How to insert multiple checkboxed answers into a database table?

Question

I'm having a huge issue with this database. It connects correctly and with the information from the form's $_POST queries that are being inserted into the table company_info within the correct fields.

Now, I have no idea what I'm doing wrong here, but I keep getting the die error of

"Error querying database".

The database version is: phpMyAdmin 2.6.4-pl3

MySQL: 5.0

Any ideas? I can provide you the rest of the code if needed.

$dbc = mysql_connect('db390590179.db.1and1.com', 'dbo390590179', '*********')
or die('Error connecting to MySQL server.');

mysql_select_db("db390590179", $dbc);


$query = "INSERT INTO company_info (company_name, company_phone,  company_contact, company_address, " .
"company_city, company_state, company_zip, " .
"state_living, vehicles, position, " .
"experience, training, hazmat, " .
"require_hazmat, load_nyc, take_truck_home, " .
"have_rider, have_pet, choose_route, " .
"fuel, cash_advance, days_before_home, " .
"log_system, slip_seat, pre_pass, " .
"ez_pass, health_insurance, retirement_plan, " .
"payment_plan, calculate_pay, freight, " .
"loads, home_on_time, idle_time, " .
"equipment_condition, canada)" .

"VALUES ('$company_name', $company_phone', '$company_contact', '$company_address', '$company_city', " .
"'$company_state', '$company_zip', " .
"'$state_living', '$vehicles', '$position', " .
"'$experience', '$training', '$hazmat', " .
"'$require_hazmat', '$load_nyc', '$take_truck_home', " .
"'$have_rider', '$have_pet', '$choose_route', " .
"'$fuel', '$cash_advance', '$days_before_home', " .
"'$log_system', '$slip_seat', '$pre_pass', " .
"'$ez_pass', '$health_insurance', '$retirement_plan', " .
"'$payment_plan', '$calculate_pay', '$freight', " .
"'$loads', '$home_on_time', '$idle_time', " .
"'$equipment_condition', '$canada')";

$result = mysql_query($query, $dbc)
or die('Error querying database.');

mysql_close($dbc);

Elias Dorneles · Accepted Answer · 2011-12-07 02:05:07Z

1

I think it's because it's missing a quote before the variable $company_phone in your INSERT statement.

answered Dec 7, 2011 at 2:05

Elias Dorneles

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Comments

Manigandan Arjunan · Accepted Answer · 2011-12-07 02:15:46Z

0

just combine the different values within a single quote.

E.g., "'$company_state , $company_zip,' " ."'$state_living , $vehicles, $position, '" ."'$experience, $training, $hazmat, '" ....

this will work perfectly and also include the missing quote in the begining of the *$company_phone* has to be included.

answered Dec 7, 2011 at 2:15

Manigandan Arjunan

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Comments

Somnath Muluk · Accepted Answer · 2011-12-07 04:29:48Z

You can remove double quote from each line and combine them. I have removed syntax errors. You can assure that result is getting or not. Try this code.

$dbc = mysql_connect('db390590179.db.1and1.com', 'dbo390590179', '*********')
        or die('Error connecting to MySQL server.');

mysql_select_db("db390590179", $dbc);


$query = "INSERT INTO company_info (company_name, company_phone,  company_contact, company_address,
        company_city, company_state, company_zip,
        state_living, vehicles, position,
       experience, training, hazmat,
       require_hazmat, load_nyc, take_truck_home,
        have_rider, have_pet, choose_route,
       fuel, cash_advance, days_before_home,
        log_system, slip_seat, pre_pass,
        ez_pass, health_insurance, retirement_plan,
        payment_plan, calculate_pay, freight,
        loads, home_on_time, idle_time,
        equipment_condition, canada)

        VALUES ('$company_name', '$company_phone', '$company_contact', '$company_address', '$company_city',
       '$company_state', '$company_zip',
        '$state_living', '$vehicles', '$position',
        '$experience', '$training', '$hazmat', 
        '$require_hazmat', '$load_nyc', '$take_truck_home',
        '$have_rider', '$have_pet', '$choose_route',
        '$fuel', '$cash_advance', '$days_before_home',
        '$log_system', '$slip_seat', '$pre_pass',
        '$ez_pass', '$health_insurance', '$retirement_plan',
        '$payment_plan', '$calculate_pay', '$freight',
        '$loads', '$home_on_time', '$idle_time',
        '$equipment_condition', '$canada')";

$result = mysql_query($query, $dbc)
        or die('Error querying database.');

mysql_close($dbc);

Andriy M · Accepted Answer · 2011-12-07 07:44:34Z

0

for (int i = 0; i < CheckBoxList1.Items.Count - 1; i++)
{
  String str = "";
  if (CheckBoxList1.Items[i].Selected)
  {
    str = CheckBoxList1.Items[i].Text;
    con.Open();
    string sql =
      "Insert into dbtable(Category,BookTitle,Feature,SubCategory)values('" +
      DDLCategory.SelectedItem.Text + "','" + TxtBooktitle.Text + "','" +
      CheckBoxList1.Items[i].Text + "','" + DDLSubcategory.SelectedItem.Text +
      "')";
    SqlCommand cmd = new SqlCommand(sql, con);
  }
}

Just use DEBUGGER and see how things are working and you should be able to resolve such issues easily.

edited Dec 7, 2011 at 7:44

Andriy M

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answered Dec 7, 2011 at 2:10

Swathi

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