计算中的每月付款计算器错误 [英] Monthly payment calculator error in computation
问题描述
我创建了一个系统，用于计算每月的贷款支付。一切似乎都没问题，但是当我计算每月并将在datagridview中显示它时，第一个月是可以的，但其余月份只是彼此相同。它应该是不同的，直到达到零余额。
I have created a system that will calculate monthly payment for loan. Everything seems to be okay but when I calculate the monthly and will display it in datagridview first months is okay but the rest months is only the same with each other. It should be different until it reach zero balance.
Months Sequence Cash Flow Interest Principal Balance Interest 2
August, 2018 1 1,555.00 93.64 229.12 7,261.88 87.64
September,2018 2 1,555.00 93.64 229.12 7,261.88 87.64
October, 2018 3 1,555.00 93.64 229.12 7,261.88 87.64
November, 2018 4 1,555.00 93.64 229.12 7,261.88 87.64
December, 2018 5 1,555.00 93.64 229.12 7,261.88 87.64
我的尝试：
这是我的代码
What I have tried:
Here's my code
'Formula Sample
P = MonthPay.Text 'Monthly dblMonthlyPayment
Num = NoTerms.Text 'Number of Terms
r = Val(IntRate.Text / 100) / 12 'Interest dblIntRate
I = r
PVIFA = 1 / I - 1 / (I * (1 + I) ^ Num)
pmt = dblPV / PVIFA
IPmt = (r * intSeq) * (Num ^ 12) * PV.Text
'Clear the ListBox
grd_order.ClearSelection()
Dim dt As New DataTable
Dim dr As DataRow
Dim result As New List(Of String)
dt.Columns.AddRange(New DataColumn() {New DataColumn("Months", GetType(String)), _
New DataColumn("Seq#", GetType(Integer)), _
New DataColumn("Cash Flow", GetType(String)), _
New DataColumn("Interest", GetType(String)), _
New DataColumn("Principal", GetType(String)), _
New DataColumn("Balance", GetType(String)), _
New DataColumn("Interest2", GetType(String))})
Do While intSeq <= intNumOfPayments
CashFlow = dblPV
PI = dblBalance * I
PP = MonthPay.Text - PI
P = dblBalance - PP
interestPaid2 = dblBalance * int2
Dim thisDay As DateTime = DateTime.Today.AddMonths(intSeq)
While thisDay = Now()
thisDay = thisDay.AddMonths(1)
End While
dr = dt.NewRow
dr("Months") = String.Format("{0:Y}", thisDay)
dr("Seq#") = intSeq
dr("Cash Flow") = String.Format("{0:N}", CashFlow)
dr("Interest") = String.Format("{0:N}", PI)
dr("Principal") = String.Format("{0:N}", PP)
dr("Balance") = String.Format("{0:N}", P)
dr("Interest2") = String.Format("{0:N}", interestPaid2)
dt.Rows.Add(dr)
intSeq += 1
Loop
grd_order.DataSource = dt
推荐答案
willfritz是对的 - 值只是在你的循环中改变，所以他们永远不会改变当你显示它们时。
当你遇到这样的问题时，不要求帮助 - 开始使用调试器。使用调试器5分钟会向您显示正在发生的事情，并且比等待其他人更好地查看它更快（并且更有价值）！哎呀，这可能比首先输入问题更快！ ：笑：
编译并不意味着你的代码是正确的！ ：笑：
将开发过程想象成编写电子邮件：成功编译意味着您使用正确的语言编写电子邮件 - 例如英语而不是德语 - 而不是电子邮件包含您的邮件想发送。
所以现在你进入第二阶段的发展（实际上它是第四或第五阶段，但你将在之后的阶段进入）：测试和调试。
首先查看它的作用，以及它与你想要的有何不同。这很重要，因为它可以为您提供有关其原因的信息。例如，如果程序旨在让用户输入一个数字并将其翻倍并打印答案，那么如果输入/输出是这样的：
willfritz is right - the values just don;t change at all inside your loop, so they never change when you display them.
When you get a problem like this, don't ask for help - start with the debugger. 5 minutes with the debugger would have shown you what was happening, and that's a lot quicker (and more rewarding) than waiting for someone else to get round to looking at it! Heck, it's probably quicker than typing the question in the first place! :laugh:
Compiling does not mean your code is right! :laugh:
Think of the development process as writing an email: compiling successfully means that you wrote the email in the right language - English, rather than German for example - not that the email contained the message you wanted to send.
So now you enter the second stage of development (in reality it's the fourth or fifth, but you'll come to the earlier stages later): Testing and Debugging.
Start by looking at what it does do, and how that differs from what you wanted. This is important, because it give you information as to why it's doing it. For example, if a program is intended to let the user enter a number and it doubles it and prints the answer, then if the input / output was like this:
Input Expected output Actual output
1 2 1
2 4 4
3 6 9
4 8 16然后很明显问题出在将它加倍的位 - 它不会将自身加到自身上，或者将它乘以2，它会将它自身相乘并返回输入的平方。
所以，你可以查看代码和很明显它在某处：
Then it's fairly obvious that the problem is with the bit which doubles it - it's not adding itself to itself, or multiplying it by 2, it's multiplying it by itself and returning the square of the input.
So with that, you can look at the code and it's obvious that it's somewhere here:
Private Function Double(ByVal value As Integer) As Integer
Return value * value
End Function
一旦你知道可能会发生什么错误，开始使用调试器找出原因。在方法的第一行放置一个断点，然后运行你的应用程序。当它到达断点时，调试器将停止，并将控制权移交给您。您现在可以逐行运行代码（称为单步执行）并根据需要查看（甚至更改）变量内容（哎呀，您甚至可以更改代码并在需要时再试一次）。 />
在执行代码之前，请考虑代码中的每一行应该做什么，并将其与使用Step over按钮依次执行每一行时实际执行的操作进行比较。它符合您的期望吗？如果是这样，请转到下一行。
如果没有，为什么不呢？它有什么不同？
希望这可以帮助你找到代码的哪个部分有问题，以及问题是什么。
这是一项技能，它是一个值得开发的，因为它可以帮助你在现实世界和发展中。和所有技能一样，它只能通过使用来改善！
Once you have an idea what might be going wrong, start using the debugger to find out why. Put a breakpoint on the first line of the method, and run your app. When it reaches the breakpoint, the debugger will stop, and hand control over to you. You can now run your code line-by-line (called "single stepping") and look at (or even change) variable contents as necessary (heck, you can even change the code and try again if you need to).
Think about what each line in the code should do before you execute it, and compare that to what it actually did when you use the "Step over" button to execute each line in turn. Did it do what you expect? If so, move on to the next line.
If not, why not? How does it differ?
Hopefully, that should help you locate which part of that code has a problem, and what the problem is.
This is a skill, and it's one which is well worth developing as it helps you in the real world as well as in development. And like all skills, it only improves by use!
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