Please Help I'm a newcomer !

[gokeeffe]

Hi,

Is it possible to keep a selected value in a drop down when

selected. For example I have a three step registration

process, with lots of drop downs from mysql tables. What I

want to do is keep the values that were selected in the drop

downs even if the user navigates back to change them.

How to you assign sessions to arrays. I can do it with

text boxes but cannot for the life of me figure it out with

drop down arrays.

Please someone help

 

[Olavxxx]

You dont need to assign the session to arrays, as you can simply echo out the session variable..

eg:

echo "<option value=\"{$_POST['value']}\" selected=\"selected\">Selected: {$_POST['value']}</option>";

ps. not tested, so it might not work.

Olav Alexander Mjelde

http://www.volvo-power.net

Admin & Webmaster

 

[gokeeffe]

Could you please show me how to do this with my code

much appreciated

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

echo'<tr>
<td colspan="2" class="labelcell"><label for="junior_subject_id">Select Subject:</label></td>';

echo'<td colspan="2" class="fieldcell2"><select name="junior_subject_id">';

$query_result = mysql_query ('SELECT * FROM junior_subjects ORDER BY junior_subject_id');

while ($row = mysql_fetch_array ($query_result, MYSQL_NUM))

{
echo "<option value=\"$row[0]\">$row[1]</option>\n";
}

echo'</select></td>';

echo'</tr>';

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

 

[Olavxxx]

 {
      $selected = "";
      if ($_SESSION[$row[0] == row[0]) {
        $selected = 'selected="selected"';
      }

      echo "<option value=\"$row[0]\" {$selected}>$row[1]</option>\n";
 }

you need to start the session on the very top of all pages which are to access the session variables:
session_start();

do this before *any*thing else, as a simple space (" ") can be enough to invoke an error.

You also need to set the session variable to something, when it's submitted.

Why do you use the field-name as row[0]?

Olav Alexander Mjelde

http://www.volvo-power.net

Admin & Webmaster

 

[gokeeffe]

Ok I've changed row[0] to reflect my table but still the code won't work, it just keeps going back to select item
and the selected item does not remain present.

{
$selected = "";
if ($_SESSION[$row['junior_subject_id'] == row['junior_subject_id'])
{
$selected = 'selected="selected"';
}

echo "<option value=\"$row['junior_subject_id']\"
{$selected}>$row[1]</option>\n";
}


Why I am I finding this so difficult !

 

[Olavxxx]

if ($_SESSION[$row['junior_subject_id'] == [b]$[/b]row['junior_subject_id'])

you forgot the $

Olav Alexander Mjelde

http://www.volvo-power.net

Admin & Webmaster

 

[gokeeffe]

I included the $ but it still doesn't work, it now just gives me the third option in the dropdown all the time , could you have another look at the updated code please

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

echo'<table align="center">';

echo'<tr>
<td colspan="2" class="labelcell"><label for="secondary_exam_id">Select Exam:</label></td>';

echo'<td colspan="2" class="fieldcell2">
<select name="secondary_exam_id">';

$query_result = mysql_query ('SELECT * FROM secondary_exam_type ORDER BY secondary_exam_id');

while ($row = mysql_fetch_array ($query_result, MYSQL_NUM))

{
$selected = "";
if ($_SESSION[$row['secondary_exam_id']] == $row['secondary_exam_id'])
{
$selected = 'selected="selected"';
}

echo "<option value=\"{$row['secondary_exam_id']}\"
{$selected}>$row[1]</option>\n";
}


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

 

[Olavxxx]

ah, wait.. I see one thing here.

I'm sorry that I have done something very faulty here

if ($_SESSION['secondary_exam_id'] == $row['secondary_exam_id'])

ps. you have to set the session somewhere too!
Also, where is your <form action="... ?

the mistake I typed, was terrible.. I was very stressed and tired today, so my brain was not o.k.

Olav Alexander Mjelde

http://www.volvo-power.net

Admin & Webmaster

 

[Olavxxx]

session_start();

   echo'<table align="center">';

      echo'<tr>
           <td colspan="2" class="labelcell"><label for="secondary_exam_id">Select Exam:</label></td>';        
        
      echo'<td colspan="2" class="fieldcell2">
           <form action="page_2.php" method="post">
           <select name="secondary_exam_id">';
 
                    $query_result = mysql_query ('SELECT * FROM secondary_exam_type ORDER BY secondary_exam_id');

while ($row = mysql_fetch_array ($query_result, MYSQL_NUM))
   
  {
       $selected = "";
      if ($_SESSION[$row['secondary_exam_id']] == $row['secondary_exam_id'])
      {
        $selected = 'selected="selected"';
      }

      echo "<option value=\"{$row['secondary_exam_id']}\"    
      {$selected}>$row[1]</option>\n";
  }

page_2.php

session_start();

?>
<form action="page_3.php" method="post">
<?php

if (isset($_POST['secondary_exam_id'])) {
$_SESSION['secondary_exam_id'] = $_POST['secondary_exam_id'];
}
?>
<input type="submit" value="submit" name="submit" />
</form>

page_3.php

session_start();
echo $_SESSION['secondary_exam_id'];

I think this should work..

Olav Alexander Mjelde

http://www.volvo-power.net

Admin & Webmaster

 

[Olavxxx]

sorry, looks like I managed to do it again
if ($_SESSION['secondary_exam_id'] == $row['secondary_exam_id'])

*hit my self in the head with a sledgehammer*

Olav Alexander Mjelde

http://www.volvo-power.net

Admin & Webmaster