VS 2008 Error Message: Index And Length Must Refer To A Location Within The String
Oct 15, 2010
I'm working on a vb.net project, when I execute the project I get the following error message: "error message: index and length must refer to a location within the string"
Public Sub New(ByVal lineIn As String)
parent = Trim(lineIn.Substring(0, colStarts(1)))
enfant = Trim(lineIn.Substring(colStarts(1), colStarts(2) - colStarts(1)))
des_F = Trim(lineIn.Substring(colStarts(2), colStarts(3) - colStarts(2)))
[Code]...
I get the error msg in the line marked bold. When i checked the input file, the length of all the fields looks fine. I dont understand the cause of this error.
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Apr 17, 2009
this is not working?
[Code]...
End WhileI am trying to read from the ": " to the end of the line. I keep getting this error: Index and length must refer to a location within the string. Parameter name: length
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Mar 1, 2009
I keep getting the error saying that index and length must refer to a location within the string.
Public Class MainForm
Private Sub btnAdd_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles btnAdd.Click
Dim description As String = String.Empty
Dim invCount As String = String.Empty
[code]....
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Sep 7, 2009
why error "Index and length must refer to a location within the string."appear when the process of the program is going to executeNonQuery, i'm looking the error for 2 day's i still can't find it.
[Code]...
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Oct 12, 2009
why when i upload less than 10 item the program do but more than it like 15 and above the program give me error saying "Index and length must refer to a location within the string."
here is my codes below
For Each myDatarow In tblLocal.Select("sync ='0'") '
' tblServer.ImportRow(myDatarow)
' dataimpt = myDatarow.Item(4).ToString
[Code].....
the error appear when the program is going to update the 14th data... meaning from 1st to 13th data program works in the 14th it's not it gives me error "Index and length must refer to a location within the string
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Sep 28, 2009
Dim Phone as string = "478-742-4050 478-256-6550"
If _Phone.Length > 25 Then
_Phone = IsNull(ReturnValue.Substring(0, 25).ToString, "")
End If
Return _Phone
Error: "Index and length must refer to a location within the string. Parameter name: length"
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Mar 23, 2009
"Index and length must refer to a location within the string. Parameter name: length" whenever I run this code
[Code]...
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Jun 23, 2009
I got this error and I don't know what should I do. Index and length must refer to a location within the string. Parameter name: length That's the line that i get the error on:
[Code]...
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Feb 17, 2011
am getting the above exception while swaping Items in the list(lstRoutePriority).PFB my code
if (lstRoutePriority.SelectedIndex > 0)
{
//Swap the two items
[Code].....
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Jul 7, 2009
why this error happens with a bit of code i was trying? The code is actually intended to be for an onclick event, so it actually works there.
But im just curious as to why it errors as an onchange event. This is 9am pre-caffeine, so the reason is probably starring me in the face! On a form i have a text box, and a combo box. The combo box has various options in it, natch. Its a search form, which then fills a gridview with the results.
Drop down list options are: home number, mobile number, post code, Name, DOB
I thought id try and be "clever" and do away with the need for the combo box. Since a mobile number begins with "07", names are just characters, postcodes are character character number, and home number is all numbers and doesent begin with "07", and date of birth has /'s in it.
So i tried this to test the idea:
Private Sub AppSearchTB_TextChanged(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles AppSearchTB.TextChanged
Dim TLength As Integer = 2
[Code]....
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Sep 13, 2011
I am trying to get the first 50 letters, so I used the subString function to get it. As you can see, I used this code to get it:
<%# Eval("BannerDescription").ToString.Substring(1, 50)%>
But unfortunately it's not working and an error message is coming up:
Index and length must refer to a location within the string.
Because the user is the one who is controlling the data entry! Some times he gonna enter 10 letters other times maybe 1000 letter?
I tried them all but can we use it this way :
<%# IIf(Eval("BannerDescription").ToString().Length > 49, Eval("BannerDescription").ToString().Substring(0, 49), Eval("BannerDescription"))%>
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Dec 21, 2008
im basically working on my final years project on the vb.net with the ilog business rules. i have trouble solving this error :Index and length must refer to a location within the string. Parameter name: length And the output show this :[size=2]
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Jul 21, 2010
If using the following in an if statement I get an error: If trg.Name.Substring(4, 6).ToUpper <> ("ABCDEF") Then I get the error: "Index and length must refer to a location within the string. Parameter name: length"
I assume this is because the string (trg.name) is too small for the 4, 6 substring. What would be the correct method of working around this problem? VB.net Studio 2008.
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Feb 13, 2012
Consider the Statement:
Dim pstr As String
FileOpen(2, pfilepath, OpenMode.Input)
pstr = LineInput(2)
pstr = pstr.Substring(13, pstr.Length)
Label1.Text = pstr
FileClose(2)
And it is Giving me the following error:
Index and length must refer to a location within the string. Parameter name: length.
