Dynamic Form Fields not inserting all data into database

Question

I am trying to insert an array into a database via a dynamic form. The form will insert the first values from the form but not any more. tried a couple of methods and have hit a wall.

Here is the PHP file:

<?php
$con = new MySQLi(CONNECTION INFORMATION);
if (!$con)
{
    die('Could not connect: ' . mysqli_error());
}


foreach($_POST['room_types'] as $cnt => $room_types);
$room_type = implode(",", $room_types);

foreach($_POST['rooms'] as $cnt => $rooms);
$room = implode(",", $rooms);

foreach($_POST['users'] as $cnt => $users);
$user = implode(",", $users);
foreach($_POST['qty'] as $cnt => $qty);


$sql="INSERT INTO ConfigForm (HotelName,ADD1,ADD2, town, county, postcode, phone, fax, hotel_email, website_url, hotel_contact_name, accounts_contact_name, accounts_email, web_dev_company, web_dev_contact, web_dev_email, tarrif_selection, website_link, payment_select_box, barclays, barclays_store_id, barclays_username, barclays_password, worldpay, worldpay_username, worldpay_password, secpay_username, secpay_password, secpay_template, securetrading, securetrading_site_ref, securetrading_username, securetrading_password, sagepay, sagepay_account, sagepay_password, welcome_secure, welcome_secure_account, no_payment, paypal, paypal_account,roomnum,roomtype) 
VALUES
('$_POST[hotel_name]','$_POST[address_line1]','$_POST[address_line2]','$_POST[town]','$_POST[county]','$_POST[postcode]','$_POST[phone]','$_POST[fax]','$_POST[hotel_email]','$_POST[website_url]','$_POST[hotel_contact_name]','$_POST[accounts_contact_name]','$_POST[accounts_email]','$_POST[web_dev_company]','$_POST[web_dev_contact]','$_POST[web_dev_email]','$_POST[tarru]','$_POST[website_link]','$_POST[payment_select_box]','$_POST[barclays]','$_POST[barclays_store_id]','$_POST[barclays_username]','$_POST[barclays_password]','$_POST[worldpay]','$_POST[worldpay_username]','$_POST[worldpay_password]','$_POST[secpay_username]','$_POST[secpay_password]','$_POST[secpay_template]','$_POST[securetrading]','$_POST[securetrading_site_ref]','$_POST[securetrading_username]','$_POST[securetrading_password]','$_POST[sagepay]','$_POST[sagepay_account]','$_POST[sagepay_password]','$_POST[welcome_secure]','$_POST[welcome_secure_account]','$_POST[no_payment]','$_POST[paypal]','$_POST[paypal_account]','$qty','".$_POST['name'][$cnt]."')";

if (!mysqli_query($con,$sql))
{
    die('Error: ' . mysqli_error($con));
}

echo "<h1>Form Submitted!<h1>";
mysqli_close($con);
?>

Here is a link to the form: http://apollowebtools.com/WAForm/config_form.html

Any help much appreciated, Thanks

no, the form runs through, gets to the form submitted. all data apart from the 'added' values are inserted. Its like it doesnt know they are there — Jack_McG
– Jack_McG, Commented Jul 17, 2014 at 10:52

Minhaz · Accepted Answer · 2014-07-17 11:19:07Z

1

its obvious What following line of codes does is

foreach($_POST['room_types'] as $cnt => $room_types);

iterate through the array and change the value of $room_types to each successive element of array. So after the foreach ends, $room_types is not an array but the last element of $_POST['room_types'] array

you should go for:

$room = '';
foreach($_POST['room_types'] as $cnt => $room_types)
    $room .= ',' .$room_types;

answered Jul 17, 2014 at 11:19

Minhaz

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Comments

Community · Accepted Answer · 2020-06-20 09:12:55Z

0

I think the mistake is that you have not selected the database.

Add something like following:

<?php
      $con=mysqli_connect('host','user','password');
    
      if (mysqli_connect_errno()) {
        echo ("conection error : %s\n" . mysqli_connect_error());
        exit();
    }

$db_selected = mysqli_select_db("//database name", $con);

if (!$db_selected)
  {
  die ("Can\'t use this databse : " . mysqli_error());
  }
  
    ?>

more help can be found from this link: mysqli_select_db

I'm not sure but hope this will work.

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Jul 17, 2014 at 10:46

Krupal Shah

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4 Comments

Jack_McG Over a year ago

('host', 'user', 'password', 'database') is what im using during the connection

Krupal Shah Over a year ago

@Jack_McG ok...so its not a problem...can you try to saperate them?

Krupal Shah Over a year ago

@Jack_McG try separating mysqli_connect('host', 'user', 'pass') and then selecting databse...If it do not work, error should be something else.

Krupal Shah Over a year ago

@mebjas sorry but I didn't understand.

Suneel Kumar · Accepted Answer · 2014-07-17 11:15:03Z

0

Check Your query, I think its not right.

Echo $sql and run this query in table manuaaly.

I can see $_POST[hotel_name], it must be $_POST['hotel_name']. You are not adding quotes.

answered Jul 17, 2014 at 11:15

Suneel Kumar

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