Question : Problem with multiple form choice and inserting into mysql

Hi,

Pleae bare with me as my php and mysql is pretty basic and i use most of the tools available to me in Dreamweaver to work with databases.

I am trying to do a simple task but can't work out how to do it. Part of a form i am making has a list of countries. I have made this list a dropdown box with multiple options selectable. How do i then go about inserting these multiple options into my database though as at the moment, the last one picked is only inserting into my database.

My code is as follows:

         
           
             1:
2:
3:
4:
5:
6:
7:
8:
9:
10:
11:
12:
13:
14:
15:
16:
17:
18:
19:
20:
21:
22:
23:
24:
25:
26:
27:
28:
29:
30:
31:
32:
33:
34:
35:
36:
37:
38:
39:
40:
41:
42:
43:
44:
45:
46:
47:
48:
49:
50:
51:
52:
53:
54:
55:
56:
57:
58:
59:
60:
61:
62:

           
           
          <?php require_once('Connections/bookedy.php'); ?>
<?php
function GetSQLValueString($theValue, $theType, $theDefinedValue = "", $theNotDefinedValue = "") 
{
  $theValue = (!get_magic_quotes_gpc()) ? addslashes($theValue) : $theValue;

  switch ($theType) {
    case "text":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;    
    case "long":
    case "int":
      $theValue = ($theValue != "") ? intval($theValue) : "NULL";
      break;
    case "double":
      $theValue = ($theValue != "") ? "'" . doubleval($theValue) . "'" : "NULL";
      break;
    case "date":
      $theValue = ($theValue != "") ? "'" . $theValue . "'" : "NULL";
      break;
    case "defined":
      $theValue = ($theValue != "") ? $theDefinedValue : $theNotDefinedValue;
      break;
  }
  return $theValue;
}

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

if ((isset($_POST["MM_insert"])) && ($_POST["MM_insert"] == "form1")) {
  $insertSQL = sprintf("INSERT INTO test (countries) VALUES (%s)",
                       GetSQLValueString($_POST['select'], "text"));

  mysql_select_db($database_bookedy, $bookedy);
  $Result1 = mysql_query($insertSQL, $bookedy) or die(mysql_error());

  $insertGoTo = "test.php";
  if (isset($_SERVER['QUERY_STRING'])) {
    $insertGoTo .= (strpos($insertGoTo, '?')) ? "&" : "?";
    $insertGoTo .= $_SERVER['QUERY_STRING'];
  }
  header(sprintf("Location: %s", $insertGoTo));
}
?>
<form action="<?php echo $editFormAction; ?>" method="POST" name="form1" id="form1">
        <label></label>
            <label>
            <select name="select" size="1" multiple="multiple">
              <option value="Country A">Country A</option>
              <option value="Country B">Country B</option>
              <option value="Country C">Country C</option>
            </select>
            <br />
            <br />
            <br />
<input type="submit" name="Submit" value="Submit" />
          </label>
            <input type="hidden" name="MM_insert" value="form1">
            </form>

           
         

Answer : Problem with multiple form choice and inserting into mysql

along the line that MelindaJohnson post,,

the subform does not have a rowsource property, it use a record source.
in the afterupdate event of the listbox, create a sring of selected items

private sub list0_afterUpdate()
dim j, sLst
with me.list0
if .itemsselected.count >0 then
for each j in .itemsselected
sLst=sLst & "," & chr(39) & .itemdata(j) & chr(39)
next
end if
sLst=mid(slst,2)
end with

me.subformname.form.recordsource="select * from tableName where [fieldname] in (" & slst & ")"

that will vary depending on the data type of the bound column of the listbox

can you post the rowsource of the listbox

also note that you have to remove the link master/child fields setting of the subform