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Aug 10, 2009
I was trying to make a simple program to search in my mysql db, but i'm seeing this error :{"Index and length must refer to a location within the string. Parameter name: length"}[code]
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Mar 29, 2011
numberofchar = timefull.Length
If timefull = "" Then
Else
If numberofchar = 11 Then
timefull = timefull.Remove(2, 9) ' isolate the hour depends on number of digits in hour parameter
Else
timefull = timefull.Remove(1, 9)
End If
End If
This is my code, it doesn't work when I copy into form load. But when I try it in another form, it works. Index and count must refer to a location within the string. Parameter name: count
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Oct 22, 2011
I got a error when I run this code:
Dim btch As String
Dim LeftPart As String
[CODE]...
It says index and location must refert to a location withing string
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Dec 9, 2010
I keep getting the error saying that index and length must refer to a location within the string. I am sure there is just something minor that I am missing but I am having a difficult time figuring it out.
CODE:
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May 4, 2009
I would like to format output to a property of a custom control I have made.For instance I have a literal control in my control and when populated I want to check if it is longer than 30 characters long, if so, then go to the next space (as to not cut the word off) after the 30th character and place a <br /> tag in to place the next 30 characters on the next line.So if there were 100 characters in a string, there would be 3 line breaks in the label.I have the looping worked out to get the label populated but cannot get how to dymaically create the array based on the string lenth property and index of method.
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Oct 27, 2009
trying to determine the index of a child node using the key string so I can add a new child node to it.I am using this code:
vNodesIndex = TreeView1.Nodes.IndexOfKey(Key)
But only works for the root node when I try to find an exiting child node the code returns -1 (Not Found)
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Mar 5, 2009
I've got the web browser with tabs in all working fine, but there's one bug, i can add tabs, delete em, etc.. but when i delete a tab then go to create a new tab, it gives me the error: InvalidArgument=Value of '6' is not valid for 'index'. Parameter name: index '6' is the number of tabs i had open.Here's the bit of code i use to create the tabs with the browser in it:
vbcode
Dim browse As New WebBrowser
browse.Name = "b1"
browse.Dock = DockStyle.Fill
[code]....
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May 9, 2011
I have lots of labels that I need to refer to.
E.g. label1 = listbox1.selecteditem.tostring & "123"
label2 = listbox1.selecteditem.tostring & "123"
I have 20 labels which i need to repeat the same process for. What I found was this code which always gives me the null reference error.
CType(Me.Controls("Label" & i), Label).Text = "what i want here"
I have stated dim i as integer = 1
And later i = i + 1 until it is 20.
I've tried i.tostring and cstr(i) when referring to it but it keeps saying it doesn't exists even though i have labels 1 to 20 named label1, label2 etc. on my Form.
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Mar 13, 2009
I am receiving the following error when trying to replace spaces with %. "String cannot be of zero length. Parameter name: oldvalue" Dim Keywords As String = "test abc" Dim sqlkeywords As String sqlkeywords = Keywords.Replace("", "%") I have also tryed sqlkeywords = Replace(Keywords, "", "%") and it just returns the original.... test abc.
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May 13, 2009
How i can set size?
[Code]...
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Dec 19, 2006
In my fresh installation of Visual Studio Professional 2005 Visual Basic I have typed this code
dim olddate as date
olddate=datevalue("01-01-2006")
I receive an error: index out of range exception
On an other pc with the same fresh installation this do not happen. Why ? I have tried with other funtions : ctype, cdate but I receive the same error, while on an other pc this code is fine.
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May 10, 2011
I have a problem to make the comparison from char to string array using Naive Algorithm in vb.net.[code]The problem is the system cannot compare the input from s1 to compare with s2.
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Jul 20, 2010
how to split a string into an array based on it's length?Say if I want the length to be 350 then every 350 chars in a textbox will be split into an array like [code]
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May 6, 2011
I am getting the error "The 'Microsoft.Jet.OLEDB.4.0' provider is not registered on the local machine." in the connection string when opening a connection to an Access database.
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Dec 15, 2011
I have to create a calculator. In order to do the calculation, when the math operator is entered, there is a space on either side, so that I can split by the white spaces.
[code]...
This was my original idea of how I could accomplish my goal, however, I get an error message saying that splitNumbers cannot be of type string. Do I have to create an array to do this or is there an easier way than what I'm seeing?
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Feb 25, 2012
I keep getting an error message in my code "text is not a member of string" on each line similiar to: TextBoxVerb1 = TextBoxVerb1.text.Trim and I can't figure it out. Here is the code:
Private Sub Buttonsubmit_Click(ByVal sender As System.Object, ByVal e As System.EventArgs) Handles Buttonsubmit.Click
Dim str As String
[Code]....
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