not passed on to the php database session

I have a page which creates an orderID of three entries of form session. I need to get the orderID of session in the database column, but it is in error "column 'orderID' cannot be null.

If ((isset($_POST["MM_insert"])) & & ($_POST ["MM_insert"] == "form1")) {}

$insertSQL = sprintf ("INSERT INTO wildOrchidRes (ID, orderID, name, address, city, County, zip code, country, email, phone, checkIn, checkOut, amount) VALUES (%s, %s, %s, %s, %s, %s, %s, %s, %s, %s, %s, %s, %s)",

GetSQLValueString ($_POST ['ID'], "int").

GetSQLValueString ($_POST ["orderID"], "text").

GetSQLValueString ($_POST ['name'], "text").

GetSQLValueString ($_POST ['address'], "text").

GetSQLValueString ($_POST ['city'], "text").

GetSQLValueString ($_POST ["count"], "text").

GetSQLValueString ($_POST ['PostalCode'], "text").

GetSQLValueString ($_POST ["country"], "text").

GetSQLValueString ($_POST ['email'], "text").

GetSQLValueString ($_POST ['telephone'], "text").

GetSQLValueString ($_POST ["checkIn"], "text").

GetSQLValueString ($_POST ['order'], "text").

GetSQLValueString ($_POST ['quantity'], "double"));

@mysql_select_db ($database_WO, $WO);

$Result1 = mysql_query ($insertSQL, $WO) or die (mysql_error ());

session_start (); order session created

$_SESSION ["orderID"] = $_POST ['name']. $_POST ["checkIn"]. $_POST ['order'];

$insertGoTo = 'confirm.php ';

If (isset {}

$insertGoTo. = (strpos ($insertGoTo, '?'))? « & » : « ? » ;

$insertGoTo. = $_SERVER ['QUERY_STRING'];

}

header (sprintf ("location: %s", $insertGoTo));

}

< do action = "<?" PHP echo $editFormAction;? ">" method = "post" name = "form1" id = "form1" >

< input type = "hidden" name = "ID" value = "" size = "32" / > "

< input type = "text" name = "name" value = "" size = "32" / > "

< input type = "text" name = "address" value = "" size = "32" / > "

< input type = "text" name = "City" = value "" size = "32" / > "

< input type = "text" name = value "County" = "" size = "32" / > "

< input type = "text" name = "PostalCode" value = "" size = "32" / > "

< input type = "text" name = "country" value = "" size = "32" / > "

< input type = "text" name = "email" value = "" size = "32" / > "

< input type = "text" name = "phone" value = "" size = "32" / > "

< input type = "text" name = "checkIn" value = "" id = "CheckIn" size = "32" / > "

< input type = "text" name = "checkOut" value = "" id = "CheckOut" size = "32" / > "

< input type = "text" name = "amount" value = "" size = "32" / > "

< input type = "submit" value = "confirm order" / >

< input type = "hidden" name = "orderID" value = "<?" PHP echo $_SESSION ["orderID"];? ">" / >

< input type = "hidden" name = "MM_insert" value = "form1" / >

< / make >

ideas please because I can think of why it does not work

Thanks in advance

> Yes, I tried to add

>

> session_start (); order session created

> $_SESSION ["orderID"] = $_POST ['name']. $_POST ["checkIn"]. $_POST ["amount"];

>

> top of page

>

> but still gives the same error

Because you always try to insert a hidden form field that will not fill. Simply insert the value of the session variable, not the field of the form displayed.

Tags: Dreamweaver

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