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hi,
declare your variables in the subroutines.
in your case this should be enougth..
sub main()
dim tmp as string
dim dLaborCost as double
' read the line
tmp = GetPrice()
' cast the value of tmp to an double
dLaborCost = CDbl(tmp)
'......
end sub
private function GetPrice() as string
GetPrice = Console.ReadLine
end function
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ok so basically i need to use byVal or byRef to pass my
data between subroutines.
i just dnt know th correct syntax
so i have
module mod1
sub main()
'where i call all my subroutines
*****I DNT KNOW HOW TO THEN CALL MY SUBROUTINE
call getData(dFeet)
???????
end sub
********THIS IS WHAT I DONT KNOW IF IT IS CORRECT
sub getData(byVal dim dFeet as double)
??????????
console.writeline("eneter feet: ")
dFeet = console.readline
'and so on for other data being asked to be entered
end sub
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I suggest you buy and study a book on VB.NET; you can't get familiar with a programming language by asking hundreds of questions one by one. Choose a book you like, study it, and in one week's time you will be a lot wiser.
Luc Pattyn
I only read code that is properly indented, and rendered in a non-proportional font; hint: use PRE tags in forum messages
Local announcement (Antwerp region): Lange Wapper? Neen!
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Try this:
Module Module1
Public Sub DoSomething(ByVal x As Integer, ByRef y As Double)
x += 5
y += 5
End Sub
Sub Main()
Dim num As Integer = 5
Dim anum As Double = 20.5
Console.WriteLine("Before Method Call:")
Console.WriteLine("Num = {0}", num.ToString)
Console.WriteLine("Anum = {0}", anum.ToString)
DoSomething(num, anum)
Console.WriteLine("After Method Call:")
Console.WriteLine("Num (After ByVal) = {0}", num.ToString)
Console.WriteLine("Anum (After ByRef) = {0}", anum.ToString)
Console.Read()
End Sub
End Module
What happens?
This is what ByVal and ByRef do.
I don't speak Idiot - please talk slowly and clearly
I don't know what all the fuss is about with America getting it's first black president. Zimbabwe's had one for years and he's sh*t. - Percy Drake , Shrewsbury
Driven to the arms of Heineken by the wife
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I am trying to filter a database with column name as "JoiningDate" with the following command:
cmd = New SqlCommand("SELECT * FROM student_details WHERE [JoiningDate] <= '" + fordate + "'", con)
Where,
Date format in my database is in "dd/MM/yyyy"
and
"fordate" is "10/01/2009", which is As string
Now while filtering, I am getting number of rows different than the exact ones ( This I know).
e.g. for the above filter of "joiningDate" less than "10/01/2009", I found that some other dates like
01/10/2009 is also get counted while filtering. I think there is something wrong in my implementation of
filtering.
I am pasting my code here :
Dim datetime_from As Date, fordate As String
' read the date from datetimepicker control "DateTimefrom"
datetime_from = DateTimefrom.Value.ToShortDateString
fordate = datetime_from.ToString("dd/MM/yyyy")
' then filtering by
cmd = New SqlCommand("SELECT * FROM student_details WHERE [JoiningDate] <= '" + fordate + "'", con)
Can Someone help me with a piece of code?
Thanks
R.S
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Why are you storing dates as strings in database? Now, you will need to perform conversions in order to calculate the dates lesser than selected dates.
It's not necessary to be so stupid, either, but people manage it. - Christian Graus, 2009 AD
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Can You please tell me how to do this?
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jeshra279 wrote: Can You please tell me how to do this?
Store dates as dates not as strings. You are comparing character strings not dates, which means the results will not be what you expect. Take a look at the documentation with particular reference to date values, rather than date formats.
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Can You or Someone Please provide me a line of code for this? I am stuck at this point.
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You really should not do it this way. If you have a DateTime datatype, use it. Don't you think so?
jeshra279 wrote: Please provide me a line of code for this?
It cannot be done in a line of code with current approach.
Still if you want to stick to this way, then since the values in your database are as strings, then break up the date, month and year part into seprate integers. Then start with checking that if the year is less than selected dates year. If the year is same, check for month and if month is same check for date. Then you can filter out the records.
Don't you think this is bit insane way of comparing dates?
PS: There can be a decent way than what I have told. But anyways, IMHO if you use strings to store dates, you are wrong no matter how you comapre them.
