I am working on this simple VB.net application that allows users to enter a decimal number (ex - 1.23) and the program lists what the digits are to the left and right of the decimal and as well as how many actual digits are to the left and right of the decimal. Here is what my code looks like so far.
Dim imput As String
imput = txtEnterNumber.Text
Dim D As Integer
D = txtEnterNumber.Text.IndexOf(".")
txtNumLeft.Text = imput.Remove(D)
txtNumRight.Text = imput.Remove(0, D)
txtDigitLeft.Text = CStr(imput.Substring(0, D).Length)
txtDigitRight.Text = CStr(imput.Substring(D).Length)
When I run my program it includes the decimal point for what the digits are to right of the decimal and as well as how many digits are right of the decimal. Why is this?
Thanks!
Try this
Dim imput As String
imput = txtEnterNumber.Text
Dim D As Integer
D = txtEnterNumber.Text.IndexOf(".")
txtNumLeft.Text = imput.Remove(D)
txtNumRight.Text = imput.Remove(0, D + 1)
txtDigitLeft.Text = CStr(imput.Substring(0, D).Length)
txtDigitRight.Text = CStr(imput.Substring(D + 1).Length)
Your substring is taking a substring from the decimal point, giving a +1 position in SUBSTRING will solve it. Also, you can simplify your code using SPLIT function:
Dim num() as string = txtEnterNumber.Text.Split(".")
txtNumLeft.Text = num(0)
txtDigitLeft.Text = num(0).Length
If(num.Length > 1)Then
txtNumRight.Text = num(1)
txtDigitRight.Text = num(1).Length
Else
txtNumRight.Text = ""
txtDigitRight.Text = 0
End If
Related
I want to extract characters from a string. However, the string doesn't have the same length every time.
Basically I get data from a database and I want extract the value I need in it. But I'm stuck on step where I have to extract the right value.
So first I get data like that :
infoDataset2 = accessRequet_odbc("select st_astext(st_snaptogrid(geom, 0.01)) from netgeo_point_tech", myConnection)
The result is something like : POINT(921021.98 6671778.45). What I need are the 2 figures, but their length is not fixed. I just want to remove POINT( and ).
Then I work on each line of the DataSet I get to cast each lines into a string with only the value needed.
For i = 0 To infoDataset2.Tables(0).Rows.Count - 1
geomPt = infoDataset2.Tables(0).Rows(i).ItemArray(0).Substring(geomPt = infoDataset2.Tables(0).Rows(i).ItemArray(0).Substring(1 + infoDataset2.Tables(0).Rows(i).ItemArray(0).LastIndexOf("(")))
Console.WriteLine(geomPt)
Next
This was my last try, where I was able to remove POINT( but I'm struggling with the length to cut ).
I want to learn from this, so, if possible, explain to me what I'm doing wrong here, or if my approach is lacking insight.
It will be horrible to debug that one long line. It makes no difference to the computer if you split it up into easy-readable parts. Here's some code to get the x- and y-coordinates from a string formatted as shown in the question:
Dim s = "POINT(921021.98 6671778.45)"
Dim b1 = s.IndexOf("("c) + 1
Dim b2 = s.IndexOf(")"c, b1) - 1
Dim parts = s.Substring(b1, b2 - b1 + 1).Split({" "c})
Dim x As Decimal = Decimal.Parse(parts(0))
Dim y As Decimal = Decimal.Parse(parts(1))
Another way of parsing the string is to use a regular expression, which can be more flexible. In this example, I used named capture groups to make it easy to see which parts are for the x and y:
Dim s = "POINT(921021.98 6671778.45)"
Dim x As Decimal
Dim y As Decimal
Dim re = New Regex("\((?<x>[0-9-.]+) (?<y>[0-9-.]+)\)")
Dim m = re.Match(s)
If m.Success Then
x = Decimal.Parse(m.Groups("x").Value)
y = Decimal.Parse(m.Groups("y").Value)
Else
' Could not parse point. Do something about it if required.
End If
Andrew Morton has given a nice answer, i upvoted that one, if you need an even easier way and that was still complicated use this
Dim s = "POINT(921021.98 6671778.45)"
Dim part1 As String = s.Remove(0, 6)
Dim part2 As String = part1.Substring(0, part1.Length - 1)
Dim split() As String = part2.Split(" ")
Dim x = split(0)
Dim y = split(1)
Here is an even probably easier to understand solution:
Dim s as String = "POINT(921021.98 6671778.45)"
Dim coordinate() as String = s.Replace("POINT(", "").Replace(")", "").Split(" ")
Enjoy!
just have a problem with visual basic. I was trying to make a simple conversion program, but the program keeps giving me the wrong answer. I've also coded this out in Java, where it gave me the answer I was expecting. I don't know what's going on here. Here is a sample of the code. The numbers I have been entering are 10 for alienYears to convert, 440 for alienYearLength and 25.5 for alienDayLength. It should come out to about 12. something but I keep getting 218.something.
