CODEWITHSUNDEEP
vb.nettrim
TheIronCheek
TheIronCheek

Reputation: 1149

How to get TrimEnd() to stop at a specific character

I have a series of percentage values saved in a database that look something like this:

Percentage
_____________
100.00000
50.00000
74.02500

When I display the values to the screen, I'd like to trim unnecessary zeroes from the end of the string along with the decimal point so the above examples become:

Percentage
_____________
100
50
74.025

I'm currently using the following code:

displayVal = rawVal.TrimEnd({"0"c, "."c})

but this code continues to trim after the decimal if there are additional zeroes. I also tried:

displayVal = rawVal.TrimEnd(New String({"0", "."}))

which almost works. It just leaves the decimal point.

Is there a way to do what I want using TrimEnd() or do I need to switch to regex?

Upvotes: 1

Views: 189

Answers (1)

Steven Doggart
Steven Doggart

Reputation: 43743

As Tim already mentioned in the comments, if the data type in the DB is already some numerical type, it would be best to keep it in that type and then use the appropriate numeric formatting when converting it to a string for output. If, however, the input data is already a string, then that's not an option. In that cast, the simplest option is to just do two trims in series, like this:

Private Function RemoveUnecessaryZeros(input As String) As String
    Return input.TrimEnd("0"c).TrimEnd("."c)
End Function

However, that doesn't give you a lot of flexibility, it doesn't remove preceding zeros, and it does nothing to reformat the string using the current culture. If that matters, you could instead parse the value into a numeric type and then use the desired string formatting options to re-output it to a string. For instance:

Private Function RemoveUnecessaryZeros(input As String) As String
    Dim result As Double
    If Double.TryParse(input, result) Then
        Return result.ToString()
    Else
        Return input
    End If
End Function

However, when you do it that way, you may potentially lose precision along the way, depending on the input numbers and the data type you choose to parse it with. If you need more control over the parsing/reformatting and you want to keep it purely in strings so no precision is lost, then you may want to consider using regex. For instance:

Private Function RemoveUnecessaryZeros(input As String) As String
    Dim m As Match = Regex.Match(input, "[1-9]\d*(\.([1-9]|0+[1-9])+)?")
    If m.Success Then
        Return m.Value
    Else
        Return input
    End If
End Function

Upvotes: 1

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