Hi:

Well, when i bring out the page with the drop down list it was able to display 
all tutors' names from "tutor_name" column. Anyway here's a review of my code
(snip) again before i continue:
-------------------------------------------------------------------------------
<snip>
$sql = "INSERT INTO class (class_code, tutor_name, edu_level, timetable_day,   
timetable_time)
        VALUES
('$class_code','$tutor_name','$edu_level','$timetable_day','$timetable_time')";


<?//retrieve data from DB & display in dynamic drop down ?>

<SELECT class="textarea" name="tutor_name" />
<?


$sql = mysql_query("SELECT DISTINCT tutor_name FROM tutor ");
while ($row = mysql_fetch_array($sql))
{
 print "<OPTION VALUE=\"$tutor_name\" SELECTED>" .$row
["tutor_name"]. "</option>";
 }
 $result = $db->query($sql);

?>
</select>

<?

while($selected_tutor_name == $tutor_name)
echo $_POST["tutor_name"];

?>

</snip>

-------------------------------------------------------------------------------

so when i submit the form, i am suppose to echo the values i have entered into 
the field and then INSERT the values into DB (Queries stated above). However i 
was able to echo all other values eg. class_code, edu_level, etc...but 
not "tutor_name"....same thing happen when i do an INSERT, all other values 
are inserted into DB but not $tutor_name....why is this so???Really need some 
help here...Anyway i have already specify a name to be reference :

<SELECT class="textarea" name="tutor_name" >

and then I also did an echo of "tutor_name" being selected:

while($selected_tutor_name == $tutor_name)
echo $_POST["tutor_name"];

All help are greatly appreciated =)

Irin.

-------------------------------------------------------------------------------
On Tue, 23 Dec 2003 16:00:04 -0500, "Aleks @ USA.net" <[EMAIL PROTECTED]> wrote:
-------------------------------------------------------------------------------
Hi, 

I am looking at your code and have a question, in your code you have 

print "<OPTION VALUE=\"$tutor_name\" SELECTED>" .$row ["tutor_name"]. 
"</option>"; 

Doesn't the SELECTED piece mark everything listed as selected?? When you 
bring up the page with the drop down list and open the source, what does it 
show?? 

I handle this  a little differently, I create my drop down list as follows: 

=================== Code used ============================================ 

// First I retrieve all customer information 

$BB = mysql_query("SELECT * FROM customer ORDER BY CID "); 



// Then I create the drop down list 

<SELECT style="WIDTH: 410px" size=1 name="Customer"> 


<? 
 // Creates the Customer dropdown with the $id number 
   while ($Site = mysql_fetch_array($BB)) 
  { 
    $Sid = $Site["CID"]; 
    $Sname = htmlspecialchars($Site["Customer"]); 
    $SCity = htmlspecialchars($Site["City"]); 
    $SState = htmlspecialchars($Site["State"]); 
    $SCountry = htmlspecialchars($Site["Country"]); 


        if($Sid == $Customer) 
        { $add = ' selected'; } else { $add = ''; } 
  
    echo("<option value='$Sid'$add>$Sname &nbsp;&nbsp;&nbsp; $SCity 
&nbsp;&nbsp;&nbsp; $SState &nbsp;&nbsp;&nbsp; $SCountry </option>\n");   

  } 
  
?> 

</select> 

=================== Code used ============================================ 

This will create a drop down list where the value of the option is the sites 
ID. In another part of this form, if there is a value set for $Customer, 
then it will set the $add to selected. This will then make the drop down 
list auto select the customer. 

The code above is passed to another form for processing and is were the 
insert data occurs. Using this method I can echo the form data easily.... 

Hope this helps... 

Aleks 



-----Original Message----- 
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] 
Sent: Tuesday, December 23, 2003 4:33 AM 
To: David Robley 
Cc: [EMAIL PROTECTED] 
Subject: Re: [PHP-DB] Re: Display Mysql Result in drop down list 

Hi: 

then now i'm trying to select a value from the drop down list, echo the 
value i have selected and lastly INSERT the value into DB with the following 
: 
Below's a snippet of code: 

<snip> 
//retrieve all "tutor_name" from DB and display in drop down list 

<SELECT class="textarea" name="tutor_name" /> <? 


$sql = mysql_query("SELECT tutor_name FROM tutor "); while ($row = 
mysql_fetch_array($sql)) {  print "<OPTION VALUE=\"$tutor_name\" SELECTED>" 
.$row ["tutor_name"]. "</option>";  }  $result = $db->query($sql); 

?> 
</select> 

</snip> 

<snip> 
***********************INSERT selected values into 
DB************************** $tutor_name = $_POST["tutor_name"]; 

$sql = "INSERT INTO class (class_code, tutor_name, edu_level, timetable_day, 
timetable_time) 
        VALUES 
('$class_code','$tutor_name','$edu_level','$timetable_day','$timetable_time' 
)"; 

//execute query statement 
$result = $db->query($sql); 

</snip> 


<snip> 
*********************echo the value i have 
selected*************************** 
<? 
if($selected_tutor_name == $tutor_name) 
echo $_POST["tutor_name"]; 
?> 

</snip> 


Problem: I was unable to echo the value i selected from drop down as well as 
INSERT into DB...wonder where have i gone wrong??? 
Reali need some help here...all help are greatly appreciated =) Thanks in 
advance. 


Irin. 

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