Thank you
Dim alienYears As Decimal = CDec(txtYearsToConvert.Text)
Dim alienYearLength As Decimal = CDec(txtDaysOnAlienPlanet.Text)
Dim alienDayLength As Decimal = CDec(txtDaysOnAlienPlanet.Text)
Dim alienHoursYears As Decimal = ((alienYears * alienYearLength) * alienDayLength)
Dim earthHourYears As Decimal = (8851.25)
Dim earthConversion As Decimal = (alienHoursYears / earthHourYears)
txtAgeOnEarth.Text = CStr(earthConversion)
Looks like a typo here (vvvv), do you mean txtYears(something)? There's no way this could be meaningful using the input from the same textbox.
Dim alienYears As Decimal = CDec(txtYearsToConvert.Text)
' VVVV
Dim alienYearLength As Decimal = CDec(txtDaysOnAlienPlanet.Text)
Dim alienDayLength As Decimal = CDec(txtDaysOnAlienPlanet.Text)
Dim alienHoursYears As Decimal = ((alienYears * alienYearLength) * alienDayLength)
Dim earthHourYears As Decimal = (8851.25)
Dim earthConversion As Decimal = (alienHoursYears / earthHourYears)
txtAgeOnEarth.Text = CStr(earthConversion)
The following Function creates multiple series for a graph.
Function createSeries(ByVal fileNames() As String, ByVal intValXAxis As Integer, ByVal intValYAxis As Integer) As Series()
Dim ChartSeries(fileNames.Count) As Series
Dim i As Integer = 0
For Each filename In fileNames
ChartSeries(i) = New Series
ChartSeries(i).ChartType = SeriesChartType.FastLine
ChartSeries(i).ChartArea = "ChartArea1"
If filename.Contains("NOK") = True Then
ChartSeries(i).Color = Color.Red
ChartSeries(i).BorderWidth = 3
Else
ChartSeries(i).Color = Color.Green
End If
Dim fileReader = My.Computer.FileSystem.OpenTextFileReader(filename)
Do
Dim strCache() As String = Split(fileReader.ReadLine(), ";")
ChartSeries(i).Points.AddXY(strCache(intValXAxis - 1), strCache(intValYAxis - 1))
ChartSeries(i).ToolTip = filename
Loop Until fileReader.EndOfStream = True
i += 1
Next
Return ChartSeries
End Function
The problem I have is, that the Y-Values of the Series I create are mostly something like that: 0,09440104 or 0,1757813. I need these Values shown on the graph as they are, but the zero's got removed and the Y-Point-Values are : 9440104 or 1757813
I tried to format them with "Globalization" before adding them to the Series, but it doesn't solved the problem.
Just to be clear: I want the numbers as shown above(0,09440104 and 0,1757813) to be the Y-values of the points.
How can i solve the problem?
Thanks in advance.
By default, En-US culture will read your comma "," as thousand separator and thus taking your data as > 0 rather than < 0.
You have two options: change the culture or change the string format. If all your numbers are less than 1000 (or, to be more precise, not having . as thousand separator), I recommend simply to replace , with .
Dim strCache() As String = Split(fileReader.ReadLine(), ";")
Dim repStrX = strCache(intValXAxis - 1).Replace(",",".")
Dim repStrY = strCache(intValYAxis - 1).Replace(",",".")
ChartSeries(i).Points.AddXY(repStrX , repStrY)
Or, if they are having value more than 1000 (or, again, to be more precise, not having . as thousand separator), without specifying the culture, you could also use Replace with some tricks: making use of non-existing character as intermediate value to flip between . and , in the original string.
Dim strCache() As String = Split(fileReader.ReadLine(), ";")
Dim repStrX = strCache(intValXAxis - 1).Replace(",","G").Replace(".",",").Replace("G",".")
Dim repStrY = strCache(intValYAxis - 1).Replace(",","G").Replace(".",",").Replace("G",".")
ChartSeries(i).Points.AddXY(repStrX , repStrY)
Examples:
Double-Number is 56.6789 result should be 4
Double-Number is 12345.67 result should be 2
Double-Number is 12345.6 result should be 1
I have a solution tinkering with strings, but I think there is an mathematical solution?