It's not necessary to be so stupid, either, but people manage it. - Christian Graus, 2009 AD
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when you have a number (say 123) you store it in a numeric field, not as a "123" string, right?
when you have a date, a time, a datetime, you store it in a field of the corresponding type, and NOT as a string; doing it properly avoids all confusion about date formats and regional settings; and it saves CPU cycles.
If your database table has a date field that is currently a string, and you want it to become a date, then this should do it:
- add a new field, with a new name;
- issue an SQL command to populate the new field based on your interpretation of the existing field;
- adapt your code base to use the new field, and no longer use the old one;
- delete the old field;
- optionally rename the new field in the database table and in your code.
That is a one time effort saving you all the trouble you would otherwise have later when your app grows, and/or when your app gets used by more people in more countries.
Luc Pattyn
I only read code that is properly indented, and rendered in a non-proportional font; hint: use PRE tags in forum messages
Local announcement (Antwerp region): Lange Wapper? Neen!
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Hey all. On a new project im working on, which is about learning different areas in visual basic. One area i havnt covered much is writing to files and retrieving data from a file.
The program includes a update script, which justs checks online reading a .txt file to see if the version is newer, if so it downloads it. I would like to record when the user last press 'checkForUpdates' button. Take the timestamp and write it into a xml file. Then using a simple label to read and display the value inside the xml file.
I have been trying a few ways, just using timestamps i got lost.
Any help would be great,
Malcom
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If you can show the code you have tried and explain the exact problem, it would be easier to help.
It's not necessary to be so stupid, either, but people manage it. - Christian Graus, 2009 AD
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Oki sorry, i asked wrong, i have tried some ways but i can't get it to work.
Im not asking for the answer, that be wrong. I am learning it afterall.
I just need say a boost. Just a quick example of generating a timestamp then i can work it from there. I did some googling and believe i can easly use xml file just need to mess around abit more.
Sorry for the bother,
Malcom
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DateTime.Now will give you the current time. And then, you can look into the classes in System.Xml namespace to write that to your XML file.
It's not necessary to be so stupid, either, but people manage it. - Christian Graus, 2009 AD
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hi, here i have some code for you.
System.Xml.XmlTextWriter xtw = new System.Xml.XmlTextWriter("C:\\timestamp.xml",System.Text.Encoding.UTF8);
xtw.Formatting == Formatting.Indented;
xtw.Indentation = 4;
xtw.WriteStartDocument();
xtw.WriteStartElement("TimeStamp");
xtw.WriteString(DateTime.Now.ToOADate());
xtw.WriteEndElement();
xtw.WriteStartElement("TimeStampAttribute");
xtw.WriteAttributeString("DateTime",DateTime.Now.ToOADate())
xtw.WriteEndElement();
xtw.WriteEndDocument();
xtw.Close();
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Thank you both. I tried your hints and i got somewhere. Just was hitting a wall. The time was coming out as 00:00:00. To the last replier before me, thanks for the code, really gave me a better idea of what i needed. Thanks 
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Can anybody tells me how to export data to excel...
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Google is your freind...
Google[^]
I don't speak Idiot - please talk slowly and clearly
I don't know what all the fuss is about with America getting it's first black president. Zimbabwe's had one for years and he's sh*t. - Percy Drake , Shrewsbury
Driven to the arms of Heineken by the wife
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From what?
Though as has already been posted, Google gives thousands of examples.
If you have knowledge, let others light their candles at it.
Margaret Fuller (1810 - 1850)
www.JacksonSoft.co.uk
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Hi All,
The problem discussed is accuracy of numeric calculation result,in vb. Namely I compare a numerical mathematic calculation, Line-Plan Intersection, in VB.Net and Matlab 2007 the intersection point result specify in below:
Matlab 2007 ---->
Xint=2.000000000000000
Yint=1.333333333333333
Zint=0.666666666666667
But
Visual Studio 2008 ---->
Xint=2.0000000596046448
Yint=1.3333333730697632
Zint=0.66666668653488159
Hand Calculated ----->
Xint=2.000000000000000
Yint=1.333333333333333
Zint=0.666666666666667
why this error occurs?
Thanks.
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What Data Type are you using for the calculation in VB?.
This looks like an issue with floating-point precision.
I don't speak Idiot - please talk slowly and clearly
I don't know what all the fuss is about with America getting it's first black president. Zimbabwe's had one for years and he's sh*t. - Percy Drake , Shrewsbury
Driven to the arms of Heineken by the wife
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story that I have been explained.
