Syntax error (missing operator) in query expression

Question

0.00/5 (No votes)

See more:

When iam trying to insert data into Access Database at that time iam showing OleDbException was unhandled actually picturebox5 and picturebox4 image not inserting

values are showing :

Syntax error (missing operator) in query expression

{"Syntax error (missing operator) in query expression '2','Atul','Wadala','24043410','collected','12/12/2015','2','12/01/2016','Active','500','50','2','100.00','550.00','@pictureBox6'."}

What I have tried:

insert code is

C#

cmd = new OleDbCommand("insert into RegistrationForm(ClientCount,Name,Address,Contact,Documents,Money_Taking_Date,Muddat,Money_Return_date,Account_status,Taking_Amout,Interest_per_month,Pending_interest_month,Pending_interst_Amount,Total_Amount,Client_image,Document_image1,Document_image2) values(" + lblcount.Text + "','" + textBox20.Text + "','" + textBox21.Text + "','" + textBox19.Text + "','" + textBox18.Text + "','" + maskedTextBox1.Text.ToString() + "','" + textBox22.Text + "','" + maskedTextBox2.Text.ToString() + "','" + textBox23.Text + "','" + textBox17.Text + "','" + textBox16.Text + "','" + textBox15.Text + "','" + textBox13.Text + "','" + textBox14.Text + "','@pictureBox6,@pictureBox4,@pictureBox5)'", con);
            conv_photo();
            //cmd.Parameters.AddWithValue("@pictureBox6", pictureBox1.Text);
            con.Open();
            int n = cmd.ExecuteNonQuery();
            con.Close();
            if (n > 0)
            {
                MessageBox.Show("record inserted");
                loaddata();
                rno++;
            }
            else
                MessageBox.Show("insertion failed");

loaddata method code is :

C#

public void loaddata()
      {

          adapter = new OleDbDataAdapter("select * from RegistrationForm", con);
           ds = new DataSet();//student-> table name in stud.accdb/stud.mdb file
           adapter.Fill(ds, "RegistrationForm");
           ds.Tables[0].Constraints.Add("ClientCount", ds.Tables[0].Columns[0], true);
      }

Posted 7-Feb-16 3:46am

Atul Rokade

Updated 7-Feb-16 4:21am

PIEBALDconsult

v4

Add a Solution

Comments

PIEBALDconsult 7-Feb-16 10:18am

I suspect it's due to the apostrophes around '@pictureBox6' I suggest you remove them.
You need to learn to use parameters, every time, especially when using data from users.

1 solution

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 · 2016-02-07T04:23:00

Don't do it like that!
Do not concatenate strings to build a SQL command. It leaves you wide open to accidental or deliberate SQL Injection attack which can destroy your entire database. Use Parametrized queries instead.
The chances are that not only will your code be more readable, your database be a lot safer, but your problem will go away at the same time.
Try something like:

C#

cmd = new OleDbCommand("INSERT INTO RegistrationForm (ClientCount) VALUES (@CC)", con);
cmd.Parameters.AddWithValue("@CC", lblcount.Text);

But if those last are images, then you are going to have to do more than that anyway: Why do I get a "Parameter is not valid." exception when I read an image from my database?[^] shows you what happens when you do that, and what to do to fix it.