Datagridview column & sql tables

Question

4.00/5 (1 vote)

See more:

Hi,
I have a sql database table with Num1, Num2, Num3 and Date columns.
In my 'SqlDataAdapter I'm NOT binding the Date column to a datagrid but only Num columns(SELECT Num1, Num2, Num3 FROM tblTestTable1). However I am adding additional column to a datagridview which will be used as a datatime picker. So basically, customer will click on that and pick the date from the calendar.

C#

CalendarColumn col = new CalendarColumn();
this.dataGridView1.Columns.Add(col);
col.Name = "Date";

Once customer picked the date we want to record that to the database. Num1 to Num1 column and so on. At the end, date from datetimepicker column to the Date column in database. And that is a problem right there.

C#

da.InsertCommand = new SqlCommand("INSERT INTO tblTestTable1 VALUES  (@Num1,@Num2,@Num3,@Date", conn);

da.InsertCommand.Parameters.Add("@Num1", SqlDbType.Int);
da.InsertCommand.Parameters.Add("@Num2", SqlDbType.Int);
da.InsertCommand.Parameters.Add("@Num3", SqlDbType.Int);
da.InsertCommand.Parameters.Add("@Date", SqlDbType.DateTime).Value =  col;

it doesn't like the last command. Any ideas?
Thanks in advance.
JML_

Posted 5-Apr-11 4:33am

JML_

Updated 5-Apr-11 4:34am

Henry Minute

v2

Add a Solution

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)

Henry Minute · Answer 1 · 2011-04-05T04:44:00

Solution 2

When you get errors either when compiling or running code and you need help, it is really helpful to the folks here on CP if you tell us what the error is. The exact wording where possible. Otherwise anything said in answer is really a guess.

Anyway, here's my guess.
Try splitting the line into two parts the first exactly like your others, just adding the parameter.
Then the second to set the value:

C#

da.InsertCommand.Parameters.Add("@Date", SqlDbType.DateTime);
da.InsertCommand.Parameters["@Date"].Value =  col;

BTW: If you could enclose any code snippets in <pre>code goes here</pre> tags it makes them much easier to read.

Posted 5-Apr-11 4:44am

Henry Minute

Comments

JML_ 5-Apr-11 11:35am

I did chamge the code to what you've got here and it gives me an error " Failed to convert parameter value from a CalendarColumn to a DateTime " :(
when I change the code to:
da.InsertCommand.Parameters["@Date"].Value = Convert.ToDateTime(col);
Unable to cast object of type 'CalendarColumn' to type 'System.IConvertible'.
So, now I'm lost. :(
JML_

Henry Minute 5-Apr-11 11:50am

You are passing a DataGridViewCalendarColumn as the value, so it is no surprise that you get an error.

You haven't shown the method that your code comes from (if it is an event handler or a stand-alone method) so it is difficult to give specific advice. What you need to do is get the cell from the row you are interested in into a suitable variable (myCalCell) and use myCalCell.Value.

Eduard Keilholz · Answer 2 · 2011-04-05T04:38:00

Solution 1

The col variable is of a type CalendarColumn and must be of type DateTime.

You must retrieve the date / time picked by the user from the control, and not insert the control in your database.

Have fun!
Eduard

Posted 5-Apr-11 4:38am

Eduard Keilholz

Comments

JML_ 5-Apr-11 11:37am

how do i do that? can you help?
JML_