Private Sub btnCalc_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Button6.Click<br />
<br />
Dim xint, yint, zint As Double<br />
Dim ob, oc, ratio As Double<br />
Dim havint, intree As Boolean<br />
Dim x0, x1, x2, xb As Double<br />
Dim y0, y1, y2, yb As Double<br />
Dim z0, z1, z2, zb As Double<br />
'3 Points of plane formed triangle too.<br />
x0 = 2<br />
x1 = 2<br />
x2 = 2<br />
y0 = 0<br />
y1 = 2<br />
y2 = 2<br />
z0 = 2<br />
z1 = 2<br />
z2 = 0<br />
'Points of line, other point is [0 0 0]<br />
xb = 1.5<br />
yb = 1<br />
zb = 0.5<br />
Line_PlaneIntersection(x0, x1, x2, xb, y0, y1, y2, yb, z0, z1, z2, _<br />
zb, xint, yint, zint, ob, oc, ratio, havint, intree)<br />
End Sub<br />
<br />
<br />
Public Shared Sub Line_PlaneIntersect(ByVal X0 As Double, ByVal X1 As Double, ByVal X2 As Double, _<br />
ByVal Xb As Double,ByVal Y0 As Double, ByVal Y1 As Double, _<br />
ByVal Y2 As Double,ByVal Yb As Double,ByVal Z0 As Double, _ <br />
ByVal Z1 As Double, ByVal Z2 As Double, ByVal Zb As Double, _<br />
ByRef Xint As Double, ByRef Yint As Double, ByRef Zint As _<br />
Double, ByRef OB As Double, ByRef OC As Double, ByRef Ratio As _<br />
Double,ByRef HaveIntersect As Boolean, ByRef InTriangle As _<br />
Boolean) <br />
<br />
<br />
'***This formulation is based on the assumption that the line starts from origin***<br />
Dim t, u, v As Single<br />
Dim A, B, C, D As Double<br />
Dim A1, A2, A3 As Double<br />
Dim B1, B2, B3 As Double<br />
Dim C1, C2, C3 As Double<br />
Dim D1, D2, D3, D4, D5, D6 As Double<br />
Dim Det As Double<br />
Dim Det1, Det2, Det3, Det4, Det5, Det6 As Double<br />
Det1 = +Xb * ((Y2 - Y0) * (Z1 - Z0))<br />
Det2 = +Yb * ((X1 - X0) * (Z2 - Z0))<br />
Det3 = +Zb * ((X2 - X0) * (Y1 - Y0))<br />
Det4 = -Xb * ((Y1 - Y0) * (Z2 - Z0))<br />
Det5 = -Yb * ((X2 - X0) * (Z1 - Z0))<br />
Det6 = -Zb * ((X1 - X0) * (Y2 - Y0))<br />
Det = Det1 + Det2 + Det3 + Det4 + Det5 + Det6<br />
If Det <> 0 Then<br />
HaveIntersect = True<br />
A1 = X0 * ((Y1 - Y0) * (Z2 - Z0) - (Y2 - Y0) * (Z1 - Z0))<br />
A2 = Y0 * ((X2 - X0) * (Z1 - Z0) - (X1 - X0) * (Z2 - Z0))<br />
A3 = Z0 * ((X1 - X0) * (Y2 - Y0) - (X2 - X0) * (Y1 - Y0))<br />
A = A1 + A2 + A3<br />
B1 = X0 * ((Y2 - Y0) * Zb - (Z2 - Z0) * Yb)<br />
B2 = Y0 * ((Z2 - Z0) * Xb - (X2 - X0) * Zb)<br />
B3 = Z0 * ((X2 - X0) * Yb - (Y2 - Y0) * Xb)<br />
B = B1 + B2 + B3<br />
C1 = X0 * ((Z1 - Z0) * Yb - (Y1 - Y0) * Zb)<br />
C2 = Y0 * ((X1 - X0) * Zb - (Z1 - Z0) * Xb)<br />
C3 = Z0 * ((Y1 - Y0) * Xb - (X1 - X0) * Yb)<br />
C = C1 + C2 + C3<br />
D1 = +(X1 - X0) * (Y2 - Y0) * Zb<br />
D2 = +(X2 - X0) * (Z1 - Z0) * Yb<br />
D3 = +(Y1 - Y0) * (Z2 - Z0) * Xb<br />
D4 = -(X1 - X0) * (Z2 - Z0) * Yb<br />
D5 = -(X2 - X0) * (Y1 - Y0) * Zb<br />
D6 = -(Y2 - Y0) * (Z1 - Z0) * Xb<br />
D = D1 + D2 + D3 + D4 + D5 + D6<br />
t = A / D<br />
u = B / D<br />
v = C / D<br />
Xint = Xb * t<br />
Yint = Yb * t<br />
Zint = Zb * t <br />
Dim Bool1, Bool2 As Boolean<br />
Bool1 = u >= 0 And u <= 1<br />
Bool2 = v >= 0 And v <= 1<br />
If Bool1 And Bool2 And (u + v) <= 1 Then<br />
'means that the intersection point lies in triangle wroughted from 3 points of the plane <br />
InTriangle = True<br />
OB = Math.Sqrt(Xb ^ 2 + Yb ^ 2 + Zb ^ 2)<br />
OC = Math.Sqrt(Xint ^ 2 + Yint ^ 2 + Zint ^ 2)<br />
Ratio = OB / OC<br />
End If<br />
Else<br />
HaveIntersect = False<br />
InTriangle = False<br />
End If<br />
End Sub
