Sending options from a drop down menu to another form

[NCC1701]

Im trying to send the result of a menu selection to another form in order for details to be edited but i have no idea where to start, can anyone steer me in the right direction, ive looked everywhere for a solution.

thanks
NCC1701

[darthmahon]

How are you doing your menu selection? Is it taking results from a database?

If so this is what I use:

<? if (!$id)

// nothing selected, show options

{ ?>

<p>Please choose an option:</p>

<form action="<? echo $PHP_SELF; ?>" method="post">

<select name="id">

<option value="">Please select an area</option>

<? 
// generate and execute query

$ids = "SELECT * FROM table ORDER BY title ASC";
$idsresult = mysql_query($ids) or die ("Error in query: $ids. " . mysql_error());

// turn results into an array
while ($idsrow = mysql_fetch_array($idsresult)) 
{

	// get results
	$catname = $idsrow["title"];
	$catid = $idsrow["sectionid"];

	// outputs result of each record in the db into the list
	echo("<option value='$catid'>$catname</option>\n");

}

?>

</select>

<input type="submit" name="submitBtn" value="Go" />

</form>

<? 

}

// option selected above
else

{

$id = $_REQUEST['id'];

$query = "SELECT * FROM table WHERE field = $id";

$result = mysql_query($query) or die ("Error in query: $query. " . mysql_error());

// form not yet submitted

// display initial form with values pre-filled

if (!$submit)

{

// if a result is returned

if (mysql_num_rows($result) > 0)

{

	// turn it into an object

	$row = mysql_fetch_object($result);

	// print form with values pre-filled

?>

This is what was in the database <? echo $row->afieldinthetable; ?>

<? } 

// form submitted

// start processing it

else

{

// enter table update information here

?>

[NCC1701]

i am using a databse to populate the list but im trying to do it all in separate functions, this is the funciton to display the form:

<?php
function display_menu_form()
{
?>
  <form method="post" action="menu.php">
  <table align="center" border="0">
  <tr>
    <td>
	<select name="menu_name">
	</select>
	</td>
  </tr>
  <td><input name="Submit" type="button" value="Submit"></td>
  </tr>
  </table>
  </form>
<?php 
}
?>

this is menu.php which im struggling with:

<?php
 require_once('tannin_level.php');
 session_start();
 do_html_header('Select Menu to Edit');

 //create short variable names
 $menu_id = $HTTP_POST_VARS['menu_id'];
 $menu_name = $HTTP_POST_VARS['menu_name'];
 $menu_text = $HTTP_POST_VARS['menu_text'];

 // create connection 
    $conn = db_connect();
    if (!$conn)
    return false;

// create SQL statement 
$query = "SELECT menu_id, menu_name, menu_text 
       FROM menu";
 // execute SQL query and get result 
$result = mysql_query($query,$conn);

// format results by row
while ($row = mysql_fetch_array($result)) 
{ 
$menu_name = $row['menu_name']; 
echo "<option>$menu_name</option>";
}

 display_menu_form();
 display_user_menu();
 do_html_footer();
?>

[darthmahon]

You will probably have to put the following in your function display_menu_form()

// format results by row
    while ($row = mysql_fetch_array($result)) 
    { 
    $menu_name = $row['menu_name']; 
    echo "<option>$menu_name</option>";
    } 

It needs to go between the select's( <select></select>. Otherwise I don't see how your code populates a list?

Also, you need to assign values to the options in your list so that when someone selects an option and it takes them to the edit page, you can select the appropriate record in the database:

// format results by row
    while ($row = mysql_fetch_array($result)) 
    { 
    $menu_name = $row['menu_name'];
    $menu_id = $row['menu_id'];
    echo "<option value='$menu_id'>$menu_name</option>";
    } 

[NCC1701]

right, thanks for the advice,

thats the whole problem though, im not sure how to populate the list between the <select></select>

[darthmahon]

Make the function look like this:

<?
function display_menu_form()
{
?>
  <form method="post" action="menu.php">
  <table align="center" border="0">
  <tr>
    <td>
    <select name="select_name">
	<?

// create connection
$conn = db_connect();
if (!$conn)
return false;

// create SQL statement
$query = "SELECT menu_id, menu_name, menu_text FROM menu";
 // execute SQL query and get result
$result = mysql_query($query,$conn);

// format results by row
while ($row = mysql_fetch_array($result)) 
{ 
$menu_name = $row['menu_name'];
$menu_id = $row['menu_id']; 
echo "<option value='menu_id'>$menu_name</option>";
}

?>
</select>
</td>
  </tr>
  <td><input name="Submit" type="button" value="Submit"></td>
  </tr>
  </table>
  </form>
<? 
}
?>

And remove the following from menu.php

 // create connection
    $conn = db_connect();
    if (!$conn)
    return false;

// create SQL statement
$query = "SELECT menu_id, menu_name, menu_text
       FROM menu";
 // execute SQL query and get result
$result = mysql_query($query,$conn);

// format results by row
while ($row = mysql_fetch_array($result)) 
{ 
$menu_name = $row['menu_name']; 
echo "<option>$menu_name</option>";
} 

[NCC1701]

thats brilliant, thank you it works now,

all i need to do now is to sumbit the value of the selected option, can anyone tell me how i would do this bearing in mind im passing the result via functions?

[NCC1701]

ive managed to pass the results but only the name of the menu in the list and i dont know how to link the id and name to display the corresponding menu text, its really stressing me out, can anyone help??😃