Please in VB.NET ...
Split the original number and get the length of the upper index (1)
myNumber = 12.3456
Dim count As Integer = Len(Split(CStr(myNumber), Application.DecimalSeparator)(1))
Debug.Print count // prints '4'
edit: replaced "." with decimal separator to ensure use across varying cultures
You can try like this:
Dim x As String = CStr(56.6789)
Dim count = x.Length - InStr(x, ".")
One way to do it is to keep knocking off the whole part, multiplying by 10, repeat until you have an integer:
Dim x As Double = 1.23456
Dim count As Integer = 0
While Math.Floor(x) <> x
x = (x - Math.Floor(x)) * 10D
count = count + 1
End While
Note this will fail if there is an infinite number of decimal places - so you could set a limit on it (If count > 100 Then Exit While)
Another way would be like this, which converts to a string but removes the need to hardcode the separator.
Dim x As Double = 1.23456
Dim x0 As Double = x - Math.Floor(x)
Dim x0String As String = x0.ToString()
Dim count As Integer = x0String.Substring(2, x0String.Length - 2).Length
Using Application.DecimalSeparator also allows a string to be used.
The method with a string will again lose information about an infinite-length fractional part, as it will truncate it.
How do I check how many decimal places a number has in VB.NET?
For example: Inside a loop I have an if statement and in that statement I want to check if a number has four decimal places (8.9659).
A similar approach that accounts for integer values.
Public Function NumberOfDecimalPlaces(ByVal number As Double) As Integer
Dim numberAsString As String = number.ToString()
Dim indexOfDecimalPoint As Integer = numberAsString.IndexOf(".")
If indexOfDecimalPoint = -1 Then ' No decimal point in number
Return 0
Else
Return numberAsString.Substring(indexOfDecimalPoint + 1).Length
End If
End Function
Dim numberAsString As String = myNumber.ToString()
Dim indexOfDecimalPoint As Integer = numberAsString.IndexOf(".")
Dim numberOfDecimals As Integer = _
numberAsString.Substring(indexOfDecimalPoint + 1).Length
Public Shared Function IsInSignificantDigits(val As Double, sigDigits As Integer)
Dim intVal As Double = val * 10 ^ sigDigits
Return intVal = Int(intVal)
End Function
For globalizations ...
Public Function NumberOfDecimalPlaces(ByVal number As Double) As Integer
Dim numberAsString As String = number.ToString(System.Globalization.CultureInfo.InvariantCulture)
Dim indexOfDecimalPoint As Integer = numberAsString.IndexOf(".")
If (indexOfDecimalPoint = -1) Then ' No decimal point in number
Return 0
Else
Return numberAsString.Substring(indexOfDecimalPoint + 1).Length
End If
End Function
Some of the other answers attached to this question suggest converting the number to a string and then using the character position of the "dot" as the indicator of the number of decimal places. But this isn't a reliable way to do it & would result in wildly inaccurate answers if the number had many decimal places, and its conversion to a string contained exponential notation.
For instance, for the equation 1 / 11111111111111111 (one divided by 17 ones), the string conversion is "9E-17", which means the resulting answer is 5 when it should be 17. One could of course extract the correct answer from the end of the string when the "E-" is present, but why do all that when it could be done mathematically instead?
Here is a function I've just cooked up to do this. This isn't a perfect solution, and I haven't tested it thoroughly, but it seems to work.
Public Function CountOfDecimalPlaces(ByVal inputNumber As Variant) As Integer
'
' This function returns the count of deciml places in a number using simple math and a loop. The
' input variable is of the Variant data type, so this function is versatile enougfh to work with
' any type of input number.
'
CountOfDecimalPlaces = 0 'assign a default value of zero
inputNumber = VBA.CDec(inputNumber) 'convert to Decimal for more working space
inputNumber = inputNumber - VBA.Fix(inputNumber) 'discard the digits left of the decimal
Do While inputNumber <> VBA.Int(inputNumber) 'when input = Int(input), it's done
CountOfDecimalPlaces = CountOfDecimalPlaces + 1 'do the counting
inputNumber = inputNumber * 10 'move the decimal one place to the right
Loop 'repeat until no decimal places left
End Function
Simple...where n are the number of digits
Dim n as integer = 2
Dim d as decimal = 100.123456
d = Math.Round(d, n);
MessageBox.Show(d.ToString())
response: 100.12