input string was not in a correct format in C# windows form

Question

1.00/5 (3 votes)

See more:

C#

private void btn_Add_Job_card_Click(object sender, EventArgs e)
       {

           Validation_Add_Job_card_order();
           if (validate == true)
           {
               DataRow r = dt.NewRow();
               // = dt.Rows.Count + 1;
               r[0] = dt.Rows.Count + 1;
               //r[0] = txt_Items_No.Text;
               r[1] = txt_Job_card.Text;
               r[2] = txt_Value_Part.Text;
               r[3] = CB_Auto_parts.Text;
               r[4] = x;
               r[5] = CB_Providers_desc.Text;
               r[6] = CB_Hand_installation.Text;
               r[7] = txt_Notes.Text;

               dt.Rows.Add(r);
               dgv_Add_job_card.DataSource = dt;
               ClearText();

               txt_Total_Values.Text = (from DataGridViewRow row in dgv_Add_job_card.Rows
                                        where row.Cells[2].FormattedValue.ToString() != string.Empty
                                        select Convert.ToDouble(row.Cells[2].FormattedValue)).Sum().ToString();
           }
       }

private void btn_Add_Click(object sender, EventArgs e)
        {
 for (int i = 0; i < dgv_Add_job_card.Rows.Count - 1; i++)
                    {
                        clsParts.ADD_AUTO_PART_DETAILS(Convert.ToInt32(dgv_Add_job_card.Rows[i].Cells[0].Value)
                                                      ,dgv_Add_job_card.Rows[i].Cells[1].Value.ToString()
                                                      ,dgv_Add_job_card.Rows[i].Cells[2].Value.ToString()
                                                      ,Convert.ToInt32(dgv_Add_job_card.Rows[i].Cells[3].Value)
                                                      ,Convert.ToBoolean(dgv_Add_job_card.Rows[i].Cells[4].Value)
                                                      ,Convert.ToInt32(dgv_Add_job_card.Rows[i].Cells[5].Value)
                                                      ,Convert.ToInt32(dgv_Add_job_card.Rows[i].Cells[6].Value).
                                                      ,dgv_Add_job_card.Rows[i].Cells[7].Value.ToString());
                    }
}

Posted 18-Jan-16 8:14am

Member 12244977

Add a Solution

Comments

Kevin Marois 18-Jan-16 14:17pm

That's nice. Now do you have a question?

Member 12244977 18-Jan-16 14:20pm

this error , input string was not in a correct format in C# windows form....
int this part:-

for (int i = 0; i < dgv_Add_job_card.Rows.Count - 1; i++)
{
clsParts.ADD_AUTO_PART_DETAILS(Convert.ToInt32(dgv_Add_job_card.Rows[i].Cells[0].Value)
,dgv_Add_job_card.Rows[i].Cells[1].Value.ToString()
,dgv_Add_job_card.Rows[i].Cells[2].Value.ToString()
,Convert.ToInt32(dgv_Add_job_card.Rows[i].Cells[3].Value)
,Convert.ToBoolean(dgv_Add_job_card.Rows[i].Cells[4].Value)
,Convert.ToInt32(dgv_Add_job_card.Rows[i].Cells[5].Value)
,Convert.ToInt32(dgv_Add_job_card.Rows[i].Cells[6].Value).
,dgv_Add_job_card.Rows[i].Cells[7].Value.ToString());
}
}

ZurdoDev 18-Jan-16 14:19pm

This is a very easy thing for you to fix on your own.

1. For us to help you, you have to tell us which line of code caused the error, otherwise we are just guessing.
2. If you debug the code you'll see what the value is and you'll see exactly why there is a problem.

Member 12244977 18-Jan-16 14:20pm

this error , input string was not in a correct format in C# windows form....
int this part:-

for (int i = 0; i < dgv_Add_job_card.Rows.Count - 1; i++)
{
clsParts.ADD_AUTO_PART_DETAILS(Convert.ToInt32(dgv_Add_job_card.Rows[i].Cells[0].Value)
,dgv_Add_job_card.Rows[i].Cells[1].Value.ToString()
,dgv_Add_job_card.Rows[i].Cells[2].Value.ToString()
,Convert.ToInt32(dgv_Add_job_card.Rows[i].Cells[3].Value)
,Convert.ToBoolean(dgv_Add_job_card.Rows[i].Cells[4].Value)
,Convert.ToInt32(dgv_Add_job_card.Rows[i].Cells[5].Value)
,Convert.ToInt32(dgv_Add_job_card.Rows[i].Cells[6].Value).
,dgv_Add_job_card.Rows[i].Cells[7].Value.ToString());
}
}

ZurdoDev 18-Jan-16 14:32pm

Probably dgv_Add_job_card.Rows[i].Cells[0].Value is not an integer. But we can't see your code. You have to run it and see if that is the actual problem.

2 solutions

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)

OriginalGriff · Answer 1 · 2016-01-18T08:32:00

We can't run your code and say "it's that bit" - because we can;t run your code in isolation and get the same results you do - we don't have access to your DGV or top the data you are filling it with.

We'd have to guess what the problem might be, where for you a simple use of the debugger will tell you exactly what the problem is.
So use the debugger. Run your app under the debugger (F5 in VS) and make the problem happen.
When the error occurs, look at the parts of the line of your code it is complaining about, and see what value does not match the data the DGV is expecting. Then you can look back in your code to work out why - we can't!

If I was to guess, I'd say it's probably that dgv_Add_job_card.Rows[i].Cells[0].Value isn't an integer or the string representation of an integer - but what value it contains and where the heck that value comes from only you can tell! :laugh:

Sergey Alexandrovich Kryukov · Answer 2 · 2016-01-18T09:21:00

You call a number of Convert.To* methods having no guarantee that the string passed as an argument really can be parsed as a value of required type. If it doesn't, you throw an exception, which you can handle and recover. But this is not a right way. Even the class name Convert is somewhat misleading. You need to understand what really happens, and this is nothing but parsing. You should better use the TryParse method for each expected type. For example, for int, you can use bool int.TryParse(string, out int), and so on. If such method returns false, you know that the input string is not in an expected format, and the valid value is not returned via the second out parameter.

Moreover, your code sample strongly suggests that you tend to work with strings representing data, instead of data itself. You should better avoid it, reserving strings mostly for output purposes, such as showing on screen. For example, for the type decimal and integer types, you can use the control System.Windows.Forms.NumericUpDown (in can also be used indirectly for floating-point type, if you need input with fixed precision, which is usually the case): NumericUpDown Class (System.Windows.Forms)[^].

Also, you can use System.Windows.Forms.DataGridView where you can use strongly-typed cell data. Some types of the cell derived from System.Windows.Forms.DataGridViewCell are already available for some commonly used data types, some you can develop yourself. Please see:
DataGridView Class (System.Windows.Forms)[^],
DataGridViewCell Class (System.Windows.Forms)[^],
DataGridViewCell.Value Property (System.Windows.Forms)[^],
DataGridViewCell.ValueType Property (System.Windows.Forms)[^].

And so on… It were just some of the idea, something to think about.

—SA

input string was not in a correct format in C# windows form

2 solutions

Solution 1

Solution 2

Add your solution here

Preview 0