.Empty;
switch(id){
case 1: typeName = "Something"; break;
default: typeName = "Unknown";
}
return typeName;
}
类中有些成员不需要公开,比如
public double AnnualPay(double amount) => 12 * amount;
可以
private double AnnualPay(double amount) => 12 * amount;
并且此方法不知何故也在 Employee
类中,它是一个模型类/POCO。 POCO 类通常不包含非属性(尽管有时包含 ToString 和空构造函数)。
关于C#更好地实现JSON文件,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/53382148/