Substring Give Error - Says Index And Location Must Refert To A Location Withing String

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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Substring - Error: Index And Length Must Refer To A Location Within The String. Parameter Name: Length

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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Output Error "Index And Length Must Refer To A Location Within The String. Parameter Name: Length" With Substring

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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SubString Function - Index And Length Must Refer To Location

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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Getting The Error Saying That Index And Length Must Refer To A Location Within The String

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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VS 2005 Error:Index And Length Must Refer To A Location Within The String?

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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Substring Size - "Index And Length Must Refer To A Location Within The String. Parameter Name: Length"

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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VS 2005 Uploading Error: Index And Length Must Refer To A Location Within The String

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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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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Index And Count Must Refer To Location Within String

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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Index And Length Must Refer To A Location Within The String?

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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Index And Length Must Refer To Location Within String?

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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Error: Index And Length Must Refer To A Location?

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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Get The Index Location Of A Child Node Using The Key String On TreeView .NET?

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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VS 2008 Index And Length Must Refer To A Location Within The String?

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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VS 2008 ArgumentOutOfRangeException - Index And Length Must Refer To A Location Within The String

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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Error - You Cannot Start Application <App_Name> From This Location Because It Is Already Installed From A Different Location

Dec 10, 2009

we have a user that my program works fine for.. and another that it just crashes for.. they get this error

"You cannot start application <App_Name> from this location because it is already installed from a different location."

then they run it up again and it comes up.. and then just closes..in reading some stuff online it says its a framework 2.0 sp 1 issue ?these pc's have 3.5, but my program uses all 2.0 stuff..so is this a 2.0 sp level issue ? or should I make my programs start to use framework 3.0 instead ?

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How To Give Location To Control T Runtime

Mar 27, 2010

I am trying to add textbox at runtime. but my problem is textbox is

added on same location but i want to add textbox at different location.

Code that i am using is :[code...]

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Give Location For Tool Strip Menu Item?

Jun 23, 2010

We have a menu strip with 4 toolstripmenuitems. In mosue right click i need to show one toolstripmenuitem as dropdown menu. here i am able to show that tool strip menu but it is populating at the top left corner of the screen. I need to set the location as cursor location.

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Index And Length Must Refer To A Location Within The String. Parameter Name: Length Exception

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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Cant Get The Index Location Of An Array?

Jan 19, 2010

just wanna get that simple problem.,but i am confused bout this i cant get the index location of an array>

my problem is have to display 5 elements so the index is 0 to 4 right?

example i entered 1 2 3 4 5 then i searched 3 the location should be index 2?right?

how can i get the index location of the searched number?if you can make me understand sub and functions too..

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C# - Add A Location To Windows 7 Search Index?

Sep 9, 2011

Is there any way by which I can add a location to Windows 7 search index and make it indexed in .Net?

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Find The Location Of Element By Index?

Jan 19, 2010

i need to find the location of element by index

example i enter 5 numbers

1 2 3 4 5

search a number, i search 3 it should display the lindex location? for me to understand more about vb,,more on arrays..

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Forms :: UnBound DataGridView Location - Grid Must Be In The Correct Location According To The Pixel Point?

Dec 19, 2011

i am using an unbound datagridview so i can dynamiclly add rows. all that is working fine. but the grid is not is the location i have coded.. i am using the defualt form as a base then coding the unbound stuff in.. should i just create a blank class file and do everything? the only problem i am having is the grid must bees in the correct location according to the pixel point that i have given it. Right now it is placing the grid at point (0,0) no matter what point is entered on the line for location. so what am i missing??????

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Where To Place Database In VB Project And How To Access With C:,D: Etc Means Open Location Not Fixed Location

May 3, 2010

dim cn as oledbconnection
cn = New OleDbConnection("Provider=Microsoft.Jet.OLEDB.4.0;Data Source=C:valid.mdb")

[code]......

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2 Group Box Components And Both Are Same Size,font,location And Visible Is False - Both Component Cannot Locate At Same Location

Mar 13, 2012

I have 2 group box components and both are same size,font,location and visible is false. When i click button2, the groupbox1 won't appear(the group box2 is on bottom n group box1 is on top).

Example:

button1

groupbox1.visible=false

groupbox2.visible=true

button2

groupbox1.visible=true

groupbox2.visible=false

Because i want to show the different,so the location have a little different(actually both are same location). I think my code is no problem. The problem i guess is both component cannot locate at same location?

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Get Index Location Of A Piece Of Data In An Array

Sep 15, 2011

I am trying to import a text file into a data table using VB.NET. I need to loop thru each line of the text file and put certain elements of that line in a new row of the data table. The line may contain 50 elements, but I only need to get 15 of them. The very first line is a header row, which contains "File Name", "Object Name", "Object Weight", etc. The trick is the element locations move from file to file. For example, "File Name" element may be located at 0 in one file and then 15 in the next file. I need to find the exact location of my desired elements and save them to a variable. Then use that variable to locate the exact piece of data in the text file line to place in my new row.

The code below obviously doesn't work, but it kind of illustrates what I need to have done.

Try
With dt
.Columns.Add("fltObjectWeight", Type.GetType("System.Double"))
.Columns.Add("vchObjectName", Type.GetType("System.String"))

[Code].....

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VS 2008 Round E.Location To Nearest Grid Square Location?

Nov 9, 2009

I was using this old method of creating a bunch of rectangles when I need to get the location of a certain point within a grid so I could draw images on the grid. The thing is, I don't want to use rectangles, I want to be able to just round the location as if I were using rectangles. I need it to be able to round the point (66,70) to (50,50) so if you can imagine a grid and the mouse position being within that square in the grid, I need to get the location of the upper left corner of that square.

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Cannot Start Application Shell From Location Because It Is Already Installed From Different Location

Jul 21, 2009

I have encounterd a problem when trying to re-enstall a program i have writen in vb 2008.I get the following message:"You cannot start application Shell from this location because it is already installed from a different location."This program is to be used on many workstation computers and i need to be able to update any changes by just reinstalling the program, no uninstalling needed first. If the program was installed from a cd first and needs to be updated from a flash disk this error will be a problem, if the update is from a cd as well then there is no problem.Is there a way to change the installation package to ignore where the program installs from and just update itself?

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Change The Location Of A Label ( Middle ) To Another Location When A Key Is Pressed Down?

Jan 27, 2010

I'm currently working on a project that has a simple game . I want to change the location of a label ( middle ) to another location when a key is pressed down. But there is something wrong with the first of the If statement .I've underlined "point " because it has the error.

Private Sub Space_Navigator_KeyDown(ByVal sender As Object, ByVal e As System.Windows.Forms.KeyEventArgs) Handles Me.KeyDown If e.KeyCode = Keys.Up Then If middle.Location = [u]Point[/u](156, 655) Then middle.Location = New Point(156, 547)
End Sub

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