'***************************************************************************************************
%MAtlab syntax
clear<br />
%Data of line's points<br />
xa=0; ya=0; za=0;<br />
xb=1.5; yb=1; zb=1/2;<br />
Ia=[xa<br />
ya<br />
za];<br />
Ib=[xb<br />
yb<br />
zb];<br />
%Data of plane<br />
x0=2; y0=0; z0=2;<br />
x1=2; y1=2; z1=2;<br />
x2=2; y2=2; z2=0;<br />
A= [xa-xb x1-x0 x2-x0<br />
ya-yb y1-y0 y2-y0<br />
za-zb z1-z0 z2-z0];<br />
B=[xa-x0<br />
ya-y0<br />
za-z0];<br />
<br />
C=A^-1*B<br />
<br />
Iint=Ia+(Ib-Ia).*C(1)<br />
after running
Iint =
2.0000
1.3333
0.6667
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I am not Maths Wizard (I was a terrible student), but this excerpt from Wikipaedia on double floating point precision may shed some light...
Why do you need to return so many digits after the decimal point?
Addition and subtraction
A simple method to add floating-point numbers is to first represent them with the same exponent. In the example below, the second number is shifted right by three digits, and we then proceed with the usual addition method:
123456.7 = 1.234567 * 10^5
101.7654 = 1.017654 * 10^2 = 0.001017654 * 10^5
Hence:
123456.7 + 101.7654 = (1.234567 * 10^5) + (1.017654 * 10^2)
= (1.234567 * 10^5) + (0.001017654 * 10^5)
= (1.234567 + 0.001017654) * 10^5
= 1.235584654 * 10^5
In detail:
e=5; s=1.234567 (123456.7)
+ e=2; s=1.017654 (101.7654)
e=5; s=1.234567
+ e=5; s=0.001017654 (after shifting)
--------------------
e=5; s=1.235584654 (true sum: 123558.4654)
This is the true result, the exact sum of the operands. It will be rounded to seven digits and then normalized if necessary. The final result is
e=5; s=1.235585 (final sum: 123558.5)
Note that the low 3 digits of the second operand (654) are essentially lost. This is round-off error. In extreme cases, the sum of two non-zero numbers may be equal to one of them:
e=5; s=1.234567
+ e=-3; s=9.876543
e=5; s=1.234567
+ e=5; s=0.00000009876543 (after shifting)
----------------------
e=5; s=1.23456709876543 (true sum)
e=5; s=1.234567 (after rounding/normalization)
Another problem of loss of significance occurs when two close numbers are subtracted. In the following example e = 5; s = 1.234571 and e = 5; s = 1.234567 are representations of the rationals 123457.1467 and 123456.659.
e=5; s=1.234571
- e=5; s=1.234567
----------------
e=5; s=0.000004
e=-1; s=4.000000 (after rounding/normalization)
The best representation of this difference is e = −1; s = 4.877000, which differs more than 20% from e = −1; s = 4.000000. In extreme cases, the final result may be zero even though an exact calculation may be several million. This cancellation illustrates the danger in assuming that all of the digits of a computed result are meaningful. Dealing with the consequences of these errors is a topic in numerical analysis; see also Accuracy problems.
I don't speak Idiot - please talk slowly and clearly
I don't know what all the fuss is about with America getting it's first black president. Zimbabwe's had one for years and he's sh*t. - Percy Drake , Shrewsbury
Driven to the arms of Heineken by the wife
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I find answer! Because I had been defined three of variable as Single.
However this variables were small but when I change those to Double this problem solved.
Dim t, u, v As Single -------->Dim t, u, v As Double
Sorry
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