can 2 form fields have same name and send data to same db field?

[jarow]

I need to create 3 separate text fields, the contents of which all need to be sent to one field in my mysql database...any ideas on how I may accomplish this in dreamweaver mx? and how I get the mysql database to accept three values?

Many thanks

Jim

[autocolismo]

$value_to_mysql = $GET['textfield1'] . $GET['textfield2'] . $_GET['textfield3'];

then

insert into your_table set value = '$value_to_mysql';

[Jim_Madrid]

Thank you for your response. I understand the concept but I am having trouble incorporating it into the Dreamweaver INSERT RECORD syntax. Let´s say I have the following code for:

Database table: session database field = animals
Form Textfield1 = animals

$editFormAction = $HTTP_SERVER_VARS['PHP_SELF'];
if (isset($HTTP_SERVER_VARS['QUERY_STRING'])) {
$editFormAction .= "?" . $HTTP_SERVER_VARS['QUERY_STRING'];
}

if ((isset($HTTP_POST_VARS["MM_insert"])) && ($HTTP_POST_VARS["MM_insert"] == "form1")) {
$insertSQL = sprintf("INSERT INTO session (animals) VALUES (%s)",
GetSQLValueString($HTTP_POST_VARS['animals'], "text"));

mysql_select_db($database_conntest, $conntest);
$Result1 = mysql_query($insertSQL, $conntest) or die(mysql_error());
}

Now, if I have two more form textfields that I want to insert into one database field I imagine I would have to create something like this:

$habitat=$GET[‘habitat1’] . =$GET[‘habitat2’]
INSERT INTO session (habitat) VALUES ('$habitat’)

How and where do I put this exactly ?

Again thanks for your reply.

Jim

[autocolismo]

$editFormAction = $HTTP_SERVER_VARS['PHP_SELF']; 
if (isset($HTTP_SERVER_VARS['QUERY_STRING'])) { 
$editFormAction .= "?" . $HTTP_SERVER_VARS['QUERY_STRING']; 
} 

if ((isset($HTTP_POST_VARS["MM_insert"])) && ($HTTP_POST_VARS["MM_insert"] == "form1")) { 


$habitat = $_POST['habitat1'] . $_POST['habitat2'] . $_POST['habitat3'];

$insertSQL = sprintf("INSERT INTO session set animals='$habitat'"; 

mysql_select_db($database_conntest, $conntest); 
$Result1 = mysql_query($insertSQL, $conntest) or die(mysql_error()); 
} 

I think this workx fine🙂

[Jim_Madrid]

muito obrigado....it works like a charm