Bug Using String.format() - Automatically Add Crlf After Char Position 1025?
Jul 6, 2011
I've found this strange behaviour, I'n using vS2010 sp1 on a x64 machine in a class library project (VB) with FM4.0, there is added automatically an crlf to the concatenated string using string.format() funtion, you can reproduce this using the following lines of code:
i want to read a character 1 by 1 from a string and a .txt, i know all about stream reader so for the .txt would it be somthing like:textbox1.text = tr.readCharacter.i would prefer to be able to read from a textbox or String array though.
I need validation for string to comply with next: no space char starts with one delimiter char ends with one delimiter char has no other char as delimiter char. Updated sorry missed that should only be one delimiter char at start and at the end
I am trying to get a character from a rich text box based on two variables (one for the x coord and one for the y coord. Everything I've tried so far has lead to errors. The get char from position option returns something about not being a part of system.draw... something
I know how to get the cursor position but how do I add a character to that position. Like I want to have a button that adds a specific character at the position of the cursor.
I don't see any richtextbox properties that allow me to add anything at the cursor position.
If TypeOf Me.ActiveControl Is RichTextBox Then Me.ActiveControl.SelectedText = Chr(176) End If
The reason this is being done is at this point we have no idea what the column contains.However, if the formatter is "{0:MM/dd/yyyy}" and the DataRow("ColumnINeed") is an integer containing 42, String.Format is returning: MM/dd/yyyy In this situation I need it to throw an exception instead of returning nonsense.Is there anyway to make String.Format throw an exception if the object does not match what the format string is expecting?
I am trying to the copy the value from string to char array using String.CopyTo() method.
Here's my code
Dim strString As String = "Hello World!" Dim strCopy(12) As Char strString.CopyTo(0, strCopy, 0, 12)
[code]....
Edit : I get the this error at runtime.ArgumentOutOfRangeException Index and count must refer to a location within the string. Parameter name: sourceIndex
I have a DLL that I am importing:Declare Function QueryInfo Lib "mydllname" (ByVal DevName As String, ByVal pcQueryParam As String, ByVal Result As String, ByVal Length As Int32) As Int32
[Code]...
Anyone have an idea what the issue could be? If I run similar code in VB6 (only diff is mtStatus = String(mtValueSize, Chr$(0)) ) it returns what I expect.
So I need a format string to pass to String.Format that would "move" the decimal point.I can't perform any math operations before doing the String.Format, so it has to work right off the bat.Basically I'm emulating a formatting string from a proprietary server. In it if I say:
"MR2"
for the value:
12345
The result is:
123.45
I'm close with this, but it's not spot on:
String.Format("{0:#0.##}", 12345)
an extra, but not necessary... there is also MR2Z, which moves the decimal 2 left, but if the value is 0 "" is returned.
I have built a DataTable from my database. Then I am looping through the rows and trying to access a string, however the value is being returned as each character in the string.
For Each theseRows In DisplayForm.MainTab.Rows If theseRows.Item("Last_Name") = userLast And theseRows.Item("First_Name") = userFirst Then
[Code]....
The Trap_Code values are two or three letter strings, the returned value is each letter once at a time. The loop seems to cycle through the individual characters of the string as an array rather than display the entire value, which is what I was hoping for.
If checkboxList.Items(i).Selected Then .Fields("DESC1").Value += checkboxList.Items(i).Text + ", " End If
should produce output such as "A, B, C,(space)", which will then be bound to a dynamically created GridView. I would like to remove the last two-char string, that is ",(space)". How can I do this?
Basically the first letter of a string must be a certain letter. At the moment the only way I know how to do it is as follows: If sConsignmentNo(0) = "J" Or sConsignmentNo(0) = "C" Or sConsignmentNo(0) = "U" Or sConsignmentNo(0) = "N" Or sConsignmentNo(0) = "H" Or sConsignmentNo(0) = "S" Or sConsignmentNo(0) = "V" Then Basically listening the same thing out over and over. Is there a way I can say, If sConsignmentNo(0) = ListOfValidEntries i.e J,C,U,N etc? Using Visual Basic 2008 and cant use regex!
I have written my own function, which in C would be declared like this, using standard Win32 calling conventions:int Thing( char * command, char * buffer, int * BufSize);I have the following amount of VB figured out, which should import the dll and call this function, wrapping it up to make it easy to call Thing("CommandHere",GetDataBackHere).UPDATE:This code is now a working solution, as shown here:
Imports Microsoft.VisualBasic Imports System.Runtime.InteropServices Imports System Imports System.Text
[Code]...
Updates: I got the code to build by following the help received here, and then I had forgot the As Return Type (which got me a MarshalDirectiveException PInvokeRestriction). Then I had an assertion failure inside my DLL, which lead to an SEHException. Once fixed, this works BEAUTIFULLY. There are newsgroups where people are saying this can not be done, that VB only loads managed dll assemblies (which I guess is the normal thing most VB users are used to).
my question is how can i check a string for a specific char (for example, in the string:"242.421" theres the char "." how can i make a rule so that char can only be typed once... (in a textbox) so when a user click the "." button on their keyboard it won't allow it again (more than once)
I want to split a string with this array of char Dim sepa As String = " '"",.;:(){}[]·#|-_<>+¿?=/&%$‚¬@!¡" The question is how do I include in this string vbcrlf? So I can do the fallowing: dim palabras as String() = RichTextBox1.Text.ToString().Split(sepa)
I tried: Dim sepa As String = " '"",.;:(){}[]·#|-_<>+¿?=/&%$‚¬@!¡\r\n" But it does not work.