Index was out of range. Must be non-negative and less than the size of the collection parameter name:index

Question

1.00/5 (1 vote)

See more:

I have been trying to import text file using VB.NET and importing it into an EXCEL FILE.

But I am having errors with my code:
- index was outside the bounds of the array
- Index was out of range. Must be non-negative and less than the size of the collection parameter name:index

Private Sub Datacalculate(ByVal files() As String)
       Try

           Dim ti() As String = New String(100000) {}
           For division As Integer = 0 To 4
               For ctr As Integer = 0 To files.Count - 1
                   Dim testarray() As String = Split(files(ctr), ",", )
                   ti(ctr) = testarray(0)
                   GetValuesPerChannel(testarray(division), division)
               Next
           Next

       Catch ex As Exception
           MsgBox(ex.Message)
       End Try
   End Sub

I cant figure out what the problem is.
The excel shows only 2 columns. and the rest was blank.

Posted 8-Sep-14 16:19pm

NekoNao

Updated 8-Sep-14 21:11pm

v4

Add a Solution

Comments

ChauhanAjay 8-Sep-14 22:53pm

Did you debug to find out about the error?
On which line do you receive this error?

NekoNao 8-Sep-14 22:55pm

I did debug it .the error is in this codes. Its like it is not reading the THIRD COLUMN

GetValuesPerChannel(testarray(division), division)

xlWorkSheet.Cells(i + 2, 4) = Column3(i).ToString
//Third Column

MuhammadUSman1 8-Sep-14 23:29pm

You are accessing an object or index which does not exists yet.

MuhammadUSman1 8-Sep-14 23:31pm

It is "GetValuesPerChannel" built in function or you created this function?

NekoNao 8-Sep-14 23:48pm

I created the getvaluesperchannel..

SRS(The Coder) 9-Sep-14 0:38am

Can you please debug and check the value of 'files.Count' it seems like sometime the value is coming less that 1 so subtracting 1 from it giving result -1 which is not a valid value for index of array.

Or you can check which particular line is failing on debug to let us know more exactly. Then check with the index on each steps also.

NekoNao 9-Sep-14 1:11am

There are values in the files.Count = 5

LINE 213: GetValuesPerChannel(testarray(division), division)

SRS(The Coder) 10-Sep-14 1:52am

Hope answer given by 'NekoNao' is helpful.

NekoNao 9-Sep-14 1:43am

I really really need to get this code done. T_T But I cant figure it our why it has errors

NekoNao 9-Sep-14 1:16am

I am really not good in debugging

2 solutions

Solution 1

Put a breakpoint within DataCalculate and go step by step, find option "step into" in your debugger (it is F11 if you're using Visual Studio)...

Dim testarray() As String = Split(files(ctr), ",", )
ti(ctr) = testarray(0)

Not sure why you have empty space behind second comma in the split, but you're definitely accesing testarray (0) without checking if it has elements (it should have the first, but then you're following with

GetValuesPerChannel(testarray(division), division)

This part is by no means guaranteed: - check the length of testarray after split.
testarray(division) <-- you probably have a file with less then division commas.

Posted 8-Sep-14 20:19pm

Sinisa Hajnal

Updated 8-Sep-14 20:20pm

v2

Comments

NekoNao 9-Sep-14 2:25am

Aw. Too Technical. Can you explain it more ? Cause I am just new in programming

Sinisa Hajnal 9-Sep-14 2:32am

Being new does not excuse you from not trying. You should always strive to find your own solution reading the forums like this one and adapting, rather then waiting for the finished code which you will copy without understanding. Also, whatever IDE you're using, learn how to follow the code with breakpoints and how to do step by step checking of the code. Never use objects without checking their validity.

SOLUTION
If testarray.Length <= 5 (or whatever number of columns you're expecting) Then
MsgBox("The file does not contain information about all the columns!")
Return
End If
(you can decorate messagebox with various styles like info or warning if you want)

NekoNao 9-Sep-14 2:34am

I did try many codes and try to debug it. but since i have lack of knowldege. I really cant see the solution. Only i know is the logic not the code. Its been my 1 week in debugging this code still dont find the right code.

Sinisa Hajnal 9-Sep-14 2:46am

"Knowing" will get you stuck. Re-check your assumptions, check the file as I mentioned. Check the length of the testarray (You have the code in the post above), check that all files have same structure etc...

NekoNao 9-Sep-14 2:36am

Guess the file does not contain the exact number . But Why do you think is that if in the TEXTFILE, it is complete.

Sinisa Hajnal 9-Sep-14 2:44am

IT is not complete! That's the whole point. If your split does not return 5 columns then either your logic doesn't work (you cannot access division 5 if there are not enough columns) or the file is NOT complete.

NekoNao 9-Sep-14 2:32am

@sinisa hajnal ,

I tried the debugging. Still has error on line ,

GetValuesPerChannel(testarray(division), division)

I cannot see what's wrong with the code.It reads, 1st column but in second column it has errors :
- index was outside the bounds of the array
- Index was out of range. Must be non-negative and less than the size of the collection parameter name:index

But when there is 5 column in one row.

Sinisa Hajnal 9-Sep-14 2:36am

That just means you don't understand what is going on. If you check the file you're using, you'll see that there are NOT five columns in the file. One of you commas got lost, is different character or you accidentally pressed enter and broke one line in two.

Use the code with length testing above it will give you a message if something isn't right...also, when the breakpoint hits the line, before going through, check the values of division and testarray (division).

NekoNao 9-Sep-14 2:38am

YOu know, I used this code the last time . but now, it does not work. I use it in reading log files. But when i change it to reading .txt files it has errors.

Sinisa Hajnal 9-Sep-14 2:45am

Be so kind and post file 5 lines of your text file.

Reusing the code is fine if you understand what it does. If you don't, you get error like this.

NekoNao 9-Sep-14 3:03am

This is just the example i am using
1,name,age,150.00,13.00
1,name1,age1,10.00,3.00

This code only works on .log file :( it does not work on strings. aw.
Does it have any difference in renaming a .txt file to .log file?

Sinisa Hajnal 9-Sep-14 3:07am

Is this from your real file you're using? Or you just typed it for me as an example?

No, type of the file makes no difference ASSUMING THAT ALL txt files in your folder are your files - if you have one that says "shopping list: groceries, bread and milk" you'll get only two columns and your code will break

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

NekoNao · Accepted Answer · 2014-09-08T21:08:00

Solution 2

The only solution here is to NEVER RENAME .txt file into a .log file.
MUST CREATE A NEW .LOG FILE ^__________^
Because i think it can't read it.

OR

CREATE A NEW TEXT FILE.

I tried using the REAL .log and it works!

Thanks for all the people who commented. ^^

Posted 8-Sep-14 21:08pm

NekoNao

Updated 8-Sep-14 21:11pm

v2

Comments

Sinisa Hajnal 9-Sep-14 3:15am

Then you have something in your program that specifically reads log file of the given format. Or you had other txt files in the folder.
I've used the code like yours for xml, txt, csv and dat files and everything works fine with those files.

No problem, glad you solved it.

NekoNao 9-Sep-14 3:21am

Its maybe because of the format of the files INSIDE the textfile is not the same as the .log file