EMI计算(数据类型中的问题) [英] EMI Calculation (Problem in DataType)

问题描述

嗨..

在我的应用程序中，我实现了EMI计算器..
在整个计算中使用十进制数据类型..
在这里我没有得到完整的十进制值.所以总值不匹配..
我用long，double替换了类型....仍然没有得到...
请给我一个好的解决方案..
在此先感谢

这是我的代码

Hi..

In my application I have implemented EMI calculator..
Decimal datatype is used throughout the calculation..
Here I am not getting the full decimal value..So the Total value getting mismatch..
I have replaced the type with long,double….Still not getting…
Please give me a good solution..
Thanks in advance

This is my code

Decimal Interest = 0;
EMICalculatorList ECL = new EMICalculatorList();
ProjectXBL.Config.EMICalculator EC;
int month = ddlMonth.SelectedValue.ToNonNullInt32() - 1;
Decimal Intr = 0; Decimal Mth = txtPrinciple.Text.ToNonNullDecimal() / txtLoanDurationInMonths.Text.ToNonNullDecimal();
Decimal principalAmt = txtPrinciple.Text.ToNonNullDecimal();
Decimal ROI = txtAnnualInterestRate.Text.ToNonNullDecimal();
Decimal months = txtLoanDurationInMonths.Text.ToNonNullDecimal();
Decimal F = 1200;
for (int i = 0; i < txtLoanDurationInMonths.Text.ToNonNullInt32(); i++)
{
    Intr = Intr + ((principalAmt * ROI) / F);
    principalAmt = principalAmt - Mth;
}
Decimal Avg = Intr.ToNonNullDecimal() / txtLoanDurationInMonths.Text.ToNonNullDecimal();
Decimal MonthlyPayment = ((txtPrinciple.Text.ToNonNullDecimal() / txtLoanDurationInMonths.Text.ToNonNullDecimal()) + Avg).ToNonNullDecimal();
lblMonthlypaymentsText.Text = string.Empty;
lblMonthlypaymentsText.Text = Math.Round(MonthlyPayment, 2).ToString();
lblAnnulLoanPaymentsText.Text = string.Empty;
lblAnnulLoanPaymentsText.Text = Math.Round((MonthlyPayment * 12), 2).ToString();
Decimal Ending = 0;
Decimal Ints = 0;
Decimal Prnc = 0;
Decimal CumPrinc = 0;
Decimal CumIntr = 0;
Decimal principal = txtPrinciple.Text.ToNonNullDecimal();
for (int i = 0; i < txtLoanDurationInMonths.Text.ToNonNullInt32(); i++)
{
    EC = new ProjectXBL.Config.EMICalculator();
    //No,Month And Year
    EC.No = i + 1;
    if (month == 0)
    {
        EC.Year = txtYear.Text;
    }
    if (month == 12)
    {
        month = 0;
        EC.Year = (txtYear.Text.ToNonNullInt32() + 1).ToString();
    }
    EC.Month = ddlMonth.Items[month].Text; month++;
    //
    EC.Payment = MonthlyPayment;
    Decimal monthly = principal / months;
    //Monthly payable without interest
    EC.Interest = (principal * ROI) / 1200;
    Ints = EC.Interest;
    principal = principal - monthly;
    Interest = Interest + EC.Interest;
    EC.Principal = MonthlyPayment - EC.Interest;
    Prnc = EC.Principal;
    if (i == 0)
    {
        EC.BeginningBalance = txtPrinciple.Text.ToNonNullDecimal();
        EC.CumulativePrinciple = EC.Principal;
        EC.CumulativeInterest = EC.Interest;
    }
    else
    {
        EC.BeginningBalance = Ending;
        EC.CumulativePrinciple = CumPrinc + Prnc;
        EC.CumulativeInterest = CumIntr + Intr;
    }
    EC.Ending = EC.BeginningBalance - EC.Principal;
    Ending = EC.Ending;
    CumPrinc = EC.CumulativePrinciple;
    CumIntr = EC.CumulativeInterest;
    ECL.Add(EC);
}
lstEMI.DataSource = ECL;
lstEMI.DataBind();

推荐答案

尝试使用BigInt.它几乎可以容纳无限数量.它是在4.0和System.Numeric命名空间中引入的

Try Using the BigInt. It can hold virtually infinite number. it is introduced with 4.0 and in the System.Numeric namespace

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